A parabola is the locus of a point which moves in such a way that its distance from a fixed point is equal to its perpendicular distance from a fixed straight line.
1.1 Focus : The fixed point is called the focus of the Parabola.
1.2 Directrix : The fixed line is called the directrix of the Parabola.
(focus)
2.3 Vertex : The point of intersection of a parabola and its axis is called the vertex of the Parabola.
NOTE: The vertex is the middle point of the focus and the point of intersection of axis and directrix
2.4 Focal Length (Focal distance) : The distance of any point P (x, y) on the parabola from the focus is called the focal length. i.e.
The focal distance of P = the perpendicular distance of the point P from the directrix.
2.5 Double ordinate : The chord which is perpendicular to the axis of Parabola or parallel to Directrix is called double ordinate of the Parabola.
2.6 Focal chord : Any chord of the parabola passing through the focus is called Focal chord.
2.7 Latus Rectum : If a double ordinate passes through the focus of parabola then it is called as latus rectum.
2.7.1 Length of latus rectum :
The length of the latus rectum = 2 x perpendicular distance of focus from the directrix.
2.1 Eccentricity : If P be a point on the parabola and PM and PS are the distances from the directrix and focus S respectively then the ratio PS/PM is called the eccentricity of the Parabola which is denoted by e.
Note: By the definition for the parabola e = 1.
If e > 1 Hyperbola, e = 0 circle, e < 1
ellipse
2.2 Axis : A straight line passes through the focus and perpendicular to the directrix is called the axis of parabola.
If we take vertex as the origin, axis as x- axis and distance between vertex and focus as 'a' then equation of the parabola in the simplest form will be-
y2 = 4ax
3.1 Parameters of the Parabola y2 = 4ax
(i) Vertex A (0, 0)
(ii) Focus S (a, 0)
(iii) Directrix x + a = 0
(iv) Axis y = 0 or x– axis
(v) Equation of Latus Rectum x = a
(vi) Length of L.R. 4a
(vii) Ends of L.R. (a, 2a), (a, – 2a)
(viii) The focal distance sum of abscissa of the point and distance between vertex and L.R.
(ix) If length of any double ordinate of parabola
y2 = 4ax is 2 𝑙 then coordinates of end points of this Double ordinate are
𝑙2 𝑙2
, 𝑙
and
, 𝑙 .
4a
4a
3.2 Other standard Parabola :
Equation of Parabola Vertex Axis Focus Directrix Equation of Latus rectum Length of Latus rectum
y2 = 4ax (0, 0) y = 0 (a, 0) x = –a x = a 4a
y2 = – 4ax (0, 0) y = 0 (–a, 0) x = a x = –a 4a
x2 = 4ay (0, 0) x = 0 (0, a) y = a y = a 4a
x2 = – 4ay (0, 0) x = 0 (0, –a) y = a y = –a 4a
Standard form of an equation of Parabola
Ex.1 If focus of a parabola is (3,–4) and directrix is x + y – 2 = 0, then its vertex is (A) (4/15, – 4/13)
(B) (–13/4, –15/4)
(C) (15/2, – 13/2)
(D) (15/4, – 13/4)
Sol. First we find the equation of axis of parabola
Distinguish equations representing the circles and the conics; use the properties of a particular geometry to sketch the graph in using the rectangular or the polar coordinate system. Furthermore, to be able to write the equation and to solve application problems involving a particular geometry.
x2 y2
Standard Equation of hyperbola is a 2 – b2 = 1
(i) Definition hyperbola : A Hyperbola is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed line (called directrix) is always constant which is always greater than unity.
The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola is called conjugate hyperbola.
Note :
(i) If e1 and e2 are the eccentricities of the
(ii) Vertices : The point A and A where the curve meets the line joining the foci S and S
hyperbola and its conjugate then
1 +
e 2 e
1 = 1
2
are called vertices of hyperbola.
(iii) Transverse and Conjugate axes : The straight line joining the vertices A and A is called transverse axes of the hyperbola. Straight line perpendicular to the transverse axes and passes through its centre called conjugate axes.
(iv) Latus Rectum : The chord of the hyperbola which passes through the focus and is perpendicular to its transverse axes is called
2b2
latus rectum. Length of latus rectum = a .
(ii) The focus of hyperbola and its1 conju2gate are concyclic.
Standard Equation and Difinitions
Ex.1 Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1,2) and
eccentricity 3 .
Sol. Let P (x,y) be any point on the hyperbola. Draw PM perpendicular from P on the directrix.
Then by definition SP = e PM
(v) Eccentricity : For the hyperbola
x2 y2
a 2 – b2
= 1,
(SP)2 = e2(PM)2
2x y 12
b2 = a2 (e2 – 1)
(x–1)2 + (y–2)2 = 3
Conjugate axes 2
5(x2 + y2 – 2x – 4y + 5} =
e = =
1
Transverse
axes
3(4x2 + y2 + 1+ 4xy – 2y – 4x)
7x2 – 2y2 + 12xy – 2x + 14y – 22 = 0
(vi) Focal distance : The distance of any point on the hyperbola from the focus is called the focal distance of the point.
Note : The difference of the focal distance of a point on the hyperbola is constant and is equal to the length
of the transverse axes. |SP – SP| = 2a (const.)
which is the required hyperbola.
Ex.2 Find the lengths of transverse axis and conjugate axis, eccentricity and the co- ordinates of foci and vertices; lengths of the latus rectum, equations of the directrices of the hyperbola 16x2 – 9y2 = –144
Sol. The equation 16x2 – 9y2 = – 144 can be
Sol. y= m1(x –a),y= m2(x + a) where m1m2 = k, given
x 2
written as 9
x2
y 2
– 16 = – 1. This is of the form
y2
In order to find the locus of their point of intersection we have to eliminate the unknown
m1 and m2. Multiplying, we get
y2 = m1m2 (x2 – a2) or y2 = k(x2–a2)
a 2 – b2 = – 1
a2 = 9, b2 = 16 a = 3, b = 4
or x – y
1 k
= a2
which represents a hyperbola.
Length of transverse axis :
The length of transverse axis = 2b = 8
Length of conjugate axis :
The length of conjugate axis = 2a = 6
5
Ex.5 T
Distinguish equations representing the circles and the conics; use the properties of a particular geometry to sketch the graph in using the rectangular or the polar coordinate system. Furthermore, to be able to write the equation and to solve application problems involving a particular geometry.
x2 y2
Standard Equation of hyperbola is a 2 – b2 = 1
(i) Definition hyperbola : A Hyperbola is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed line (called directrix) is always constant which is always greater than unity.
The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola is called conjugate hyperbola.
Note :
(i) If e1 and e2 are the eccentricities of the
(ii) Vertices : The point A and A where the curve meets the line joining the foci S and S
hyperbola and its conjugate then
1 +
e 2 e
1 = 1
2
are called vertices of hyperbola.
(iii) Transverse and Conjugate axes : The straight line joining the vertices A and A is called transverse axes of the hyperbola. Straight line perpendicular to the transverse axes and passes through its centre called conjugate axes.
(iv) Latus Rectum : The chord of the hyperbola which passes through the focus and is perpendicular to its transverse axes is called
2b2
latus rectum. Length of latus rectum = a .
(ii) The focus of hyperbola and its1 conju2gate are concyclic.
Standard Equation and Difinitions
Ex.1 Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1,2) and
eccentricity 3 .
Sol. Let P (x,y) be any point on the hyperbola. Draw PM perpendicular from P on the directrix.
Then by definition SP = e PM
(v) Eccentricity : For the hyperbola
x2 y2
a 2 – b2
= 1,
(SP)2 = e2(PM)2
2x y 12
b2 = a2 (e2 – 1)
(x–1)2 + (y–2)2 = 3
Conjugate axes 2
5(x2 + y2 – 2x – 4y + 5} =
e = =
1
Transverse
axes
3(4x2 + y2 + 1+ 4xy – 2y – 4x)
7x2 – 2y2 + 12xy – 2x + 14y – 22 = 0
(vi) Focal distance : The distance of any point on the hyperbola from the focus is called the focal distance of the point.
Note : The difference of the focal distance of a point on the hyperbola is constant and is equal to the length
of the transverse axes. |SP – SP| = 2a (const.)
which is the required hyperbola.
Ex.2 Find the lengths of transverse axis and conjugate axis, eccentricity and the co- ordinates of foci and vertices; lengths of the latus rectum, equations of the directrices of the hyperbola 16x2 – 9y2 = –144
Sol. The equation 16x2 – 9y2 = – 144 can be
Sol. y= m1(x –a),y= m2(x + a) where m1m2 = k, given
x 2
written as 9
x2
y 2
– 16 = – 1. This is of the form
y2
In order to find the locus of their point of intersection we have to eliminate the unknown
m1 and m2. Multiplying, we get
y2 = m1m2 (x2 – a2) or y2 = k(x2–a2)
a 2 – b2 = – 1
a2 = 9, b2 = 16 a = 3, b = 4
or x – y
1 k
= a2
which represents a hyperbola.
Length of transverse axis :
The length of transverse axis = 2b = 8
Length of conjugate axis :
The length of conjugate axis = 2a = 6
5
Ex.5 T
ATOMIC STRUCTURE
1. ATOM & MOLECULES
(a) The smallest particle of a matter that takes part in a chemical reaction is called an atom. The atom of all gases except those of noble gases, cannot exist in free state. These exist in molecular form. The molecules of hydrogen, nitrogen, oxygen and halogens are diatomic (H2, N2). Phosphorus molecule is tetratomic and that of
sulphur is octa atomic.
(b) The smallest particle of a matter that can exist in free state in nature, is known as a molecule.
(c) Some molecules are composed of homoatomic atom, e.g., H2, O2, N2, Cl2, O3 etc., while the molecules of compounds are made up of two or more heteroatomic atoms e.g., HCl, NaOH, HNO3, CaCO3, etc.
2. DALTON’S ATOMIC THEORY
The concepts put forward by John Dalton regarding the composition of matter are known as Dalton’s atomic theory. Its important points are as follows.
(a) Every matter is composed of very minute particles, called atoms that take part in chemical reactions.
(b) Atoms cannot be further subdivided.
(c) The atoms of different elements differ from each other in their properties and masses, while the atoms of the same element are identical in all respects.
(d) The atoms of different elements can combine in simple ratio to form compounds. The masses of combining elements represent the masses of combining atoms.
(e) Atom can neither be created nor destroyed.
2.1 Modern Concept :
Many of the concepts of Dalton’s atomic theory cannot be explained. Therefore, foundation of modern atomic theory was laid down by the end of nineteenth century. The modern theory is substantiated by the existence of isotopes, radioactive disintegration, etc. The important points of the modern atomic theory are as follows.
(a) Prof. Henri Bacquerel discovered the phenomenon of radioactivity and found that an atom is divisible.
(b) An atom is mainly composed of three fundamental particles, viz. electron, proton and neutron.
(c) Apart from the aforesaid three fundamental particles, many others have also been identified, viz. positron, meson, neutrino, antiproton, etc.
(d) Soddy discovered the existence of isotopes, which were atom of the same element having different masses. For example, protium, deuterium and tritium are atoms of hydrogen having atomic masses 1, 2 and 3 a.m.u. respectively.
(e) Atoms having same mass may have different atomic numbers. These are known as isobars. For example,
40 Ar and 40 Ca .
18 20
(f) Atoms of elements combines to form molecules.
(g) It is not necessary that the atoms should combine in simple ratio for the formation of compounds. The atoms in non-stoichiometric compounds are not present in simple ratio. For example, in ferrous sulphide crystals, iron and sulphur atoms are present in the ratio of 0.86 : 1.00.
(h) Atoms participate in chemical reactions.
3. CATHODE RAYS (DISCOVERY OF ELECTRON)
Dry gases are normally bad conductors of electricity. But under low pressure, i.e., 0.1 mm of mercury or lower, electric current can pass thro
The process of finding area of some plane region is called Quadrature. In this chapter we shall find the area bounded by some simple plane curves with the help of definite integral. For solving
(i) The area bounded by a cartesian curve y = f(x), x-axis and ordinates x = a and x = b is given by,
the problems on quadrature easily, if possible first draw the rough sketch of the required area.
b
Area = y dx
a
b
= f(x) dx
a
In chapter function, we have seen graphs of some simple elementary curves. Here we introduce some essential steps for curve tracing which will enable us to determine the required area.
(i) Symmetry
The curve f(x, y) = 0 is symmetrical
about x-axis if all terms of y contain even powers.
about y-axis if all terms of x contain even
powers.
about the origin if f (– x, – y) = f (x, y).
Examples
based on
Area Bounded by A Curve
For example, y2 = 4ax is symmetrical about x-axis, and x2 = 4ay is symmetrical about y- axis and the curve y = x3 is symmetrical about the origin.
(ii) Origin
Ex.1 Find the area bounded by the curve y = x3, x-axis and ordinates x = 1 and x = 2.
2
Sol. Required Area = ydx
1
If the equation of the curve contains no constant
2
= x3 dx =
Lx4 O2 15
MN PQ=
Ans.
term then it passes through the origin.
For example x2 + y2 + 2ax = 0 passes through origin.
(iii) Points of intersection with the axes
If we get real values of x on putting y = 0 in the equation of the curve, then real values of x and y = 0 give those points where the curve cuts the x-axis. Similarly by putting x = 0, we can get the points of intersection of the curve and y-axis.
1 4 1 4
Ex.2 Find the area bounded by the curve y = sec2x,
x-axis and the line x = 4
/4
Sol. Required Area = ydx
x0
For example, the curve x2/a2 + y2 /b2 = 1 intersects the axes at points (± a, 0) and (0, ± b) .
(iv) Region
/4
= sec2
0
xdx = tan x /4 = 1 Ans.
Write the given equation as y = f(x) , and find minimum and maximum values of x which determine the region of the curve.
For example for the curve xy2 = a2 (a – x)
Ex.3 Find the area bounded by the curve y = mx, x-axis and ordinates x = 1 and x = 2
2
Sol. Required area = y dx .
1
y = a
a x
x
2
= mxdx =
LMmx2 2
Now y is real, if 0 < x a , so its region lies between the lines x = 0 and x = a.
1 MN2
= m ( 4 – 1) =
2
PQ1
FGH3 JKm Ans.
Ex.4 Find the area bounded by the curve y = x (1– x)2 and x-axis.
Sol. Clearly the given curve meets the x-axis at (0,0) and (1,0) and for x = 0 to 1, y is positive
Ex.7 Find the area bounded between the curve y2 = 2y – x and y-axis.
Sol. The area between the given curve x = 2y – y2 and y-axis will be as shown in diagram.
so required area- Y
= xb1 xg2 dx
0
1
= ex 2x2 x3 jdx
0
O (0,0) (1,0) X
Lx2
2x3
x4 O1
1 2 1 1
= M P=
– + =
MN2 3 4 PQ0
2 3 4 12
Ans.
Required Area =
e2y y2 jdy
(i
/2
Ex.5 Evaluate : sin2
x dx .
d
If dx
[f(x)] =
(x) and a and b, are two values
0
/2
Sol. sin2 x dx
independent of variable x, then 0
b /2 FG1 cos 2xIJ
(x) dx = f(x) a = f(b) – f(a)
a
= zH 2 dx
1 LM sin 2x O /2
is called Definite Integral of (x) within limits
= x
2
2 PQ
a and b. Here a is called the lower limit and b is called the upper limit of the integral. The interval [a,b] is known as range of integration. It should be noted that every definite integral has
a unique value.
= 1 LM 0
0
= / 4 Ans.
1
x2
Ex.6 Evaluate : xe dx.
0
1
2 x2
Ex.1 Evaluate : x4 dx.
1
Sol. xe dx
0
2 Lx5 O2
1 ex2 1
Sol.
zx4 dx = MP= 32 – 1 = 31
=
Ans. 2 0
MN5 PQ1 5 5 5
/4
= 1 (e –1) Ans.
2
Ex.2 Evaluate : sec2 x dx.
0
1 x3
/4
Sol. sec2 x .dx = tan x /4 = tan / 4 – tan 0 = 1
0
Ex.7 Find the value of
0
1 x8
dx.
0
Ans.
2 1
Sol. Let x4 = t, then 4x3 dx = dt
Ex.3 Evaluate :
1
4 x2
dx.
I =
1 1 dt
4 =
1 [ sin–1 t] 1 =
4 8
2 1 L 2 0
Sol. z dx = Msin GJP
Ans.
1 4 x2
N H2KQ1
= sin–1 (1) – sin–1 (1/2)
z/3 cos x
= – =
Ans.
Ex.8 Evaluate :
0
3 4 sin x dx.
2 6 3
Sol. I =
/3 cos x 3 4 sin x dx.
Ex.4 Evaluate :
z2 1
2 dx
0 4 x2
0
Let 3 + 4 sin x = t 4 cos x. dx = dt
cos x dx = dt/4
Now in the given integral x lies between the
Sol.
4 x2 dx
limit x = 0 to x = / 3 . Now we will decide the limit of t.
1
= tan
2
1 x OP2
0
In 3 + 4 sin x = t, by putting lower limit of x as x = 0; and upper limit as x = / 3 . We
= 1 tan11 0 = / 8 Ans.
2
get lower and upper limit of t respectively.
Putting x = 0 3 + 4 sin 0 = t t = 3
z3 z2
z3 (x) dx
/3 cos x
dx =
zt3 2 3 1 dt
= 2 x2dx +
0
z3b3x 4gdx
0 3 4 sin x
t3 t 4
Fx3 I2 F3x2 I3
= 1 zt3 2 3 1 dt
= GJ+ G
4xJ
4 t3 t
H3 K H2
= 1 log t 32
4
= 8 +
3
27 – 12 – 6 + 8
2
= 1 [ log (3 + 2
4
) – log 3] Ans.
= 37/6 Ans.
Ex.9
sin(tan1 x)
2
dx equals-
Ex.11 Evaluate : |1 x|dx.
0
0 1 x
Sol. Put tan
x = t, then 1 dx = dt
Sol. |1 x| =
RST1 x, when
0 x 1
–1
(1 x2 )
I =
x 1, when 1 x 2
1b1 xgdx + 2 bx 1gdx
/ 2
I = sin t dt [– cos t] /2 = 1 Ans.
0 1
L x2 O1 Lx2 O2
0 Mx
P M xP
0 = MN
2 PQ+
MN2
PQ1
= b1/ 2 0 + b0 1/ 2 = 1 Ans.
z z
i.e. the value of a definite integral remains unchanged if its variable is placed by any other symbol.
[P-4] f(x) dx = f(a x) dx .
0 0
Note :
[P-2]
b
f(x) dx
a
a
= – f(x) dx
b
This property can be used only when lower limit is zero. It is generally used for those complicated
i.e. the interchange of limits of a definite integral
changes only its sign.
zb zc zb
integrals whose denominators are unchanged when x is replaced by a– x. With the hel
Integration is a reverse process of differentiation. The integral or primitive of a function f(x) with
respect to x is that function (x) whose derivative with respect to x is the given function f(x). It is
i. 0. dx = c
ii. 1.dx = x + c
iii. k.dx = kx + c (k R)
xn1
expressed symbolically as -
zf (x) dx (x)
iv. xn dx =
n 1
+ c (n –1)
v. z1 dx = log
x + c
Thus x e
vi. ex dx = ex + c
ax
The process of finding the integral of a function is called Integration and the given function is
vii. ax dx =
loge
a + c = ax loga e + c
called Integrand. Now, it is obvious that the operation of integration is inverse operation of differentiation. Hence integral of a function is also named as anti-derivative of that function.
Further we observe that-
viii. sin x dx = – cos x + c
ix. cos x dx = sin x + c
x. tan x dx = log sec x + c = – log cos x + c
d (x2 )
dx
2 x
xi. cot x dx = log sin x + c
d (x2 2) 2xV| 2xdx x2 constant
xii. sec x dx = log(secx + tanx) + c
dx = – log (sec x –tan x) + c
d 2
dx (x k) 2x|
= log tan
FGH xIJ+ c
So we always add a constant to the integral of function, which is called the constant of
xiii. cosec x dx = – log (cosec x + cot x) + c
Integration. It is generally denoted by c. Due to presence of this constant such an integral is called an Indefinite integral.
= log (cosec x – cot x) + c = log tan
xiv. sec x tan x dx = sec x + c
FGHxIJK+ c
If f(x), g(x) are two functions of a variable x and k is a constant, then-
(i) k f(x) dx = k f(x) dx.
(ii) [f(x) g(x)] dx = f(x)dx ± g(x) dx
(iii) d/dx ( f(x) dx) = f(x)
(iv) f(x)KJdx = f(x)
The following integrals are directly obtained from the derivatives of standard functions.
xv. cosec x cot x dx = – cosec x + c
xvi. sec2 x dx = tan x + c
xvii. cosec2 x dx = – cot x + c xviii. sinh x dx = cosh x + c
xix. cosh x dx = sinh x + c
xx. sech2 x dx = tanh x + c
xxi. cosech2 x dx = – coth x + c
xxii. sech x tanh x dx = – sech x + c
xxiii. cosech x coth x = – cosech x + c
1 1
FxI
eax
R 1FbI
xxiv. xxiv.
x2 + a2 dx =
a tan–1
GHa + c
= a2 b2
sin
STbx tan
GHaJK+ c
xxv. z 1
1
dx = log
FGx a + c
xxxv. zeax cos bx dx
x2 a2
2 a Hx aK
eax
xxvi. z 1
dx = 1 log FGa xIJ + c
= a2 b2
(a cos bx + b sin bx) + c
a2 x2
1
2 a Ha xK
FxI
= cos
STbx tan
1 b V+ c
xxvii. za2 x2 dx = sin–1
GHaJK+ c
FxI
Examples Integration of Function
xxviii. xxviii.
= – cos–1
1
dx = sinh–1
x2 a2
GHaJK+ c
FGxIJ+ c
Ex.1 Evaluate : zx–55 dx
Sol. x–55 dx
x54
= log (x +
) + c
= 54
+ c Ans.
xxix. z 1
dx = cosh–1
FGxIJ+ c
Ex.2 Evaluate :
zex2 1j2
x2 a2
= log (x +
HaK
) + c
Sol.
x
x4 2 x2 1
dx
x
xxx. xxx.
2 2 dx
= zx3 2x 1IJdx
za x
H xK
x4
= x +
2
a . sin–1
2
x + c
a
INDEFINITE INTEGRATION
Preface 1
1. Integration of a Function 2
2. Basic Theorems on Integration 2
3. Standard Integrals 2
4. Methods of Integration 4
4.1 Integration by Substitution
4.2 Integration by Parts
4.3 Integration of Rational Function
4.4 Integration of Irrational Function
4.5 Integration of Trigonometric Function
5. Some Integrates of Different Expression of ex 12
6. Solved Examples 14
DEFINITE INTEGRATION
Preface 24
1. Defination 25
2. Properties Of Definite Integral 26
3. Some Important Formulae 29
4. Summation of Series by Integration 30
5. Solved Examples 32
QUADRATURE
Preface 41
1. Introduction 42
2. Curve Tracing 42
3. Area Bounded by A Curve 42
4. Symmetrical Area 43
5. Positive and Negative Area 44
6. Area Between Two Curves 45
7. Solved Examples 47
DIFFERENTIAL EQUATION
Preface 52
1. Introduction 53
2. Differential Equation 53
2.1 Order of Differential Equation
2.2 Degree of Differential Equation
3. Linear And Non- Linear Differential Equation 53
4. Formation of Differential Equation 53
5. Solution of Differential Equation 54
6. Methods of Solving A first Order & A first Degree Differential EqN 55
6.1 Differential Equation of the form dy/dx = f(x)
6.2 Differential Equation of the form dy/dx = f(x) g(y)
6.3 Differential Equation of the form dy/dx = f(ax+by+ c)
6.4 Differential Equation of homogeneous type
6.5 Differential Equation reducible to homogeneous form
6.6 Linear Differential Equation
6.7 Equation reducible to linear form
6.8 Differential Equation of the form of d2y / dx2 = f(x)
7. Solved Examples 62
All Rights Reserved with CAREER POINT Revised Edition
LEVEL # 1
Order and degree of differential equation
Q.7 The degree of the differential equation
d2y
Q.1 A differential equation of first order and first degree is-
dy 2
dx2 +
(A) 1
= 0 is-
(B) 2 (C) 3 (D) 6
(A) x
d2y
– x + a = 0
Q.8 The order of the differential equation whose solution is y = a cos x + b sin x + c e–x is-
(A) 3 (B) 2
(B) (B)
dx2 + xy = 0
(C) 1 (D) None of these
(C) dy + dx = 0
(D) None of these
Q.2 The order and degree of differential equation
Q.9 The differential equation of all circles of radius a is of order-
(A) 2 (B) 3
(C) 4 (D) None of these
dx + y
dy = 0 are respectively-
Q.10 The order of the differential equation of all
(A) 1,2 (B) 1,1
(C) 2,1 (D) 2,2
Q.3 The order and degree of the differential
circles of radius r, having centre on y-axis and passing through the origin is-
(A) 1 (B) 2 (C) 3 (D) 4
Q.11 The degree of the differential equation
dy d2 y
dy 2
d2 y
equation y = x dx +
is -
+ 3 = x2 log
2 is-
dx2
dx
dx
(A) 1,2 (B) 2,1
(C) 1,1 (D) 2, 2
Q.4 The order and degree of the differential
(A) 1 (B) 2
(C) 3 (D) None of these
Q.12 The differential equation
dy 2 2 / 3 2
equation
4
d y
= are-
d2 y 2
dy 4
dx
dx2
x +
+ y = x2 is of -
dx2
dx
(A) 2, 2 (B) 3, 3
(C) 2, 3 (D) 3, 2
Q.5 The order and the degree of differential
(A) Degree 2 and order 2
(B) Degree 1 and order 1
(C) Degree 4 and order 3
d4 y d3 y
d2y dy
(D) Degree 4 and order 4
equation
dx 4 – 4 dx3 + 8
dx2 – 8 dx
+ 4y = 0 are respectively-
(A) 4,1 (B) 1,4
(C) 1,1 (D) None of these
Q.13
Linear and non linear differential equation
Which of the following equation is linear-
Q.6 The order and degree of differential equation (xy2 + x) dx + (y – x2 y) dy = 0 are-
(A) 1, 2 (B) 2,1
(C) 1,1 (D) 2, 2
(A)
(C)
dy + xy2 = 1 (B) x2
dx
dy + 3y = xy2 (D) x
dx
dy + y = ex
dx
dy + y2 = sinx
dx
Q.14 Which of the following equation is non- linear-
dy
(A) dx = cos x
Q.18 The differential equation of the family of curves y2 = 4a (x + a) , where a is an arbitrary constant, is-
dy 2 dy
d2y
(A) y
1
dx
= 2x dx
(B)
dx2 + y = 0
(C) dx + dy = 0
dy 2 dy
dy 3
(B) y
1
dx
= 2x dx
(D) x
dx +
dy = y2
dx
Q.15 Which of the following equation is linear-
(C)
d2y dx2
dy
+ 2 dx = 0
d2 y 2
dy 2
dy 3 dy
(A) 2 + x2
= 0
(D) + 3
+ y = 0
dx
dx
dx dx
dy
(B) y = dx +
Q.19 The differential equation of all the lines in the xy- plane is-
dy y
dy d2y dy
(C) dx + x = log x
dy
(A) dx – x = 0 (B) dx2 – x dx = 0
d2y d2y
(D) y dx – 4 = x
(C) = 0 (D) + x = 0
dx2 dx2
Formation of differential equation
Q.20 The differential equation of the family
LEVEL # 1
Area bounded by a curve
Q.1 The area between the curves y = 6 – x – x2
and x-axis is -
(A) 125/6 (B) 125/2
(C) 25/6 (D) 25/2
Q.2 The area between the curve y =ex and x-axis which lies between x = – 1 and x = 1 is-
(A) e2 – 1 (B) (e2 –1)/e
(C) (1–e)/e (D) ( e– 1)/e2
Q.3 The area bounded by the curve y = sin 2x,
Q.9 The area bounded by the curve y = 1 + 8/x2, x-axis, x = 2 and x = 4 is-
(A) 2 (B) 3 (C) 4 (D) 5
Q.10 The area between the curve y = log x and x-axis which lies between x = a and x = b (a > 1, b > 1) is-
(A) b log (b/e) – a log (a/e)
(B) b log (b/e) + a log (a/e)
(C) log ab
(D) log (b/a)
Q.11 Area bounded by the curve y = xex2 , x- axis and the ordinates x = 0, x = is-
x- axis and the ordinate x = /4 is- (A) /4 (B) /2 (C) 1 (D) 1/2
(A)
e2 1
2
sq. units (B)
e2 1
2
sq.units
Q.4 The area between the curve xy = a2, x-axis, x = a and x = 2a is-
(A) a log 2 (B) a2 log 2
(C)
e 2
1 sq. units (D)
e 2
1sq.units
(C) 2a log 2 (D) None of these
Q.5 Area under the curve y = sin 2x + cos 2x
between x = 0 and x = 4 , is-
(A) 2 sq. units (B) 1 sq. units
(C) 3 sq. units (D) 4 sq. units
Q.6 The area bounded by the curve y = 4x2 ; x = 0, y = 1 and y = 4 in the first quadrant is-
Q.12 The area bounded between the curve y = 2x2 + 5, x-axis and ordinates x = – 2 and x = 1 is-
(A) 21 (B) 29/5 (C) 23 (D) 24
Q.13 Area bounded by curve xy = c, x-axis between x = 1 and x = 4, is-
(A) c log 3 sq. units
(B) 2 log c sq. units
(C) 2c log 2 sq. units
(D) 2c log 5 sq. units
2 1
(A) 2 3 (B) 3 3
Q.14 The area bounded by the curve y = x sin x2,
(C) 2 1
3
(D) 3 1
2
x-axis and x = 0 and x =
is-
Q.7 The area between the curve y = sec x and y-axis when 1 y 2 is-
(A) 1/2 (B) 1/
(C) 1/4 (D) /2
(A)
2 – log ( 2 + )
3
Q.15 The area bounded between the curve
x – y + 1 = 0, x = – 2, x = 3 and x-axis is-
2 4 2
(B) 3 + log ( 2 + )
(C) – 1 log (2 + )
(A) 45/4 (B) 45/2
(C) 15 (D) 25/2
Q.16 The area bounded by curves y = tan x,
3 2
(D) None of these
Q.8 The area bounded by the lines y = x, y = 0
x- axis and x =
3 is-
and x = 2 is-
(A) 2 log 2 (B) log 2
(A) 1 (B) 2
(C) log
FG2
J (D) 0
(C) 4 (D) None of these
H3 K
Q.17 The area between the curve x2 = 4ay, x-axis, and ordinate x = d is-
(A) d3 /12a (B) d3/a
(C) d3/2a (D) d3/6a
Q.18 Area bounded by the curve y = x (x – 1)2 and x-axis is-
(A) 4 (B) 1/3 (C) 1/12 (D) 1/2
Q.19 The area bounded by the curve y = loge x, x-axis and ordinate x = e is-
(A) loge2 (B) 1/2 unit
(C) 1 unit (D) e unit
1
Q.20 The area bounded by the curve y = cos 2 x ,
Q.28 The area of a loop bounded by the curve y = a sin x and x-axis is-
(A) a (B) 2a2 (C) 0 (D) 2a
Q.29 The area between the curves x = 2 – y – y2 and y-axis is-
(A) 9 (B) 9/2 (C) 9/4 (D) 3
Q.30 The area bounded by y = 4x – x2 and the x-axis is-
(A) 30/7 (B) 31/7 (C
LEVEL # 1
Q.1
Integration of function
1 sin2x dx equals-
dx
Q.7 4x2 9
1
dx is equal to -
3x 1
2x
(A) sin x + cos x + c
(B) sin x – cos x + c
(A) tan–1 + c (B) tan–1 + c
(C) cos x – sin x + c
(D) None of these
(C)
2x
tan–1 3 + c (D)
2x
tan–1 3 + c
2 2
4 5 sin x
Q.2 cos2 x
dx equals-
Q.8 cos2x sin4x dx is equal to -
(A) 4 tan x – sec x + c
(B) 4 tan x + 5 sec x + c
(A)
1 (cos 6x + 3 cos 2x ) + c
12
1
(C) 9 tan x + c
(D) None of these
(B) 6 (cos 6x + 3 cos 2x) + c
1
(C) – 12 (cos 6x + 3 cos 2x) + c
Q.3 (tan x + cot x) dx equals-
(A) log (c tan x)
(B) log (sin x + cosx) + c
(C) log (cx)
(D) None of these
(D) None of these
2dx
Q.9 x2 – 1 equals -
1 x 1
1 x – 1
(A) 2 log x – 1 + c (B)
log x 1 + c
Q.4
e5 loge x e4 loge x
e3 loge x e2 loge x dx equals-
x 1
2
x – 1
(C) log
x – 1
+ c (D) log
x 1
(A)
x + c (B)
2
4
x + c
3
Q.10
(ax bx )2 axbx
dx equals-
(C) x
4
+ c (D) None of these
(A) (a/b)x + 2x + c (B) (b/a)x + 2x + c
(C) (a/b)x – 2x + c (D) None of these
Q.5 1 cos 2x 1 cos 2x
dx equals-
Q.11
dx
sin2 xcos2 x
equals-
(A) tan x + x + c (B) tan x – x + c
(C) sin x – x + c (D) sin x + x + c
(A) tan x – cot x + c (B) tan x + cot x + c
(C) cot x – tan x + c (D) None of these
Q.6 The value of (1 x)
(A) x – x2 – x3 + c
(B) x + x2 – x3 + c
(C) x + x2 + x3 + c
(D) None of these
(1 + 3x) dx equal to -
sin x
Q.12 dx equals-
1 cos x
(A) 2 cos (x/2) + c
(B) 2 sin(x/2) + c (C)2 2 cos (x/2) + c
(D) –2 cos(x/2)+c
Q.13 sec x (tan x + sec x) dx equals-
Q.20
2x 3 x
5x
dx equals-
(A) tan x – sec x + c
(B) sec x – tan x+ c
(C) tan x + sec x + c
(A)
b2 / 5gx log 2 / 5 +
b3 / 5gx
log 3 / 5 + c
(D) None of these
Q.14 The value of sin x cos x
1 sin 2x
dx is-
(B) loge (2x/5) + loge (3x/5) + c
(C) x + c
(D) None of these
sin2 x
(A) sin x + c (B) x + c
Q.21 The value of
dx is-
1 cos x
(C) cos x + c (D) 1 (sin x + cos x)
2
1
(A) x – sin x + c
(B) x + sin x + c
(C) – x – sin x + c
Q.15 The value of
1 cos x dx is-
(D) None of these
(A) 1 2
cot (x/2) + c
1
(B) – 2 cot (x/2) + c
Q.22 e2x+3 dx equals-
(C) – cot (x/2) + c (D) – tan (x/2) + c
(A) 1 e2x+3 + c (B) 1 e2x+5 + c
cos 2x 2 sin2 x 2 2
Q.16
cos2 x
dx equals-
(C) 1 e2x+3 + c (D) 1 e2x+4 – c
(A) cot x + c (B) sec x + c
(C) tan x + c (D) cosec x + c
Q.23
3 2
1 tan x
1 tan x dx equals-
Q.17 cosx FG1
sin x IJ
dx equals-
(A) log (cos x + sin x) + c
Hsin2 x
cos3 xK
(B) log (cos x – sin x) + c
(A) sec x – cosec x + c
(B) cosec x – sec x + c
(C) sec x + cosec x + c
(D) None of these
(C) log (sin x – cos x) + c
(D) None of these
sin4 x cos4 x
Q.24
sin2 xcos2 x
dx equals-
Q.18
sin3 x cos3 x
In this chapter, we shall study the nature of a
function which is governed by the sign of its derivative. If the graph of a function is in upward going direction or in downward coming direction then it is called as monotonic function, and this property of the function is called Monotonicity. If a function is defined in any interval, and if in some part of the interval, graph moves upwards
A function is said to be monotonic function in a domain if it is either monotonic increasing or monotonic decreasing in that domain.
NOTE : If x < x f(x ) < f(x ) x , x D, then
and in the remaining part moves downward then
1 2 1
2 1 2
function is not monotonic in that interval.
f(x) is called strictly increasing in domain D.
These are of two types –
2.1 Monotonic Increasing :
A function f(x) defined in a domain D is said to be monotonic increasing function if the value of f(x) does not decrease (increase) by increasing (decreasing) the value of x or
We can say that the value of f(x) should increase (decrease) or remain equal by increasing
(Decreasing) the value of x.
Similarly if x1 < x2 f(x1) > f(x2), x1, x2 D then it is called strictly decreasing in domain D.
R
If Tor x1 x2 f(x1) f(x2 )
, x1 , x2 D
or SR x1 x2 f(x1) f(x2 ) , x1 , x2 D
2.2 Monotonic Decreasing :
A function f(x) defined in a domain D is said to be monotonic decreasing function if the value of f(x) does not increase (decrease) by increasing (decreasing) the value of x or
We can say that the value of f(x) should decrease (increase) or remain equal by increasing
(Decreasing) the value of x.
For Example
(i) f(x) = ex is a monotonic increasing function where as g(x) = 1/x is monotonic decreasing function.
(ii) f(x) = x2 and g(x) = | x | are monotonic increasing for x > 0 and monotonic decreasing for x < 0. In general they are not monotonic functions.
(iii) Sin x, cos x are not monotonic function whereas tan x, cot x are monotonic.
(i) At a Point : A function f(x) is said to be monotonic increasing (decreasing) at a point x = a of its domain if it is monotonic increasing
R
If Tor x1 x2 f(x1) f(x2 )
, x1 , x2 D
(decreasing) in the interval (a – h, a + h) where h is a small positive number. Hence we may
or SR x1 x2 f(x1) f(x2 ) , x1 , x2 D
observe that if f(x) is monotonic increasing at x = a, then at this point tangent to its graph will make an acute angle with the x–axis where as if the function is monotonic decreasing these tangent will make
an obtuse angle with x–axis. Consequently f' (a) will be positive or negative according as f(x) is monotonic increasing or decreasing at x = a.
Ex. Function f(x) = x2 + 1 is monotonically decreasing in [ –1, 0] because
f' (x) = 2x < 0, x (–1, 0)
Ex. Function f(x) = x2 is not a monotonic function in the interval [–1, 1] because
f' (x) > 0, when x = 1/2
Ex. Function f(x) = sin2x + cos2x is constant
In this chapter we shall study those points of the domain of a function where its graph changes its direction from upwards to downwards or from downwards to upwards. At such points the derivative of the function, if it exists, is necessarily zero.
The value of a function f (x) is said to be maximum at x = a, if there exists a very small positive number h, such that
f(x) < f(a) x (a – h,a + h) , x a
In this case the point x = a is called a point of maxima for the function f(x).
Simlarly, the value of f(x) is said to the minimum at x = b, If there exists a very small positive number, h, such that
f(x) > f(b), x (b – h,b + h), x b
In this case x = b is called the point of minima for the function f(x).
Hene we find that,
(i) x = a is a maximum point of f(x)
f(a) f(a h) 0
f(a) f(a h) 0
(ii) x = b is a minimum point of f(x)
Rf(b) f(b h) 0
f(b) f(b h) 0
(iii) x = c is neither a maximum point nor a minimum point
Note :
(i) The maximum and minimum points are also known as extreme points.
(ii) A function may have more than one maximum and minimum points.
(iii) A maximum value of a function f(x) in an interval [a,b] is not necessarily its greatest value in that interval. Similarly, a minimum value may not be the least value of the function. A minimum value may be greater than some maximum value for a function.
(iv) If a continuous function has only one maximum (minimum) point, then at this point function has its greatest (least) value.
(v) Monotonic functions do not have extreme points.
Ex. Function y = sin x, x (0, ) has a maximum point at x = /2 because the value of sin /2 is greatest in the given interval for sin x.
Clearly function y = sin x is increasing in the interval (0, /2) and decreasing in the interval ( /2, ) for that reason also it has maxima at x = /2. Similarly we can see from the graph of cos x which has a minimum point at x = .
Ex. f(x) = x2 , x (–1,1) has a minimum point at x = 0 because at x = 0, the value of x2 is 0, which is
less than the all the values of function at different points of the interval.
Clearly function y = x2 is decreasing in the interval
Rf(c) f(c h)
Sand
Tf(c) f(c h) 0
|V| have opposite signs.
(–1, 0) and increasing in the interval (0,1) So it has minima at x = 0.
Ex. f(x) = |x| has a minimum point at x = 0. It can be easily observed from its graph.
A. Necessary Condition : A point x = a is an extreme point of a function f(x) if f’(a) = 0, provided f’(a) exists. Thus if f’ (a) exists, then
x = a is an extreme point f’(a) = 0
or
f’ (a) 0 x = a is not an extreme point.
But its converse is not true i.e.
f’ (a) = 0 x = a is an extreme point.
For example if f(x) = x3 , then f’ (0) = 0 but x = 0 is not an extreme point.
B. Sufficient Condition :
(i) The value of the function f(x) at x = a is maximum, if f’ (a) = 0 and f” (a) < 0.
(ii) The value of
CONTINUITY AND DIFFERENTIABILITY
Total No. of questions in Continuity and Differentiability are-
In Chapter Examples 14
Solved Examples 17
Total No. of questions 31
Ex.
The word 'Continuous' means without any break or gap. If the graph of a function has no break or gap or jump, then it is said to be continuous.
A function which is not continuous is called a
discontinuous function. In other words,
If there is slight (finite) change in the value of a function by slightly changing the value of x then function is continuous, otherwise discontinuous, while studying graphs of functions, we see that graphs of functions sin x, x, cos x, ex etc. are continuous but greatest integer function [x] has break at every integral point, so it is not continu- ous. Similarly tan x, cot x, secx, 1/x etc. are also discontinuous function.
(Discontinuous)
For examining continuity of a function at a point, we find its limit and value at that point, If these two exist and are equal, then function is continu- ous at that point.
A function f(x) is said to be continuous at a point x = a if
(i) f (a) exists
(ii) x Lim
f(x) exists and finite
Lim Lim
so x a f(x) = x a f(x)
(iii) x Lim
f(x) = f(a) .
or function f(x) is continuous at x = a.
Lim Lim
(Continuous function)
If x a f(x) = x a f(x) = f(a).
i.e. If right hand limit at 'a' = left hand limit at 'a'= value of the function at 'a'.
If x Lim
f(x) does not exist or
Lim
x a
f(x) f(a),
then f(x) is said to be discontinuous at x= a.
Ex.1 Examine the continuity of the function
X R|x2 9
f (x) =
S| x 3 , when x 3
at x = 3.
T6, when x 3
Sol. f (3) = 6 ( given)
bx 3gbx 3
Lim
x 3
f(x) =
Lim
x 3
bx 3g = 6
(Discontinuous at x = 0)
Lim
x 3
f(x) = f(3)
f (x) is continuous at x = 3.
Ex.2 If f(x) =
log (1 2ax) log 1 bx
x 0
x
Thus a function f(x) is continuous at a point x = a if it is left continuous as well as right continuous at x = a.
Tk ,
x 0
If function is continuous at x = 0 then the value of k is –
(A) a + b (B) 2a +b
Ex.4 Examine the continuity of the function
(C) a – b (D) 0
x2 1, when x 2
f(x) =
|T2x, when x 2
Sol.
Lim
x 0
logFGH1 2axIJ
x
at the point x = 2.
Sol. f(2) = 22 + 1 = 5
= x Lim
FG1 bx J.
b1 bxgb2ag b1 2axgbb
2
f(2– 0) =
Lim
h 0
Lim
b2 hg2 1 = 5
0 H1 2ax
2a b
0 b1 bxgb1 2 axg
b1 bxg
b2a b
b1gb1g
f (2+ 0) = h 0 2( 2+ h) = 4
f(2– 0) f(2+ 0) f(2)
x Lim
= = 2a + b
Ans.[B]
f(x) is not continuous at x = 2.
Ex.5 Check the continuity of the function
1 cos 4x, x 0
R|x 2
x 3
Ex.3 If f(x) = S| x2
is continuous then
f(x) = 5 x 3
at x = 3.
Ta,
x 0
T8 x x 3
the value of a is equal to –
(A) 0 (B) 1
(C) 4 (D) 8
Sol. Since the given function is continuous at x= 0
Sol.
02Application of Derivative # 1 (Tangent & Normal)~1 Module-4.pdfRajuSingh806014
If y = f(x) be a given function, then the differential coefficient f' (x) or dy at the point P (x , y ) is
(i) If the tangent at P (x1,y1) of the curve y = f(x) is parallel to the x- axis (or perpendicular to y- axis) then = 0 i.e. its slope will be
zero.
FGdy J
dx 1 1
m = H K = 0
the trigonometrical tangent of the angle (say) which the positive direction of the tangent to the curve at P makes with the positive direction of x- axis Gdy J, therefore represents the slope of the
tangent. Thus
dx (x1,y1)
The converse is also true. Hence the tangent at (x1,y1) is parallel to x- axis.
GJ = 0
(x1,y1)
(ii) If the tangent at P (x , y ) of the curve y =
1 1
f (x) is parallel to y - axis (or perpendicular
to x-axis) then = / 2 , and its slope will be infinity i.e.
dy
m = =
dx (x ,y )
The converse is also true. Hence the tangent at (x1, y1) is parallel to y- axis
Fdy
(x1,y1)
Thus
(i) The inclination of tangent with x- axis.
dy
(iii) If at any point P (x1, y1) of the curve y = f(x), the tangent makes equal angles with the
axes, then at the point P, = / 4 or 3 / 4 ,
= tan–1
GHdxJK
Hence at P, tan = dy/dx = 1. The
(ii) Slope of tangent = dy
dx
(iii) Slope of the normal = – dx/dy
Ex.1 Find the following for the curve y2 = 4x at point (2,–2)
(i) Inclination of the tangent
(ii) Slope of the tangent
(iii) Slope of the normal
Sol. Differentiating the given equation of curve, we get dy/dx = 2/y = –1 at (2,–2)
so at the given point.
(i) Inclination of the tangent = tan–1(–1) = 135º
(ii) Slope of the tangent = –1
(iii) Slope of the normal = 1
converse of the result is also true. thus at
(x1,y1) the tangent line makes equal angles with the axes.
GJ = 1
(x1,y1)
Ex.2 The equation of tangent to the curve y2 = 6x at (2, – 3).
(A) x + y – 1 = 0 (B) x + y + 1 = 0 (C) x – y + 1 = 0 (D) x + y + 2 = 0
Sol. Differentiating equation of the curve with respect to x
(a) Equation of tangent to the curve y = f(x) at A (x1,y1) is
2y dy = 6
dx
FGdy J
dx (2,3)
= 3 = –1
3
y – y1 =
FGdy J
(x1,y1)
(x–x1)
Therefore equation of tangent is y + 3 = – (x – 2)
x + y + 1 = 0 Ans. [B]
Ex.3 The equation of tangent at any of the curve x = at2, y = 2at is -
(A) x = ty + at2 (B) ty + x + at2 = 0
(C) ty = x + at2 (D) ty = x + at3
2 a 1
Sol. dy/dx = (dy/dt)/(dx/dt) = 2 at = t
equation of the tangent at (x,y) point is
(y – 2 at) = 1 (x – at2)
t
ty = x + at2 Ans.[C]
Ex.4 The equation of the tangent to the curve x2 (x – y) + a2 (x + y) = 0 at origin is-
(A) x + y + 1 = 0 (B) x + y + 2 = 0 (C) x + y = 0 (D) 2x – y = 0
Sol. Differentiating equation of the curve w.r.t. x
dy/dx = – y x
(i) If tangent line is parallel to x - axis, then dy/dx = 0 y = 0 and x = a
Thus the point is (a,0)
(ii) If tangent is parallel to y – axis , then dy/dx = x = 0 and y = a
Thus the point is (0,a)
(iii) If tangent line makes equal angles with both axis , then dy/dx = 1
y =
(a) Natural Numbers : N = {1,2,3,4,...}
(b) Whole Numbers : W = {0,1,2,3,4, }
(c) Integer Numbers :
or Z = {...–3,–2,–1, 0,1,2,3, },
Z+ = {1,2,3,....}, Z– = {–1,–2,–3, }
Z0 = {± 1, ± 2, ± 3, }
(d) Rational Numbers :
p
Q = { q ; p, q z, q 0 }
(i) R0 : all real numbers except 0 (Zero).
(j) Imaginary Numbers : C = {i,, }
(k) Prime Numbers :
These are the natural numbers greater than 1 which is divisible by 1 and itself only, called prime numbers.
Ex. 2,3,5,7,11,13,17,19,23,29,31,37,41,...
(l) Even Numbers : E = {0,2,4,6, }
(m) Odd Numbers : O = {1,3,5,7, }
Ex. {1,
Note :
5
, –10, 105,
3
22 20
7 , 3
, 0 ....}
The set of the numbers between any two real numbers is called interval.
(a) Close Interval :
(i) In rational numbers the digits are repeated after decimal.
(ii) 0 (zero) is a rational number.
(e) Irrational numbers: The numbers which are not rational or which can not be written in the form of p/q ,called irrational numbers
Ex. { , ,21/3, 51/4, ,e, }
Note:
(i) In irrational numbers, digits are not repeated after decimal.
(ii) and e are called special irrational quantities.
(iii) is neither a rational number nor a irrational number.
(f) Real Numbers : {x, where x is rational and irrational number}
20
[a,b] = { x, a x b }
(b) Open Interval:
(a, b) or ]a, b[ = { x, a < x < b }
(c) Semi open or semi close interval:
[a,b[ or [a,b) = {x; a x < b}
]a,b] or (a,b] = {x ; a < x b}
Let A and B be two given sets and if each element a A is associated with a unique element b B under a rule f , then this relation is called function.
Here b, is called the image of a and a is called the pre- image of b under f.
Note :
(i) Every element of A should be associated with
Ex. R = { 1,1000, 20/6, ,
, –10, –
,.....}
3
B but vice-versa is not essential.
(g) Positive Real Numbers: R+ = (0,)
(h) Negative Real Numbers : R– = (– ,0)
(ii) Every element of A should be associated with a unique (one and only one) element of but
any element of B can have two or more rela- tions in A.
3.1 Representation of Function :
It can be done by three methods :
(a) By Mapping
(b) By Algebraic Method
(c) In the form of Ordered pairs
(A) Mapping :
It shows the graphical aspect of the relation of the elements of A with the elements of B .
Ex. f1:
f2 :
f3 :
f4 :
In the above given mappings rule f1 and f2
shows a function because each element of A is
associated with a unique element of B. Whereas
f3 and f4 are not function because in f 3, element c is associated with two elements of B, and in f4 , b is not associated with any element
of B, which do not follow the definition of function. In f2, c and d are associated with same element, still it obeys the rule of definition of function because it does not tell that every element of A should be associated with different elements of B.
(B) Algebraic Method :
It shows the relation between the elem
01. Differentiation-Theory & solved example Module-3.pdfRajuSingh806014
Total No. of questions in Differentiation are-
In Chapter Examples 31
Solved Examples 32
The rate of change of one quantity with respect to some another quantity has a great importance. For example the rate of change of displacement of a particle with respect to time is called its velocity and the rate of change of velocity is
called its acceleration.
The following results can easily be established using the above definition of the derivative–
d
(i) dx (constant) = 0
The rate of change of a quantity 'y' with respect to another quantity 'x' is called the derivative or differential coefficient of y with respect to x.
Let y = f(x) be a continuous function of a variable quantity x, where x is independent and y is
(ii)
(iii)
(iv)
(v)
d
dx (ax) = a
d (xn) = nxn–1
dx
d ex =ex
dx
d (ax) = ax log a
dependent variable quantity. Let x be an arbitrary small change in the value of x and y be the
dx
d
(vi) dx
e
(logex) = 1/x
corresponding change in y then lim
y
if it exists, d 1
x0 x
is called the derivative or differential coefficient of y with respect to x and it is denoted by
(vii) dx
(logax) =
x log a
dy . y', y
dx 1
or Dy.
d
(viii) dx (sin x) = cos x
So, dy dx
dy
dx
lim
x0
lim
x0
y
x
f (x x) f (x)
x
(ix) (ix)
(x) (x)
d
dx (cos x) = – sin x
d (tan x) = sec2x
dx
The process of finding derivative of a function is called differentiation.
If we again differentiate (dy/dx) with respect to x
(xi)
d (cot x) = – cosec2x
dx
d
then the new derivative so obtained is called second derivative of y with respect to x and it is
Fd2 y
(xii) dx
d
(xiii) dx
(secx)= secx tan x
(cosec x) = – cosec x cot x
denoted by
HGdx2 Jor y" or y2 or D2y. Similarly,
d 1
we can find successive derivatives of y which
(xiv) dx
(sin–1 x) = , –1< x < 1
1 x2
may be denoted by
d –1 1
d3 y d4 y
dn y
(xv) dx (cos x) = –
,–1 < x < 1
dx3 ,
dx4 , ........, dxn , ......
d
(xvi) dx
(tan–1 x) = 1
1 x2
Note : (i)
y is a ratio of two quantities y and
x
(xvii) (xvii)
d (cot–1 x) = – 1
where as dy
dx
dy
is not a ratio, it is a single
dx
d
(xviii) (xviii)
(sec–1 x) =
1 x2
1
|x| > 1
quantity i.e.
dx dy÷ dx
dx x x2 1
(ii)
dy is
dx
d (y) in which d/dx is simply a symbol
dx
(xix)
d (cosec–1 x) = – 1
dx
of operation and not 'd' divided by dx.
d
(xx) dx
(sinh x) = cosh x
d
(xxi) dx
d
(cosh x) = sinh x
Theorem V Derivative of the function of the function. If 'y' is a function of 't' and t' is a function of 'x' then
(xxii) dx
d
(tanh x) = sech2 x
dy =
dx
dy . dt
dt dx
(xxiii) dx
d
(xxiv) dx
d
(coth x) = – cosec h2 x (sech x) = – sech x tanh x
Theorem VI Derivative of parametric equations If x = (t) , y = (t) then
dy dy / dt
=
(xxv) dx
(cosech x) = – cosec hx coth x
dx dx / dt
(xxvi) (xxvi)
(xxvii) (xxvii)
d (sin h–1 x) =
Level # 1 ........................................ 48
Level # 2 ........................................ 16
Level # 3 ........................................ 12
Level # 4 ........................................ 16
LEVEL # 1
Q.1 When x < 0, function f(x) = x2 is
(A) Decreasing
(B) Increasing
(C) Constant
(D) Not monotonic
Q.2 When x > 1, function f(x) = x3 is
(A) Increasing (B) Decreasing
(C) Constant (D) not monotonic
Q.3 In the interval (0, 1), f(x) = x2 – x + 1 is
(A) Monotonic (B) Not monotonic
(C) Decreasing (D) Increasing
Q.4 f(x) = x + 1/x, x 0 is increasing when
(A) | x | < 1 (B) | x | > 1
(C) | x | < 2 (D) | x | > 2
|x|
Q.11 For which value of x, the function f(x) = x2 –2x is decreasing
(A) x > 1 (B) x > 2
(C) x < 1 (D) x < 2
x 2
Q.12 Function f(x) = x 1 , x –1 is
(A) Increasing
(B) Decreasing
(C) Not monotonic
(D) None of these
Q.13 Function f(x) = x3 is
(A) Increasing in (0, ) and decreasing in (–, 0)
(B) Decreasing in (0, ) and increasing in (–, 0)
(C) Decreasing throughout
(D) Increasing throughout
Q.5 The function f(x) =
(A) Decreasing
(B) Increasing
x (x 0), x > 0 is
Q.14 Function f(x) = x | x | is
(A) Monotonic increasing
(B) Monotonic decreasing
(C) Constant function
(D) None of these
Q.6 When x (0, 1), function f(x) = 1 / is
(A) Increasing
(B) Decreasing
(C) Neither increasing nor decreasing
(D) Constant
Q.7 Function f(x) = 3x4 + 7x2 + 3 is
(A) Monotonically increasing
(B) Monotonically decreasing
(C) Not monotonic
(D) Odd function
Q.8 For what values of x, the function
(C) Not monotonic
(D) None of these
Q.15 If f and g are two decreasing functions such that fog is defined then fog is
(A) Decreasing (B) Increasing
(C) Can't say (D) None of these
Q.16 For the function f(x) = | x |, x > 0 is
(A) Decreasing
(B) Increasing
(C) Constant function
(D) None of these
Q.17 In the following , monotonic increasing
4
f(x) = x + x2
is monotonically decreasing
fucntion is
(A) x + | x | (B) x – | x |
(A) x < 0 (B) x > 2
(C) x < 2 (D) 0 < x < 2
(C) | x | (D) x | x |
Q.9 If f(x) = x 2
for –7 x 7, then f(x) is
x 1
Q.18 At x = 0, f(x) = is
2 x x 2
increasing function of x in the interval (A) [7, 0] (B) (2, 7]
(C) [–2, 2] (D) [0, 7]
(A) Increasing (B) Decreasing
(C) Not monotonic (D) Constant
x
Q.10 The function y = 1 x2
interval
decreases in the
Q.19 If f(x) = 2x3 – 9x2 + 12x – 6, then in which interval f(x) is monotonically increasing
(A) (1, 2) (B) (–, 1)
(A) (–, –) (B) (–1, –1)
(C) (0, ) (D) (–, –1)
(C) (2, ) (D) (–, 1) or (2, )
Q.20 For the function f(x) = x3 – 6x2 – 36x + 7 which of the following statement is false
(A) f(x) is decareasing, if –2 < x < 6
(B) f(x) is increasing, if –3 < x < 5
(C) f(x) is increasing, if x < –2
(D) f(x) is increasing, if x > 6
Q.28 Which of the following function is not monotonic
(A) ex – e–x (B) ex + e–x
(C) e–1/x (D) None of these
Q.29 In the following, decreasing function
Level # 1 ........................................ 92
Level # 2 ........................................ 27
Level # 3 ........................................ 30
Level # 4 ........................................ 26
LEVEL # 1
R|sin1 ax
continuity of a function at a point
Q.7 If f(x) =
S x
Tk,
, x 0
x 0
is continuous at
Q.1 Function f(x) =
R1 x,
T
when x 2
x = 2 is
x = 0, then k is equal to-
(A) 0 (B) 1
5 x, when x 2
continuous at x = 2, if f(2) equals-
(A) 0 (B) 1
(C) 2 (D) 3
Rx cos1/ x, x 0
(C) a (D) None of these
Q.8 What is the value of (cos x)1/x at x = 0 so that it becomes continuous at x = 0-
(A) 0 (B) 1
(C) –1 (D) e
Q.2 If f(x) = k,
x 0
is continuous at
R|k
cos x
x = 0, then
(A) k > 0 (B) k< 0
Q.9 If f(x) =
S| 2x
, x / 2 x / 2
is a continuous
(C) k = 0 (D) k 0
function at x = / 2 , then the value of k is-
Q.3 If function f(x) = S|x2 2, x 1
is continuous
(A) –1 (B) 1
(C) –2 (D) 2
x3 a3
at x = 1, then value of f(x) for x< 1 is-
Q.10 If function f(x) =
x a
, is continuous at
(A) 3 (B) 1–2x
(C) 1–4x (D) None of these
Q.4 Which of the following function is continuous at x= 0-
x= a, then the value of f(a) is -
(A) 2a (B) 2a2
(C) 3a (D) 3a2
Rsin 1/ x, x 0
(A) f(x) =
RSsin 2x / x, x 0
Q.11 If f(x) = k,
x 0
is continuous at
(B) f(x) =
T1,
R|1 x1/ x,
T1,
x 0
x 0
x 0
x = 0, then k is equal to-
(A) 8 (B) 1
(C) –1 (D) None of these
Q.12 Function f(x) =
G1
xIJ1/ x
is continuous at
(C) f(x) =
R| 1/ x
|T1,
x 0
x 0
H aK
x= 0 if f(0) equals-
(D) None of these
R|6 5 x,x 0
(A) ea (B) e–a
(C) 0 (D) e1/a
1 cos 7bx
Q.5 If f(x) =
S|T2a x, x 0
is continuous at
Q.13 If f(x) =
x
, (x )
is continu-
x = 0, then the value of a is -
(A) 1 (B) 2
(C) 3 (D) None of these
ous at x= , then f( ) equals-
(A) 0 (B) 1
(C) –1 (D) 7
x2 ba 2gx a
R tan x
Q.6 If f(x) = S|
x 2
, x 2
is continu-
Q.14 If f(x) =
S|sin
x
x 0
, then f(x) is -
T2,
x 2
(A) Continuous everywhere
ous at x = 2, then a is equal to-
(A) 0 (B) 1
(C) –1 (D) 2
(B) Continuous nowhere
(C) Continuous at x= 0
(D) Continuous only at x = 0
Q.15 If f(x) =
2x tan x x
is continuous at x = 0,
Q.22 Function f(x) = [x] is discontinuous at-
(A) Every real number
then f(0) equals-
(A) 0 (B) 1
(C) 2 (D) 3
1 x 3 1 x
(B) Every natural number
(C) Every integer
(D) No where
Q.23 Function f(x) = 3x2–x is-
(A) Discontinuous at x = 1
(B) Discontinuous at x = 0
Q.16 If f(x) =
x , (x 0) is continuous
(C) Continuous only at x = 0
at x= 0, then the value of f(0) is- (A) 1/6 (B) 1/4
(C) 2 (D) 1/3
(D) Continuous at x = 0
R|x2, when x 0
Q.24 If f(x) = S|1, when 0 x 1 , then f(x) is-
Q.17 If f (x) =
Rax2 b when 0 x 1 S2 when x 1 is x 1 when 1 x 2
1 / x, when x 1
(A
LEVEL # 1
Questions
based on
inequation
Q.8 If x2 – 1 0 and x2 – x – 2 0, then x line in the interval/set
(A) (–1, 2) (B) (–1, 1)
Q.1 The inequality
2 < 3 is true, when x belongs to-
x
(C) (1, 2) (D) {– 1}
2
2
Questions Definition of function
(A) 3 ,
(B) 3
based on
2 ,
Q.9 Which of the following relation is a function ?
(C)
x 4
(–, 0) (D) none of these
(A) {(1,4), (2,6), (1,5), (3,9)}
(B) {(3,3), (2,1), (1,2), (2,3)}
(C) {(1,2), (2,2,), (3,2), (4,2)}
(D) {(3,1), (3,2), (3,3), (3,4)}
Q.2
x 3 < 2 is satisfied when x satisfies-
(A) (–, 3) (10, ) (B) (3, 10)
(C) (–, 3) [10, ) (D) none of these
Q.10 If x, y R, then which of the following rules is not a function-
(A) y = 9 –x2 (B) y = 2x2
x 7
(C) y = – |x| (D) y = x2 + 1
Q.3 Solution of x 3 > 2 is-
Questions Even and odd function
(A) (–3, ) (B) (–, –13)
(C) (–13, –3) (D) none of these
2x 3
based on
Q.11 Which one of the following is not an odd function -
Q.4 Solution of
3x 5
3 is-
(A) sin x (B) tan x
(C) tanh x (D) None of these
12 5 12
(A) 1, 7
(B) , 4 4
3 7
Q.12 The function f(x) = sin x cos x
is -
, 5
12 ,
x tanx
(C)
3
(D) 7
(A) odd
(B) Even
Q.5 Solution of (x – 1)2 (x + 4) < 0 is-
(A) (–, 1) (B) (–, –4)
(C) (–1, 4) (D) (1, 4)
Q.6 Solution of (2x + 1) (x – 3) (x + 7) < 0 is-
(C) neither even nor odd
(D) odd and periodic
Q.13 A function is called even function if its graph is symmetrical w.r.t.-
(A) origin (B) x = 0
(C) y = 0 (D) line y = x
1 ,3
1 ,3
(A) (– , –7)
2
(B) (– , – 7)
Q.14 A function is called odd function if its graph is symmetrical w.r.t.-
(C) (–, 7) 1 ,3
2
(D) (–, –7) (3, )
(A) Origin (B) x = 0
(C) y = 0 (D) line y = x
Q.15 The even function is-
Q.7 If x2 + 6x – 27 > 0 and x2 – 3x – 4 < 0, then-
(A) x > 3 (B) x < 4
(A) f(x) = x2 (x2 +1) (B) f(x) = sin3 x + 2
(C) f(x) = x (x +1) (D) f(x) = tan x + c
(C) 3 < x < 4 (D) x = 7 2
Q.16 A function whose graph is symmetrical about the y-axis is given by-
Q.25 In the following which function is not periodic-
(A) f(x) = loge
(x + )
(A) tan 4x (B) cos 2x
(C) cos x2 (D) cos2x
(B) f(x + y) = f(x) + f(y) for all x, y R
(C) f(x) = cos x + sin x
(D) None of these
Q.17 Which of the following is an even function ?
ax 1
1
Q.26 Domain of the function f(x) = x 2
is-
(A) x
ax 1
(B) tan x
(A) R (B) (–2, )
(C) [2, ] (D) [0, ]
(C) (C)
ax ax (D)
2
ax 1
ax 1
Q.27 The domain where function f(x) = 2x2 – 1 and g(x) = 1 – 3x are equal, is-
Q.18 In the following, odd function is -
(A) cos x2 (B) (ex + 1)/(ex – 1)
(C) x2 – |x| (D) None of these
Q.19 The function f(x) = x2 – |x| is -
(A) an odd function
(B) a rational function
(C) an even func
POINTS
Preface 1
1. System of Co - ordinates 2
1.1 Cartesian Co-ordinates
1.2 Polar Co-ordinates
2. Distance Formula 2
3. Applications of Distance Formula 3
3.1 Position of three points
3.2 Position of four points
4. Section Formula 4
5. Co-ordinate of some particular points 4
5.1 Centroid
5.2 Incentre
5.3 Circumcentre
5.4 Ortho Centre
6. Area of triangle and quadrilateral 6
6.1 Area of Triangle
6.2 Area of Quadrilateral
7. Transformation of axes 7
7.1 Parallel Transformations
7.2 Rotational Transformation
7.3 Reflection of a Point
8. Locus 8
9. Some important points 9
10. Solved examples 10
STRAIGHT LINE
Preface 13
1. Equation of Straight line 14
2. Equation of Straight line parallel to axes 14
3. Slope of a line 14
4. Different forms of the equation of straight line 14
4.1 Slope – Intercept Form
4.2 Slope Point Form
4.3 Two Point Form
4.4 Intercept Form
4.5 Normal Form
4.6 Parametric Form
5. Reduction of General form of Equations into Standard Forms 15
6. Position of a point relative to a line 17
7. Angle between two straight lines 17
7.1 Parallel lines
7.2 Perpendicular lines
7.3 Coincident lines
8. Equation of parallel & perpendicular lines 18
9. Equation of Straight lines through (x , y ) making an angle α with y = mx + c 18
10. Length of perpendicular 19
10.1 Distance between two Parallel Lines
11. Condition of concurrency 19
12. Bisectors of angles between two lines 20
13. Line passing through the point of intersection of two lines 20
14. Homogeneous equation 20
15. General Equation of Second degree 21
16. Equation of lines joining the intersection points of a line & a curve to the origin 21
17. Some important points 22
18. Solved examples 23
CIRCLE
Preface 29
1. Definition 30
2. Standard Form of equation of a circle 31
2.1 General Form
2.2 Central Form
2.3 Diametral Form
2.4 Parametric Form
3. Equation of a circle in some special cases 31
4. Position of a point with respect to a circle 32
4.1 The least and the greatest distance of a point from a circle
5. Line and circle 33
5.1 Condition of Tangency
5.2 Intercepts made on coordinate axes by the circle
6. Equation of tangent and normal 34
6.1 Equation of Tangent
6.2 Equation of Normal
6.3 Length of Tangent
6.4 Pair of Tangents
7. Chord of contact 35
8. Director circle 35
9. Position of two circles 36
10. Equation of a chord whose middle point is given 38
11. Circle through the point of intersection 38
12. Common chord of two circles 39
13. Angle of intersection of two circles 39
14.1 Condition of Orthogonality
14. Some important points 40
15. Pole and Polar 40
9.1 Equation of Polar
9.2 Co-ordinates of Pole
9.3 Conjugate Points and Conjugate Lines
16. Radical axis and radical centre 41
16.1 Radical Axis
16.2 Radical Centre
17. Solved examples 42
PARABOLA
Preface 50
1. Definition 51
2. Terms related to parabola 51
3. Standard form of equation of parabola 51
3.1 Parameters of the Parabola y2 = 4ax
3.2 Other Standard Parabola
4. Reduction to Standard Equation 53
.
5. General equat
An ellipse is the locus of a point which moves in such a way that its distance form a fixed point is in constant ratio to its distance from a fixed line. The fixed point is called the focus and fixed line is called the directrix and the constant ratio is called the eccentricity of a ellipse denoted by (e).
In other word, we can say an ellipse is the locus of a point which moves in a plane so that the sum of it distances from fixed points is constant.
2.1 Standard Form of the equation of ellipse
Let the distance between two fixed points S and S' be 2ae and let C be the mid point of SS.
Taking CS as x- axis, C as origin.
Let P(h,k) be the moving point Let SP+ SP = 2a (fixed distance) then
(ii) Major & Minor axis : The straight line AA is called major axis and BB is called minor axis. The major and minor axis taken together are called the principal axes and its length will be given by
Length of major axis 2a Length of minor axis 2b
(iii) Centre : The point which bisect each chord of an ellipse is called centre (0,0) denoted by 'C'.
(iv) Directrix : ZM and Z M are two directrix and their equation are x= a/e and x = – a/e.
(v) Focus : S (ae, 0) and S (–ae,0) are two foci of an ellipse.
(vi) Latus Rectum : Such chord which passes through either focus and perpendicular to the major axis is called its latus rectum.
Length of Latus Rectum :
If L is (ae, 𝑙 ) then 2𝑙 is the length of
SP+S'P=
{(h ae)2 k 2} +
= 2a
Latus Rectum.
Length of Latus rectum is given by
2b2
.
h2(1– e2) + k2 = a2(1– e2)
Hence Locus of P(h, k) is given by. x2(1– e2) + y2 = a2(1– e2)
2
a
(vii) Relation between constant a, b, and e
a 2 b2
b2 = a2(1– e2) e2 =
a 2
x2
a 2 +
y
a 2 (1 e 2 ) = 1
e =
a 2
Result :
Major Axis
(a) Centre C is the point of intersection of the axes of an ellipse. Also C is the mid point of AA.
(b) Another form of standard equation of ellipse
x 2 y2
a 2 b2
1 when a < b.
Directrix Minor Axis Directrix x = -a/e x = a/e
Let us assume that a2(1– e2 )= b2
The standard equation will be given by
x2 y2
a2 b2
2.1.1 Various parameter related with standard ellipse :
In this case major axis is BB= 2b which is along y- axis and minor axis is AA= 2a along x- axis. Focus S(0,be) and S(0,–be) and directrix are y = b/e and y = –b/e.
2.2 General equation of the ellipse
The general equation of an ellipse whose focus is (h,k) and the directrix is the line ax + by + c = 0 and the eccentricity will be e. Then let P(x1,y1) be any point on the ellipse which moves such that SP = ePM
Let the equation of the ellipse x
y2
a > b
(x –h)2 + (y –k)2 =
e 2 (ax1 by1 c) 2
a 2 b2
1 1 a 2 b2
(i) Vertices of an ellipse : The point of which ellipse cut the axis x-axis at (a,0) & (– a, 0) and y- axis at (0, b) & (0, – b) is called the vertices of an ellipse.
Hence the locus of (x1,y1) will be given by (a2 + b
ATOMIC STRUCTURE
1. ATOM & MOLECULES
(a) The smallest particle of a matter that takes part in a chemical reaction is called an atom. The atom of all gases except those of noble gases, cannot exist in free state. These exist in molecular form. The molecules of hydrogen, nitrogen, oxygen and halogens are diatomic (H2, N2). Phosphorus molecule is tetratomic and that of
sulphur is octa atomic.
(b) The smallest particle of a matter that can exist in free state in nature, is known as a molecule.
(c) Some molecules are composed of homoatomic atom, e.g., H2, O2, N2, Cl2, O3 etc., while the molecules of compounds are made up of two or more heteroatomic atoms e.g., HCl, NaOH, HNO3, CaCO3, etc.
2. DALTON’S ATOMIC THEORY
The concepts put forward by John Dalton regarding the composition of matter are known as Dalton’s atomic theory. Its important points are as follows.
(a) Every matter is composed of very minute particles, called atoms that take part in chemical reactions.
(b) Atoms cannot be further subdivided.
(c) The atoms of different elements differ from each other in their properties and masses, while the atoms of the same element are identical in all respects.
(d) The atoms of different elements can combine in simple ratio to form compounds. The masses of combining elements represent the masses of combining atoms.
(e) Atom can neither be created nor destroyed.
2.1 Modern Concept :
Many of the concepts of Dalton’s atomic theory cannot be explained. Therefore, foundation of modern atomic theory was laid down by the end of nineteenth century. The modern theory is substantiated by the existence of isotopes, radioactive disintegration, etc. The important points of the modern atomic theory are as follows.
(a) Prof. Henri Bacquerel discovered the phenomenon of radioactivity and found that an atom is divisible.
(b) An atom is mainly composed of three fundamental particles, viz. electron, proton and neutron.
(c) Apart from the aforesaid three fundamental particles, many others have also been identified, viz. positron, meson, neutrino, antiproton, etc.
(d) Soddy discovered the existence of isotopes, which were atom of the same element having different masses. For example, protium, deuterium and tritium are atoms of hydrogen having atomic masses 1, 2 and 3 a.m.u. respectively.
(e) Atoms having same mass may have different atomic numbers. These are known as isobars. For example,
40 Ar and 40 Ca .
18 20
(f) Atoms of elements combines to form molecules.
(g) It is not necessary that the atoms should combine in simple ratio for the formation of compounds. The atoms in non-stoichiometric compounds are not present in simple ratio. For example, in ferrous sulphide crystals, iron and sulphur atoms are present in the ratio of 0.86 : 1.00.
(h) Atoms participate in chemical reactions.
3. CATHODE RAYS (DISCOVERY OF ELECTRON)
Dry gases are normally bad conductors of electricity. But under low pressure, i.e., 0.1 mm of mercury or lower, electric current can pass thro
The process of finding area of some plane region is called Quadrature. In this chapter we shall find the area bounded by some simple plane curves with the help of definite integral. For solving
(i) The area bounded by a cartesian curve y = f(x), x-axis and ordinates x = a and x = b is given by,
the problems on quadrature easily, if possible first draw the rough sketch of the required area.
b
Area = y dx
a
b
= f(x) dx
a
In chapter function, we have seen graphs of some simple elementary curves. Here we introduce some essential steps for curve tracing which will enable us to determine the required area.
(i) Symmetry
The curve f(x, y) = 0 is symmetrical
about x-axis if all terms of y contain even powers.
about y-axis if all terms of x contain even
powers.
about the origin if f (– x, – y) = f (x, y).
Examples
based on
Area Bounded by A Curve
For example, y2 = 4ax is symmetrical about x-axis, and x2 = 4ay is symmetrical about y- axis and the curve y = x3 is symmetrical about the origin.
(ii) Origin
Ex.1 Find the area bounded by the curve y = x3, x-axis and ordinates x = 1 and x = 2.
2
Sol. Required Area = ydx
1
If the equation of the curve contains no constant
2
= x3 dx =
Lx4 O2 15
MN PQ=
Ans.
term then it passes through the origin.
For example x2 + y2 + 2ax = 0 passes through origin.
(iii) Points of intersection with the axes
If we get real values of x on putting y = 0 in the equation of the curve, then real values of x and y = 0 give those points where the curve cuts the x-axis. Similarly by putting x = 0, we can get the points of intersection of the curve and y-axis.
1 4 1 4
Ex.2 Find the area bounded by the curve y = sec2x,
x-axis and the line x = 4
/4
Sol. Required Area = ydx
x0
For example, the curve x2/a2 + y2 /b2 = 1 intersects the axes at points (± a, 0) and (0, ± b) .
(iv) Region
/4
= sec2
0
xdx = tan x /4 = 1 Ans.
Write the given equation as y = f(x) , and find minimum and maximum values of x which determine the region of the curve.
For example for the curve xy2 = a2 (a – x)
Ex.3 Find the area bounded by the curve y = mx, x-axis and ordinates x = 1 and x = 2
2
Sol. Required area = y dx .
1
y = a
a x
x
2
= mxdx =
LMmx2 2
Now y is real, if 0 < x a , so its region lies between the lines x = 0 and x = a.
1 MN2
= m ( 4 – 1) =
2
PQ1
FGH3 JKm Ans.
Ex.4 Find the area bounded by the curve y = x (1– x)2 and x-axis.
Sol. Clearly the given curve meets the x-axis at (0,0) and (1,0) and for x = 0 to 1, y is positive
Ex.7 Find the area bounded between the curve y2 = 2y – x and y-axis.
Sol. The area between the given curve x = 2y – y2 and y-axis will be as shown in diagram.
so required area- Y
= xb1 xg2 dx
0
1
= ex 2x2 x3 jdx
0
O (0,0) (1,0) X
Lx2
2x3
x4 O1
1 2 1 1
= M P=
– + =
MN2 3 4 PQ0
2 3 4 12
Ans.
Required Area =
e2y y2 jdy
(i
/2
Ex.5 Evaluate : sin2
x dx .
d
If dx
[f(x)] =
(x) and a and b, are two values
0
/2
Sol. sin2 x dx
independent of variable x, then 0
b /2 FG1 cos 2xIJ
(x) dx = f(x) a = f(b) – f(a)
a
= zH 2 dx
1 LM sin 2x O /2
is called Definite Integral of (x) within limits
= x
2
2 PQ
a and b. Here a is called the lower limit and b is called the upper limit of the integral. The interval [a,b] is known as range of integration. It should be noted that every definite integral has
a unique value.
= 1 LM 0
0
= / 4 Ans.
1
x2
Ex.6 Evaluate : xe dx.
0
1
2 x2
Ex.1 Evaluate : x4 dx.
1
Sol. xe dx
0
2 Lx5 O2
1 ex2 1
Sol.
zx4 dx = MP= 32 – 1 = 31
=
Ans. 2 0
MN5 PQ1 5 5 5
/4
= 1 (e –1) Ans.
2
Ex.2 Evaluate : sec2 x dx.
0
1 x3
/4
Sol. sec2 x .dx = tan x /4 = tan / 4 – tan 0 = 1
0
Ex.7 Find the value of
0
1 x8
dx.
0
Ans.
2 1
Sol. Let x4 = t, then 4x3 dx = dt
Ex.3 Evaluate :
1
4 x2
dx.
I =
1 1 dt
4 =
1 [ sin–1 t] 1 =
4 8
2 1 L 2 0
Sol. z dx = Msin GJP
Ans.
1 4 x2
N H2KQ1
= sin–1 (1) – sin–1 (1/2)
z/3 cos x
= – =
Ans.
Ex.8 Evaluate :
0
3 4 sin x dx.
2 6 3
Sol. I =
/3 cos x 3 4 sin x dx.
Ex.4 Evaluate :
z2 1
2 dx
0 4 x2
0
Let 3 + 4 sin x = t 4 cos x. dx = dt
cos x dx = dt/4
Now in the given integral x lies between the
Sol.
4 x2 dx
limit x = 0 to x = / 3 . Now we will decide the limit of t.
1
= tan
2
1 x OP2
0
In 3 + 4 sin x = t, by putting lower limit of x as x = 0; and upper limit as x = / 3 . We
= 1 tan11 0 = / 8 Ans.
2
get lower and upper limit of t respectively.
Putting x = 0 3 + 4 sin 0 = t t = 3
z3 z2
z3 (x) dx
/3 cos x
dx =
zt3 2 3 1 dt
= 2 x2dx +
0
z3b3x 4gdx
0 3 4 sin x
t3 t 4
Fx3 I2 F3x2 I3
= 1 zt3 2 3 1 dt
= GJ+ G
4xJ
4 t3 t
H3 K H2
= 1 log t 32
4
= 8 +
3
27 – 12 – 6 + 8
2
= 1 [ log (3 + 2
4
) – log 3] Ans.
= 37/6 Ans.
Ex.9
sin(tan1 x)
2
dx equals-
Ex.11 Evaluate : |1 x|dx.
0
0 1 x
Sol. Put tan
x = t, then 1 dx = dt
Sol. |1 x| =
RST1 x, when
0 x 1
–1
(1 x2 )
I =
x 1, when 1 x 2
1b1 xgdx + 2 bx 1gdx
/ 2
I = sin t dt [– cos t] /2 = 1 Ans.
0 1
L x2 O1 Lx2 O2
0 Mx
P M xP
0 = MN
2 PQ+
MN2
PQ1
= b1/ 2 0 + b0 1/ 2 = 1 Ans.
z z
i.e. the value of a definite integral remains unchanged if its variable is placed by any other symbol.
[P-4] f(x) dx = f(a x) dx .
0 0
Note :
[P-2]
b
f(x) dx
a
a
= – f(x) dx
b
This property can be used only when lower limit is zero. It is generally used for those complicated
i.e. the interchange of limits of a definite integral
changes only its sign.
zb zc zb
integrals whose denominators are unchanged when x is replaced by a– x. With the hel
Integration is a reverse process of differentiation. The integral or primitive of a function f(x) with
respect to x is that function (x) whose derivative with respect to x is the given function f(x). It is
i. 0. dx = c
ii. 1.dx = x + c
iii. k.dx = kx + c (k R)
xn1
expressed symbolically as -
zf (x) dx (x)
iv. xn dx =
n 1
+ c (n –1)
v. z1 dx = log
x + c
Thus x e
vi. ex dx = ex + c
ax
The process of finding the integral of a function is called Integration and the given function is
vii. ax dx =
loge
a + c = ax loga e + c
called Integrand. Now, it is obvious that the operation of integration is inverse operation of differentiation. Hence integral of a function is also named as anti-derivative of that function.
Further we observe that-
viii. sin x dx = – cos x + c
ix. cos x dx = sin x + c
x. tan x dx = log sec x + c = – log cos x + c
d (x2 )
dx
2 x
xi. cot x dx = log sin x + c
d (x2 2) 2xV| 2xdx x2 constant
xii. sec x dx = log(secx + tanx) + c
dx = – log (sec x –tan x) + c
d 2
dx (x k) 2x|
= log tan
FGH xIJ+ c
So we always add a constant to the integral of function, which is called the constant of
xiii. cosec x dx = – log (cosec x + cot x) + c
Integration. It is generally denoted by c. Due to presence of this constant such an integral is called an Indefinite integral.
= log (cosec x – cot x) + c = log tan
xiv. sec x tan x dx = sec x + c
FGHxIJK+ c
If f(x), g(x) are two functions of a variable x and k is a constant, then-
(i) k f(x) dx = k f(x) dx.
(ii) [f(x) g(x)] dx = f(x)dx ± g(x) dx
(iii) d/dx ( f(x) dx) = f(x)
(iv) f(x)KJdx = f(x)
The following integrals are directly obtained from the derivatives of standard functions.
xv. cosec x cot x dx = – cosec x + c
xvi. sec2 x dx = tan x + c
xvii. cosec2 x dx = – cot x + c xviii. sinh x dx = cosh x + c
xix. cosh x dx = sinh x + c
xx. sech2 x dx = tanh x + c
xxi. cosech2 x dx = – coth x + c
xxii. sech x tanh x dx = – sech x + c
xxiii. cosech x coth x = – cosech x + c
1 1
FxI
eax
R 1FbI
xxiv. xxiv.
x2 + a2 dx =
a tan–1
GHa + c
= a2 b2
sin
STbx tan
GHaJK+ c
xxv. z 1
1
dx = log
FGx a + c
xxxv. zeax cos bx dx
x2 a2
2 a Hx aK
eax
xxvi. z 1
dx = 1 log FGa xIJ + c
= a2 b2
(a cos bx + b sin bx) + c
a2 x2
1
2 a Ha xK
FxI
= cos
STbx tan
1 b V+ c
xxvii. za2 x2 dx = sin–1
GHaJK+ c
FxI
Examples Integration of Function
xxviii. xxviii.
= – cos–1
1
dx = sinh–1
x2 a2
GHaJK+ c
FGxIJ+ c
Ex.1 Evaluate : zx–55 dx
Sol. x–55 dx
x54
= log (x +
) + c
= 54
+ c Ans.
xxix. z 1
dx = cosh–1
FGxIJ+ c
Ex.2 Evaluate :
zex2 1j2
x2 a2
= log (x +
HaK
) + c
Sol.
x
x4 2 x2 1
dx
x
xxx. xxx.
2 2 dx
= zx3 2x 1IJdx
za x
H xK
x4
= x +
2
a . sin–1
2
x + c
a
INDEFINITE INTEGRATION
Preface 1
1. Integration of a Function 2
2. Basic Theorems on Integration 2
3. Standard Integrals 2
4. Methods of Integration 4
4.1 Integration by Substitution
4.2 Integration by Parts
4.3 Integration of Rational Function
4.4 Integration of Irrational Function
4.5 Integration of Trigonometric Function
5. Some Integrates of Different Expression of ex 12
6. Solved Examples 14
DEFINITE INTEGRATION
Preface 24
1. Defination 25
2. Properties Of Definite Integral 26
3. Some Important Formulae 29
4. Summation of Series by Integration 30
5. Solved Examples 32
QUADRATURE
Preface 41
1. Introduction 42
2. Curve Tracing 42
3. Area Bounded by A Curve 42
4. Symmetrical Area 43
5. Positive and Negative Area 44
6. Area Between Two Curves 45
7. Solved Examples 47
DIFFERENTIAL EQUATION
Preface 52
1. Introduction 53
2. Differential Equation 53
2.1 Order of Differential Equation
2.2 Degree of Differential Equation
3. Linear And Non- Linear Differential Equation 53
4. Formation of Differential Equation 53
5. Solution of Differential Equation 54
6. Methods of Solving A first Order & A first Degree Differential EqN 55
6.1 Differential Equation of the form dy/dx = f(x)
6.2 Differential Equation of the form dy/dx = f(x) g(y)
6.3 Differential Equation of the form dy/dx = f(ax+by+ c)
6.4 Differential Equation of homogeneous type
6.5 Differential Equation reducible to homogeneous form
6.6 Linear Differential Equation
6.7 Equation reducible to linear form
6.8 Differential Equation of the form of d2y / dx2 = f(x)
7. Solved Examples 62
All Rights Reserved with CAREER POINT Revised Edition
LEVEL # 1
Order and degree of differential equation
Q.7 The degree of the differential equation
d2y
Q.1 A differential equation of first order and first degree is-
dy 2
dx2 +
(A) 1
= 0 is-
(B) 2 (C) 3 (D) 6
(A) x
d2y
– x + a = 0
Q.8 The order of the differential equation whose solution is y = a cos x + b sin x + c e–x is-
(A) 3 (B) 2
(B) (B)
dx2 + xy = 0
(C) 1 (D) None of these
(C) dy + dx = 0
(D) None of these
Q.2 The order and degree of differential equation
Q.9 The differential equation of all circles of radius a is of order-
(A) 2 (B) 3
(C) 4 (D) None of these
dx + y
dy = 0 are respectively-
Q.10 The order of the differential equation of all
(A) 1,2 (B) 1,1
(C) 2,1 (D) 2,2
Q.3 The order and degree of the differential
circles of radius r, having centre on y-axis and passing through the origin is-
(A) 1 (B) 2 (C) 3 (D) 4
Q.11 The degree of the differential equation
dy d2 y
dy 2
d2 y
equation y = x dx +
is -
+ 3 = x2 log
2 is-
dx2
dx
dx
(A) 1,2 (B) 2,1
(C) 1,1 (D) 2, 2
Q.4 The order and degree of the differential
(A) 1 (B) 2
(C) 3 (D) None of these
Q.12 The differential equation
dy 2 2 / 3 2
equation
4
d y
= are-
d2 y 2
dy 4
dx
dx2
x +
+ y = x2 is of -
dx2
dx
(A) 2, 2 (B) 3, 3
(C) 2, 3 (D) 3, 2
Q.5 The order and the degree of differential
(A) Degree 2 and order 2
(B) Degree 1 and order 1
(C) Degree 4 and order 3
d4 y d3 y
d2y dy
(D) Degree 4 and order 4
equation
dx 4 – 4 dx3 + 8
dx2 – 8 dx
+ 4y = 0 are respectively-
(A) 4,1 (B) 1,4
(C) 1,1 (D) None of these
Q.13
Linear and non linear differential equation
Which of the following equation is linear-
Q.6 The order and degree of differential equation (xy2 + x) dx + (y – x2 y) dy = 0 are-
(A) 1, 2 (B) 2,1
(C) 1,1 (D) 2, 2
(A)
(C)
dy + xy2 = 1 (B) x2
dx
dy + 3y = xy2 (D) x
dx
dy + y = ex
dx
dy + y2 = sinx
dx
Q.14 Which of the following equation is non- linear-
dy
(A) dx = cos x
Q.18 The differential equation of the family of curves y2 = 4a (x + a) , where a is an arbitrary constant, is-
dy 2 dy
d2y
(A) y
1
dx
= 2x dx
(B)
dx2 + y = 0
(C) dx + dy = 0
dy 2 dy
dy 3
(B) y
1
dx
= 2x dx
(D) x
dx +
dy = y2
dx
Q.15 Which of the following equation is linear-
(C)
d2y dx2
dy
+ 2 dx = 0
d2 y 2
dy 2
dy 3 dy
(A) 2 + x2
= 0
(D) + 3
+ y = 0
dx
dx
dx dx
dy
(B) y = dx +
Q.19 The differential equation of all the lines in the xy- plane is-
dy y
dy d2y dy
(C) dx + x = log x
dy
(A) dx – x = 0 (B) dx2 – x dx = 0
d2y d2y
(D) y dx – 4 = x
(C) = 0 (D) + x = 0
dx2 dx2
Formation of differential equation
Q.20 The differential equation of the family
LEVEL # 1
Area bounded by a curve
Q.1 The area between the curves y = 6 – x – x2
and x-axis is -
(A) 125/6 (B) 125/2
(C) 25/6 (D) 25/2
Q.2 The area between the curve y =ex and x-axis which lies between x = – 1 and x = 1 is-
(A) e2 – 1 (B) (e2 –1)/e
(C) (1–e)/e (D) ( e– 1)/e2
Q.3 The area bounded by the curve y = sin 2x,
Q.9 The area bounded by the curve y = 1 + 8/x2, x-axis, x = 2 and x = 4 is-
(A) 2 (B) 3 (C) 4 (D) 5
Q.10 The area between the curve y = log x and x-axis which lies between x = a and x = b (a > 1, b > 1) is-
(A) b log (b/e) – a log (a/e)
(B) b log (b/e) + a log (a/e)
(C) log ab
(D) log (b/a)
Q.11 Area bounded by the curve y = xex2 , x- axis and the ordinates x = 0, x = is-
x- axis and the ordinate x = /4 is- (A) /4 (B) /2 (C) 1 (D) 1/2
(A)
e2 1
2
sq. units (B)
e2 1
2
sq.units
Q.4 The area between the curve xy = a2, x-axis, x = a and x = 2a is-
(A) a log 2 (B) a2 log 2
(C)
e 2
1 sq. units (D)
e 2
1sq.units
(C) 2a log 2 (D) None of these
Q.5 Area under the curve y = sin 2x + cos 2x
between x = 0 and x = 4 , is-
(A) 2 sq. units (B) 1 sq. units
(C) 3 sq. units (D) 4 sq. units
Q.6 The area bounded by the curve y = 4x2 ; x = 0, y = 1 and y = 4 in the first quadrant is-
Q.12 The area bounded between the curve y = 2x2 + 5, x-axis and ordinates x = – 2 and x = 1 is-
(A) 21 (B) 29/5 (C) 23 (D) 24
Q.13 Area bounded by curve xy = c, x-axis between x = 1 and x = 4, is-
(A) c log 3 sq. units
(B) 2 log c sq. units
(C) 2c log 2 sq. units
(D) 2c log 5 sq. units
2 1
(A) 2 3 (B) 3 3
Q.14 The area bounded by the curve y = x sin x2,
(C) 2 1
3
(D) 3 1
2
x-axis and x = 0 and x =
is-
Q.7 The area between the curve y = sec x and y-axis when 1 y 2 is-
(A) 1/2 (B) 1/
(C) 1/4 (D) /2
(A)
2 – log ( 2 + )
3
Q.15 The area bounded between the curve
x – y + 1 = 0, x = – 2, x = 3 and x-axis is-
2 4 2
(B) 3 + log ( 2 + )
(C) – 1 log (2 + )
(A) 45/4 (B) 45/2
(C) 15 (D) 25/2
Q.16 The area bounded by curves y = tan x,
3 2
(D) None of these
Q.8 The area bounded by the lines y = x, y = 0
x- axis and x =
3 is-
and x = 2 is-
(A) 2 log 2 (B) log 2
(A) 1 (B) 2
(C) log
FG2
J (D) 0
(C) 4 (D) None of these
H3 K
Q.17 The area between the curve x2 = 4ay, x-axis, and ordinate x = d is-
(A) d3 /12a (B) d3/a
(C) d3/2a (D) d3/6a
Q.18 Area bounded by the curve y = x (x – 1)2 and x-axis is-
(A) 4 (B) 1/3 (C) 1/12 (D) 1/2
Q.19 The area bounded by the curve y = loge x, x-axis and ordinate x = e is-
(A) loge2 (B) 1/2 unit
(C) 1 unit (D) e unit
1
Q.20 The area bounded by the curve y = cos 2 x ,
Q.28 The area of a loop bounded by the curve y = a sin x and x-axis is-
(A) a (B) 2a2 (C) 0 (D) 2a
Q.29 The area between the curves x = 2 – y – y2 and y-axis is-
(A) 9 (B) 9/2 (C) 9/4 (D) 3
Q.30 The area bounded by y = 4x – x2 and the x-axis is-
(A) 30/7 (B) 31/7 (C
LEVEL # 1
Q.1
Integration of function
1 sin2x dx equals-
dx
Q.7 4x2 9
1
dx is equal to -
3x 1
2x
(A) sin x + cos x + c
(B) sin x – cos x + c
(A) tan–1 + c (B) tan–1 + c
(C) cos x – sin x + c
(D) None of these
(C)
2x
tan–1 3 + c (D)
2x
tan–1 3 + c
2 2
4 5 sin x
Q.2 cos2 x
dx equals-
Q.8 cos2x sin4x dx is equal to -
(A) 4 tan x – sec x + c
(B) 4 tan x + 5 sec x + c
(A)
1 (cos 6x + 3 cos 2x ) + c
12
1
(C) 9 tan x + c
(D) None of these
(B) 6 (cos 6x + 3 cos 2x) + c
1
(C) – 12 (cos 6x + 3 cos 2x) + c
Q.3 (tan x + cot x) dx equals-
(A) log (c tan x)
(B) log (sin x + cosx) + c
(C) log (cx)
(D) None of these
(D) None of these
2dx
Q.9 x2 – 1 equals -
1 x 1
1 x – 1
(A) 2 log x – 1 + c (B)
log x 1 + c
Q.4
e5 loge x e4 loge x
e3 loge x e2 loge x dx equals-
x 1
2
x – 1
(C) log
x – 1
+ c (D) log
x 1
(A)
x + c (B)
2
4
x + c
3
Q.10
(ax bx )2 axbx
dx equals-
(C) x
4
+ c (D) None of these
(A) (a/b)x + 2x + c (B) (b/a)x + 2x + c
(C) (a/b)x – 2x + c (D) None of these
Q.5 1 cos 2x 1 cos 2x
dx equals-
Q.11
dx
sin2 xcos2 x
equals-
(A) tan x + x + c (B) tan x – x + c
(C) sin x – x + c (D) sin x + x + c
(A) tan x – cot x + c (B) tan x + cot x + c
(C) cot x – tan x + c (D) None of these
Q.6 The value of (1 x)
(A) x – x2 – x3 + c
(B) x + x2 – x3 + c
(C) x + x2 + x3 + c
(D) None of these
(1 + 3x) dx equal to -
sin x
Q.12 dx equals-
1 cos x
(A) 2 cos (x/2) + c
(B) 2 sin(x/2) + c (C)2 2 cos (x/2) + c
(D) –2 cos(x/2)+c
Q.13 sec x (tan x + sec x) dx equals-
Q.20
2x 3 x
5x
dx equals-
(A) tan x – sec x + c
(B) sec x – tan x+ c
(C) tan x + sec x + c
(A)
b2 / 5gx log 2 / 5 +
b3 / 5gx
log 3 / 5 + c
(D) None of these
Q.14 The value of sin x cos x
1 sin 2x
dx is-
(B) loge (2x/5) + loge (3x/5) + c
(C) x + c
(D) None of these
sin2 x
(A) sin x + c (B) x + c
Q.21 The value of
dx is-
1 cos x
(C) cos x + c (D) 1 (sin x + cos x)
2
1
(A) x – sin x + c
(B) x + sin x + c
(C) – x – sin x + c
Q.15 The value of
1 cos x dx is-
(D) None of these
(A) 1 2
cot (x/2) + c
1
(B) – 2 cot (x/2) + c
Q.22 e2x+3 dx equals-
(C) – cot (x/2) + c (D) – tan (x/2) + c
(A) 1 e2x+3 + c (B) 1 e2x+5 + c
cos 2x 2 sin2 x 2 2
Q.16
cos2 x
dx equals-
(C) 1 e2x+3 + c (D) 1 e2x+4 – c
(A) cot x + c (B) sec x + c
(C) tan x + c (D) cosec x + c
Q.23
3 2
1 tan x
1 tan x dx equals-
Q.17 cosx FG1
sin x IJ
dx equals-
(A) log (cos x + sin x) + c
Hsin2 x
cos3 xK
(B) log (cos x – sin x) + c
(A) sec x – cosec x + c
(B) cosec x – sec x + c
(C) sec x + cosec x + c
(D) None of these
(C) log (sin x – cos x) + c
(D) None of these
sin4 x cos4 x
Q.24
sin2 xcos2 x
dx equals-
Q.18
sin3 x cos3 x
In this chapter, we shall study the nature of a
function which is governed by the sign of its derivative. If the graph of a function is in upward going direction or in downward coming direction then it is called as monotonic function, and this property of the function is called Monotonicity. If a function is defined in any interval, and if in some part of the interval, graph moves upwards
A function is said to be monotonic function in a domain if it is either monotonic increasing or monotonic decreasing in that domain.
NOTE : If x < x f(x ) < f(x ) x , x D, then
and in the remaining part moves downward then
1 2 1
2 1 2
function is not monotonic in that interval.
f(x) is called strictly increasing in domain D.
These are of two types –
2.1 Monotonic Increasing :
A function f(x) defined in a domain D is said to be monotonic increasing function if the value of f(x) does not decrease (increase) by increasing (decreasing) the value of x or
We can say that the value of f(x) should increase (decrease) or remain equal by increasing
(Decreasing) the value of x.
Similarly if x1 < x2 f(x1) > f(x2), x1, x2 D then it is called strictly decreasing in domain D.
R
If Tor x1 x2 f(x1) f(x2 )
, x1 , x2 D
or SR x1 x2 f(x1) f(x2 ) , x1 , x2 D
2.2 Monotonic Decreasing :
A function f(x) defined in a domain D is said to be monotonic decreasing function if the value of f(x) does not increase (decrease) by increasing (decreasing) the value of x or
We can say that the value of f(x) should decrease (increase) or remain equal by increasing
(Decreasing) the value of x.
For Example
(i) f(x) = ex is a monotonic increasing function where as g(x) = 1/x is monotonic decreasing function.
(ii) f(x) = x2 and g(x) = | x | are monotonic increasing for x > 0 and monotonic decreasing for x < 0. In general they are not monotonic functions.
(iii) Sin x, cos x are not monotonic function whereas tan x, cot x are monotonic.
(i) At a Point : A function f(x) is said to be monotonic increasing (decreasing) at a point x = a of its domain if it is monotonic increasing
R
If Tor x1 x2 f(x1) f(x2 )
, x1 , x2 D
(decreasing) in the interval (a – h, a + h) where h is a small positive number. Hence we may
or SR x1 x2 f(x1) f(x2 ) , x1 , x2 D
observe that if f(x) is monotonic increasing at x = a, then at this point tangent to its graph will make an acute angle with the x–axis where as if the function is monotonic decreasing these tangent will make
an obtuse angle with x–axis. Consequently f' (a) will be positive or negative according as f(x) is monotonic increasing or decreasing at x = a.
Ex. Function f(x) = x2 + 1 is monotonically decreasing in [ –1, 0] because
f' (x) = 2x < 0, x (–1, 0)
Ex. Function f(x) = x2 is not a monotonic function in the interval [–1, 1] because
f' (x) > 0, when x = 1/2
Ex. Function f(x) = sin2x + cos2x is constant
In this chapter we shall study those points of the domain of a function where its graph changes its direction from upwards to downwards or from downwards to upwards. At such points the derivative of the function, if it exists, is necessarily zero.
The value of a function f (x) is said to be maximum at x = a, if there exists a very small positive number h, such that
f(x) < f(a) x (a – h,a + h) , x a
In this case the point x = a is called a point of maxima for the function f(x).
Simlarly, the value of f(x) is said to the minimum at x = b, If there exists a very small positive number, h, such that
f(x) > f(b), x (b – h,b + h), x b
In this case x = b is called the point of minima for the function f(x).
Hene we find that,
(i) x = a is a maximum point of f(x)
f(a) f(a h) 0
f(a) f(a h) 0
(ii) x = b is a minimum point of f(x)
Rf(b) f(b h) 0
f(b) f(b h) 0
(iii) x = c is neither a maximum point nor a minimum point
Note :
(i) The maximum and minimum points are also known as extreme points.
(ii) A function may have more than one maximum and minimum points.
(iii) A maximum value of a function f(x) in an interval [a,b] is not necessarily its greatest value in that interval. Similarly, a minimum value may not be the least value of the function. A minimum value may be greater than some maximum value for a function.
(iv) If a continuous function has only one maximum (minimum) point, then at this point function has its greatest (least) value.
(v) Monotonic functions do not have extreme points.
Ex. Function y = sin x, x (0, ) has a maximum point at x = /2 because the value of sin /2 is greatest in the given interval for sin x.
Clearly function y = sin x is increasing in the interval (0, /2) and decreasing in the interval ( /2, ) for that reason also it has maxima at x = /2. Similarly we can see from the graph of cos x which has a minimum point at x = .
Ex. f(x) = x2 , x (–1,1) has a minimum point at x = 0 because at x = 0, the value of x2 is 0, which is
less than the all the values of function at different points of the interval.
Clearly function y = x2 is decreasing in the interval
Rf(c) f(c h)
Sand
Tf(c) f(c h) 0
|V| have opposite signs.
(–1, 0) and increasing in the interval (0,1) So it has minima at x = 0.
Ex. f(x) = |x| has a minimum point at x = 0. It can be easily observed from its graph.
A. Necessary Condition : A point x = a is an extreme point of a function f(x) if f’(a) = 0, provided f’(a) exists. Thus if f’ (a) exists, then
x = a is an extreme point f’(a) = 0
or
f’ (a) 0 x = a is not an extreme point.
But its converse is not true i.e.
f’ (a) = 0 x = a is an extreme point.
For example if f(x) = x3 , then f’ (0) = 0 but x = 0 is not an extreme point.
B. Sufficient Condition :
(i) The value of the function f(x) at x = a is maximum, if f’ (a) = 0 and f” (a) < 0.
(ii) The value of
CONTINUITY AND DIFFERENTIABILITY
Total No. of questions in Continuity and Differentiability are-
In Chapter Examples 14
Solved Examples 17
Total No. of questions 31
Ex.
The word 'Continuous' means without any break or gap. If the graph of a function has no break or gap or jump, then it is said to be continuous.
A function which is not continuous is called a
discontinuous function. In other words,
If there is slight (finite) change in the value of a function by slightly changing the value of x then function is continuous, otherwise discontinuous, while studying graphs of functions, we see that graphs of functions sin x, x, cos x, ex etc. are continuous but greatest integer function [x] has break at every integral point, so it is not continu- ous. Similarly tan x, cot x, secx, 1/x etc. are also discontinuous function.
(Discontinuous)
For examining continuity of a function at a point, we find its limit and value at that point, If these two exist and are equal, then function is continu- ous at that point.
A function f(x) is said to be continuous at a point x = a if
(i) f (a) exists
(ii) x Lim
f(x) exists and finite
Lim Lim
so x a f(x) = x a f(x)
(iii) x Lim
f(x) = f(a) .
or function f(x) is continuous at x = a.
Lim Lim
(Continuous function)
If x a f(x) = x a f(x) = f(a).
i.e. If right hand limit at 'a' = left hand limit at 'a'= value of the function at 'a'.
If x Lim
f(x) does not exist or
Lim
x a
f(x) f(a),
then f(x) is said to be discontinuous at x= a.
Ex.1 Examine the continuity of the function
X R|x2 9
f (x) =
S| x 3 , when x 3
at x = 3.
T6, when x 3
Sol. f (3) = 6 ( given)
bx 3gbx 3
Lim
x 3
f(x) =
Lim
x 3
bx 3g = 6
(Discontinuous at x = 0)
Lim
x 3
f(x) = f(3)
f (x) is continuous at x = 3.
Ex.2 If f(x) =
log (1 2ax) log 1 bx
x 0
x
Thus a function f(x) is continuous at a point x = a if it is left continuous as well as right continuous at x = a.
Tk ,
x 0
If function is continuous at x = 0 then the value of k is –
(A) a + b (B) 2a +b
Ex.4 Examine the continuity of the function
(C) a – b (D) 0
x2 1, when x 2
f(x) =
|T2x, when x 2
Sol.
Lim
x 0
logFGH1 2axIJ
x
at the point x = 2.
Sol. f(2) = 22 + 1 = 5
= x Lim
FG1 bx J.
b1 bxgb2ag b1 2axgbb
2
f(2– 0) =
Lim
h 0
Lim
b2 hg2 1 = 5
0 H1 2ax
2a b
0 b1 bxgb1 2 axg
b1 bxg
b2a b
b1gb1g
f (2+ 0) = h 0 2( 2+ h) = 4
f(2– 0) f(2+ 0) f(2)
x Lim
= = 2a + b
Ans.[B]
f(x) is not continuous at x = 2.
Ex.5 Check the continuity of the function
1 cos 4x, x 0
R|x 2
x 3
Ex.3 If f(x) = S| x2
is continuous then
f(x) = 5 x 3
at x = 3.
Ta,
x 0
T8 x x 3
the value of a is equal to –
(A) 0 (B) 1
(C) 4 (D) 8
Sol. Since the given function is continuous at x= 0
Sol.
02Application of Derivative # 1 (Tangent & Normal)~1 Module-4.pdfRajuSingh806014
If y = f(x) be a given function, then the differential coefficient f' (x) or dy at the point P (x , y ) is
(i) If the tangent at P (x1,y1) of the curve y = f(x) is parallel to the x- axis (or perpendicular to y- axis) then = 0 i.e. its slope will be
zero.
FGdy J
dx 1 1
m = H K = 0
the trigonometrical tangent of the angle (say) which the positive direction of the tangent to the curve at P makes with the positive direction of x- axis Gdy J, therefore represents the slope of the
tangent. Thus
dx (x1,y1)
The converse is also true. Hence the tangent at (x1,y1) is parallel to x- axis.
GJ = 0
(x1,y1)
(ii) If the tangent at P (x , y ) of the curve y =
1 1
f (x) is parallel to y - axis (or perpendicular
to x-axis) then = / 2 , and its slope will be infinity i.e.
dy
m = =
dx (x ,y )
The converse is also true. Hence the tangent at (x1, y1) is parallel to y- axis
Fdy
(x1,y1)
Thus
(i) The inclination of tangent with x- axis.
dy
(iii) If at any point P (x1, y1) of the curve y = f(x), the tangent makes equal angles with the
axes, then at the point P, = / 4 or 3 / 4 ,
= tan–1
GHdxJK
Hence at P, tan = dy/dx = 1. The
(ii) Slope of tangent = dy
dx
(iii) Slope of the normal = – dx/dy
Ex.1 Find the following for the curve y2 = 4x at point (2,–2)
(i) Inclination of the tangent
(ii) Slope of the tangent
(iii) Slope of the normal
Sol. Differentiating the given equation of curve, we get dy/dx = 2/y = –1 at (2,–2)
so at the given point.
(i) Inclination of the tangent = tan–1(–1) = 135º
(ii) Slope of the tangent = –1
(iii) Slope of the normal = 1
converse of the result is also true. thus at
(x1,y1) the tangent line makes equal angles with the axes.
GJ = 1
(x1,y1)
Ex.2 The equation of tangent to the curve y2 = 6x at (2, – 3).
(A) x + y – 1 = 0 (B) x + y + 1 = 0 (C) x – y + 1 = 0 (D) x + y + 2 = 0
Sol. Differentiating equation of the curve with respect to x
(a) Equation of tangent to the curve y = f(x) at A (x1,y1) is
2y dy = 6
dx
FGdy J
dx (2,3)
= 3 = –1
3
y – y1 =
FGdy J
(x1,y1)
(x–x1)
Therefore equation of tangent is y + 3 = – (x – 2)
x + y + 1 = 0 Ans. [B]
Ex.3 The equation of tangent at any of the curve x = at2, y = 2at is -
(A) x = ty + at2 (B) ty + x + at2 = 0
(C) ty = x + at2 (D) ty = x + at3
2 a 1
Sol. dy/dx = (dy/dt)/(dx/dt) = 2 at = t
equation of the tangent at (x,y) point is
(y – 2 at) = 1 (x – at2)
t
ty = x + at2 Ans.[C]
Ex.4 The equation of the tangent to the curve x2 (x – y) + a2 (x + y) = 0 at origin is-
(A) x + y + 1 = 0 (B) x + y + 2 = 0 (C) x + y = 0 (D) 2x – y = 0
Sol. Differentiating equation of the curve w.r.t. x
dy/dx = – y x
(i) If tangent line is parallel to x - axis, then dy/dx = 0 y = 0 and x = a
Thus the point is (a,0)
(ii) If tangent is parallel to y – axis , then dy/dx = x = 0 and y = a
Thus the point is (0,a)
(iii) If tangent line makes equal angles with both axis , then dy/dx = 1
y =
(a) Natural Numbers : N = {1,2,3,4,...}
(b) Whole Numbers : W = {0,1,2,3,4, }
(c) Integer Numbers :
or Z = {...–3,–2,–1, 0,1,2,3, },
Z+ = {1,2,3,....}, Z– = {–1,–2,–3, }
Z0 = {± 1, ± 2, ± 3, }
(d) Rational Numbers :
p
Q = { q ; p, q z, q 0 }
(i) R0 : all real numbers except 0 (Zero).
(j) Imaginary Numbers : C = {i,, }
(k) Prime Numbers :
These are the natural numbers greater than 1 which is divisible by 1 and itself only, called prime numbers.
Ex. 2,3,5,7,11,13,17,19,23,29,31,37,41,...
(l) Even Numbers : E = {0,2,4,6, }
(m) Odd Numbers : O = {1,3,5,7, }
Ex. {1,
Note :
5
, –10, 105,
3
22 20
7 , 3
, 0 ....}
The set of the numbers between any two real numbers is called interval.
(a) Close Interval :
(i) In rational numbers the digits are repeated after decimal.
(ii) 0 (zero) is a rational number.
(e) Irrational numbers: The numbers which are not rational or which can not be written in the form of p/q ,called irrational numbers
Ex. { , ,21/3, 51/4, ,e, }
Note:
(i) In irrational numbers, digits are not repeated after decimal.
(ii) and e are called special irrational quantities.
(iii) is neither a rational number nor a irrational number.
(f) Real Numbers : {x, where x is rational and irrational number}
20
[a,b] = { x, a x b }
(b) Open Interval:
(a, b) or ]a, b[ = { x, a < x < b }
(c) Semi open or semi close interval:
[a,b[ or [a,b) = {x; a x < b}
]a,b] or (a,b] = {x ; a < x b}
Let A and B be two given sets and if each element a A is associated with a unique element b B under a rule f , then this relation is called function.
Here b, is called the image of a and a is called the pre- image of b under f.
Note :
(i) Every element of A should be associated with
Ex. R = { 1,1000, 20/6, ,
, –10, –
,.....}
3
B but vice-versa is not essential.
(g) Positive Real Numbers: R+ = (0,)
(h) Negative Real Numbers : R– = (– ,0)
(ii) Every element of A should be associated with a unique (one and only one) element of but
any element of B can have two or more rela- tions in A.
3.1 Representation of Function :
It can be done by three methods :
(a) By Mapping
(b) By Algebraic Method
(c) In the form of Ordered pairs
(A) Mapping :
It shows the graphical aspect of the relation of the elements of A with the elements of B .
Ex. f1:
f2 :
f3 :
f4 :
In the above given mappings rule f1 and f2
shows a function because each element of A is
associated with a unique element of B. Whereas
f3 and f4 are not function because in f 3, element c is associated with two elements of B, and in f4 , b is not associated with any element
of B, which do not follow the definition of function. In f2, c and d are associated with same element, still it obeys the rule of definition of function because it does not tell that every element of A should be associated with different elements of B.
(B) Algebraic Method :
It shows the relation between the elem
01. Differentiation-Theory & solved example Module-3.pdfRajuSingh806014
Total No. of questions in Differentiation are-
In Chapter Examples 31
Solved Examples 32
The rate of change of one quantity with respect to some another quantity has a great importance. For example the rate of change of displacement of a particle with respect to time is called its velocity and the rate of change of velocity is
called its acceleration.
The following results can easily be established using the above definition of the derivative–
d
(i) dx (constant) = 0
The rate of change of a quantity 'y' with respect to another quantity 'x' is called the derivative or differential coefficient of y with respect to x.
Let y = f(x) be a continuous function of a variable quantity x, where x is independent and y is
(ii)
(iii)
(iv)
(v)
d
dx (ax) = a
d (xn) = nxn–1
dx
d ex =ex
dx
d (ax) = ax log a
dependent variable quantity. Let x be an arbitrary small change in the value of x and y be the
dx
d
(vi) dx
e
(logex) = 1/x
corresponding change in y then lim
y
if it exists, d 1
x0 x
is called the derivative or differential coefficient of y with respect to x and it is denoted by
(vii) dx
(logax) =
x log a
dy . y', y
dx 1
or Dy.
d
(viii) dx (sin x) = cos x
So, dy dx
dy
dx
lim
x0
lim
x0
y
x
f (x x) f (x)
x
(ix) (ix)
(x) (x)
d
dx (cos x) = – sin x
d (tan x) = sec2x
dx
The process of finding derivative of a function is called differentiation.
If we again differentiate (dy/dx) with respect to x
(xi)
d (cot x) = – cosec2x
dx
d
then the new derivative so obtained is called second derivative of y with respect to x and it is
Fd2 y
(xii) dx
d
(xiii) dx
(secx)= secx tan x
(cosec x) = – cosec x cot x
denoted by
HGdx2 Jor y" or y2 or D2y. Similarly,
d 1
we can find successive derivatives of y which
(xiv) dx
(sin–1 x) = , –1< x < 1
1 x2
may be denoted by
d –1 1
d3 y d4 y
dn y
(xv) dx (cos x) = –
,–1 < x < 1
dx3 ,
dx4 , ........, dxn , ......
d
(xvi) dx
(tan–1 x) = 1
1 x2
Note : (i)
y is a ratio of two quantities y and
x
(xvii) (xvii)
d (cot–1 x) = – 1
where as dy
dx
dy
is not a ratio, it is a single
dx
d
(xviii) (xviii)
(sec–1 x) =
1 x2
1
|x| > 1
quantity i.e.
dx dy÷ dx
dx x x2 1
(ii)
dy is
dx
d (y) in which d/dx is simply a symbol
dx
(xix)
d (cosec–1 x) = – 1
dx
of operation and not 'd' divided by dx.
d
(xx) dx
(sinh x) = cosh x
d
(xxi) dx
d
(cosh x) = sinh x
Theorem V Derivative of the function of the function. If 'y' is a function of 't' and t' is a function of 'x' then
(xxii) dx
d
(tanh x) = sech2 x
dy =
dx
dy . dt
dt dx
(xxiii) dx
d
(xxiv) dx
d
(coth x) = – cosec h2 x (sech x) = – sech x tanh x
Theorem VI Derivative of parametric equations If x = (t) , y = (t) then
dy dy / dt
=
(xxv) dx
(cosech x) = – cosec hx coth x
dx dx / dt
(xxvi) (xxvi)
(xxvii) (xxvii)
d (sin h–1 x) =
Level # 1 ........................................ 48
Level # 2 ........................................ 16
Level # 3 ........................................ 12
Level # 4 ........................................ 16
LEVEL # 1
Q.1 When x < 0, function f(x) = x2 is
(A) Decreasing
(B) Increasing
(C) Constant
(D) Not monotonic
Q.2 When x > 1, function f(x) = x3 is
(A) Increasing (B) Decreasing
(C) Constant (D) not monotonic
Q.3 In the interval (0, 1), f(x) = x2 – x + 1 is
(A) Monotonic (B) Not monotonic
(C) Decreasing (D) Increasing
Q.4 f(x) = x + 1/x, x 0 is increasing when
(A) | x | < 1 (B) | x | > 1
(C) | x | < 2 (D) | x | > 2
|x|
Q.11 For which value of x, the function f(x) = x2 –2x is decreasing
(A) x > 1 (B) x > 2
(C) x < 1 (D) x < 2
x 2
Q.12 Function f(x) = x 1 , x –1 is
(A) Increasing
(B) Decreasing
(C) Not monotonic
(D) None of these
Q.13 Function f(x) = x3 is
(A) Increasing in (0, ) and decreasing in (–, 0)
(B) Decreasing in (0, ) and increasing in (–, 0)
(C) Decreasing throughout
(D) Increasing throughout
Q.5 The function f(x) =
(A) Decreasing
(B) Increasing
x (x 0), x > 0 is
Q.14 Function f(x) = x | x | is
(A) Monotonic increasing
(B) Monotonic decreasing
(C) Constant function
(D) None of these
Q.6 When x (0, 1), function f(x) = 1 / is
(A) Increasing
(B) Decreasing
(C) Neither increasing nor decreasing
(D) Constant
Q.7 Function f(x) = 3x4 + 7x2 + 3 is
(A) Monotonically increasing
(B) Monotonically decreasing
(C) Not monotonic
(D) Odd function
Q.8 For what values of x, the function
(C) Not monotonic
(D) None of these
Q.15 If f and g are two decreasing functions such that fog is defined then fog is
(A) Decreasing (B) Increasing
(C) Can't say (D) None of these
Q.16 For the function f(x) = | x |, x > 0 is
(A) Decreasing
(B) Increasing
(C) Constant function
(D) None of these
Q.17 In the following , monotonic increasing
4
f(x) = x + x2
is monotonically decreasing
fucntion is
(A) x + | x | (B) x – | x |
(A) x < 0 (B) x > 2
(C) x < 2 (D) 0 < x < 2
(C) | x | (D) x | x |
Q.9 If f(x) = x 2
for –7 x 7, then f(x) is
x 1
Q.18 At x = 0, f(x) = is
2 x x 2
increasing function of x in the interval (A) [7, 0] (B) (2, 7]
(C) [–2, 2] (D) [0, 7]
(A) Increasing (B) Decreasing
(C) Not monotonic (D) Constant
x
Q.10 The function y = 1 x2
interval
decreases in the
Q.19 If f(x) = 2x3 – 9x2 + 12x – 6, then in which interval f(x) is monotonically increasing
(A) (1, 2) (B) (–, 1)
(A) (–, –) (B) (–1, –1)
(C) (0, ) (D) (–, –1)
(C) (2, ) (D) (–, 1) or (2, )
Q.20 For the function f(x) = x3 – 6x2 – 36x + 7 which of the following statement is false
(A) f(x) is decareasing, if –2 < x < 6
(B) f(x) is increasing, if –3 < x < 5
(C) f(x) is increasing, if x < –2
(D) f(x) is increasing, if x > 6
Q.28 Which of the following function is not monotonic
(A) ex – e–x (B) ex + e–x
(C) e–1/x (D) None of these
Q.29 In the following, decreasing function
Level # 1 ........................................ 92
Level # 2 ........................................ 27
Level # 3 ........................................ 30
Level # 4 ........................................ 26
LEVEL # 1
R|sin1 ax
continuity of a function at a point
Q.7 If f(x) =
S x
Tk,
, x 0
x 0
is continuous at
Q.1 Function f(x) =
R1 x,
T
when x 2
x = 2 is
x = 0, then k is equal to-
(A) 0 (B) 1
5 x, when x 2
continuous at x = 2, if f(2) equals-
(A) 0 (B) 1
(C) 2 (D) 3
Rx cos1/ x, x 0
(C) a (D) None of these
Q.8 What is the value of (cos x)1/x at x = 0 so that it becomes continuous at x = 0-
(A) 0 (B) 1
(C) –1 (D) e
Q.2 If f(x) = k,
x 0
is continuous at
R|k
cos x
x = 0, then
(A) k > 0 (B) k< 0
Q.9 If f(x) =
S| 2x
, x / 2 x / 2
is a continuous
(C) k = 0 (D) k 0
function at x = / 2 , then the value of k is-
Q.3 If function f(x) = S|x2 2, x 1
is continuous
(A) –1 (B) 1
(C) –2 (D) 2
x3 a3
at x = 1, then value of f(x) for x< 1 is-
Q.10 If function f(x) =
x a
, is continuous at
(A) 3 (B) 1–2x
(C) 1–4x (D) None of these
Q.4 Which of the following function is continuous at x= 0-
x= a, then the value of f(a) is -
(A) 2a (B) 2a2
(C) 3a (D) 3a2
Rsin 1/ x, x 0
(A) f(x) =
RSsin 2x / x, x 0
Q.11 If f(x) = k,
x 0
is continuous at
(B) f(x) =
T1,
R|1 x1/ x,
T1,
x 0
x 0
x 0
x = 0, then k is equal to-
(A) 8 (B) 1
(C) –1 (D) None of these
Q.12 Function f(x) =
G1
xIJ1/ x
is continuous at
(C) f(x) =
R| 1/ x
|T1,
x 0
x 0
H aK
x= 0 if f(0) equals-
(D) None of these
R|6 5 x,x 0
(A) ea (B) e–a
(C) 0 (D) e1/a
1 cos 7bx
Q.5 If f(x) =
S|T2a x, x 0
is continuous at
Q.13 If f(x) =
x
, (x )
is continu-
x = 0, then the value of a is -
(A) 1 (B) 2
(C) 3 (D) None of these
ous at x= , then f( ) equals-
(A) 0 (B) 1
(C) –1 (D) 7
x2 ba 2gx a
R tan x
Q.6 If f(x) = S|
x 2
, x 2
is continu-
Q.14 If f(x) =
S|sin
x
x 0
, then f(x) is -
T2,
x 2
(A) Continuous everywhere
ous at x = 2, then a is equal to-
(A) 0 (B) 1
(C) –1 (D) 2
(B) Continuous nowhere
(C) Continuous at x= 0
(D) Continuous only at x = 0
Q.15 If f(x) =
2x tan x x
is continuous at x = 0,
Q.22 Function f(x) = [x] is discontinuous at-
(A) Every real number
then f(0) equals-
(A) 0 (B) 1
(C) 2 (D) 3
1 x 3 1 x
(B) Every natural number
(C) Every integer
(D) No where
Q.23 Function f(x) = 3x2–x is-
(A) Discontinuous at x = 1
(B) Discontinuous at x = 0
Q.16 If f(x) =
x , (x 0) is continuous
(C) Continuous only at x = 0
at x= 0, then the value of f(0) is- (A) 1/6 (B) 1/4
(C) 2 (D) 1/3
(D) Continuous at x = 0
R|x2, when x 0
Q.24 If f(x) = S|1, when 0 x 1 , then f(x) is-
Q.17 If f (x) =
Rax2 b when 0 x 1 S2 when x 1 is x 1 when 1 x 2
1 / x, when x 1
(A
LEVEL # 1
Questions
based on
inequation
Q.8 If x2 – 1 0 and x2 – x – 2 0, then x line in the interval/set
(A) (–1, 2) (B) (–1, 1)
Q.1 The inequality
2 < 3 is true, when x belongs to-
x
(C) (1, 2) (D) {– 1}
2
2
Questions Definition of function
(A) 3 ,
(B) 3
based on
2 ,
Q.9 Which of the following relation is a function ?
(C)
x 4
(–, 0) (D) none of these
(A) {(1,4), (2,6), (1,5), (3,9)}
(B) {(3,3), (2,1), (1,2), (2,3)}
(C) {(1,2), (2,2,), (3,2), (4,2)}
(D) {(3,1), (3,2), (3,3), (3,4)}
Q.2
x 3 < 2 is satisfied when x satisfies-
(A) (–, 3) (10, ) (B) (3, 10)
(C) (–, 3) [10, ) (D) none of these
Q.10 If x, y R, then which of the following rules is not a function-
(A) y = 9 –x2 (B) y = 2x2
x 7
(C) y = – |x| (D) y = x2 + 1
Q.3 Solution of x 3 > 2 is-
Questions Even and odd function
(A) (–3, ) (B) (–, –13)
(C) (–13, –3) (D) none of these
2x 3
based on
Q.11 Which one of the following is not an odd function -
Q.4 Solution of
3x 5
3 is-
(A) sin x (B) tan x
(C) tanh x (D) None of these
12 5 12
(A) 1, 7
(B) , 4 4
3 7
Q.12 The function f(x) = sin x cos x
is -
, 5
12 ,
x tanx
(C)
3
(D) 7
(A) odd
(B) Even
Q.5 Solution of (x – 1)2 (x + 4) < 0 is-
(A) (–, 1) (B) (–, –4)
(C) (–1, 4) (D) (1, 4)
Q.6 Solution of (2x + 1) (x – 3) (x + 7) < 0 is-
(C) neither even nor odd
(D) odd and periodic
Q.13 A function is called even function if its graph is symmetrical w.r.t.-
(A) origin (B) x = 0
(C) y = 0 (D) line y = x
1 ,3
1 ,3
(A) (– , –7)
2
(B) (– , – 7)
Q.14 A function is called odd function if its graph is symmetrical w.r.t.-
(C) (–, 7) 1 ,3
2
(D) (–, –7) (3, )
(A) Origin (B) x = 0
(C) y = 0 (D) line y = x
Q.15 The even function is-
Q.7 If x2 + 6x – 27 > 0 and x2 – 3x – 4 < 0, then-
(A) x > 3 (B) x < 4
(A) f(x) = x2 (x2 +1) (B) f(x) = sin3 x + 2
(C) f(x) = x (x +1) (D) f(x) = tan x + c
(C) 3 < x < 4 (D) x = 7 2
Q.16 A function whose graph is symmetrical about the y-axis is given by-
Q.25 In the following which function is not periodic-
(A) f(x) = loge
(x + )
(A) tan 4x (B) cos 2x
(C) cos x2 (D) cos2x
(B) f(x + y) = f(x) + f(y) for all x, y R
(C) f(x) = cos x + sin x
(D) None of these
Q.17 Which of the following is an even function ?
ax 1
1
Q.26 Domain of the function f(x) = x 2
is-
(A) x
ax 1
(B) tan x
(A) R (B) (–2, )
(C) [2, ] (D) [0, ]
(C) (C)
ax ax (D)
2
ax 1
ax 1
Q.27 The domain where function f(x) = 2x2 – 1 and g(x) = 1 – 3x are equal, is-
Q.18 In the following, odd function is -
(A) cos x2 (B) (ex + 1)/(ex – 1)
(C) x2 – |x| (D) None of these
Q.19 The function f(x) = x2 – |x| is -
(A) an odd function
(B) a rational function
(C) an even func
POINTS
Preface 1
1. System of Co - ordinates 2
1.1 Cartesian Co-ordinates
1.2 Polar Co-ordinates
2. Distance Formula 2
3. Applications of Distance Formula 3
3.1 Position of three points
3.2 Position of four points
4. Section Formula 4
5. Co-ordinate of some particular points 4
5.1 Centroid
5.2 Incentre
5.3 Circumcentre
5.4 Ortho Centre
6. Area of triangle and quadrilateral 6
6.1 Area of Triangle
6.2 Area of Quadrilateral
7. Transformation of axes 7
7.1 Parallel Transformations
7.2 Rotational Transformation
7.3 Reflection of a Point
8. Locus 8
9. Some important points 9
10. Solved examples 10
STRAIGHT LINE
Preface 13
1. Equation of Straight line 14
2. Equation of Straight line parallel to axes 14
3. Slope of a line 14
4. Different forms of the equation of straight line 14
4.1 Slope – Intercept Form
4.2 Slope Point Form
4.3 Two Point Form
4.4 Intercept Form
4.5 Normal Form
4.6 Parametric Form
5. Reduction of General form of Equations into Standard Forms 15
6. Position of a point relative to a line 17
7. Angle between two straight lines 17
7.1 Parallel lines
7.2 Perpendicular lines
7.3 Coincident lines
8. Equation of parallel & perpendicular lines 18
9. Equation of Straight lines through (x , y ) making an angle α with y = mx + c 18
10. Length of perpendicular 19
10.1 Distance between two Parallel Lines
11. Condition of concurrency 19
12. Bisectors of angles between two lines 20
13. Line passing through the point of intersection of two lines 20
14. Homogeneous equation 20
15. General Equation of Second degree 21
16. Equation of lines joining the intersection points of a line & a curve to the origin 21
17. Some important points 22
18. Solved examples 23
CIRCLE
Preface 29
1. Definition 30
2. Standard Form of equation of a circle 31
2.1 General Form
2.2 Central Form
2.3 Diametral Form
2.4 Parametric Form
3. Equation of a circle in some special cases 31
4. Position of a point with respect to a circle 32
4.1 The least and the greatest distance of a point from a circle
5. Line and circle 33
5.1 Condition of Tangency
5.2 Intercepts made on coordinate axes by the circle
6. Equation of tangent and normal 34
6.1 Equation of Tangent
6.2 Equation of Normal
6.3 Length of Tangent
6.4 Pair of Tangents
7. Chord of contact 35
8. Director circle 35
9. Position of two circles 36
10. Equation of a chord whose middle point is given 38
11. Circle through the point of intersection 38
12. Common chord of two circles 39
13. Angle of intersection of two circles 39
14.1 Condition of Orthogonality
14. Some important points 40
15. Pole and Polar 40
9.1 Equation of Polar
9.2 Co-ordinates of Pole
9.3 Conjugate Points and Conjugate Lines
16. Radical axis and radical centre 41
16.1 Radical Axis
16.2 Radical Centre
17. Solved examples 42
PARABOLA
Preface 50
1. Definition 51
2. Terms related to parabola 51
3. Standard form of equation of parabola 51
3.1 Parameters of the Parabola y2 = 4ax
3.2 Other Standard Parabola
4. Reduction to Standard Equation 53
.
5. General equat
An ellipse is the locus of a point which moves in such a way that its distance form a fixed point is in constant ratio to its distance from a fixed line. The fixed point is called the focus and fixed line is called the directrix and the constant ratio is called the eccentricity of a ellipse denoted by (e).
In other word, we can say an ellipse is the locus of a point which moves in a plane so that the sum of it distances from fixed points is constant.
2.1 Standard Form of the equation of ellipse
Let the distance between two fixed points S and S' be 2ae and let C be the mid point of SS.
Taking CS as x- axis, C as origin.
Let P(h,k) be the moving point Let SP+ SP = 2a (fixed distance) then
(ii) Major & Minor axis : The straight line AA is called major axis and BB is called minor axis. The major and minor axis taken together are called the principal axes and its length will be given by
Length of major axis 2a Length of minor axis 2b
(iii) Centre : The point which bisect each chord of an ellipse is called centre (0,0) denoted by 'C'.
(iv) Directrix : ZM and Z M are two directrix and their equation are x= a/e and x = – a/e.
(v) Focus : S (ae, 0) and S (–ae,0) are two foci of an ellipse.
(vi) Latus Rectum : Such chord which passes through either focus and perpendicular to the major axis is called its latus rectum.
Length of Latus Rectum :
If L is (ae, 𝑙 ) then 2𝑙 is the length of
SP+S'P=
{(h ae)2 k 2} +
= 2a
Latus Rectum.
Length of Latus rectum is given by
2b2
.
h2(1– e2) + k2 = a2(1– e2)
Hence Locus of P(h, k) is given by. x2(1– e2) + y2 = a2(1– e2)
2
a
(vii) Relation between constant a, b, and e
a 2 b2
b2 = a2(1– e2) e2 =
a 2
x2
a 2 +
y
a 2 (1 e 2 ) = 1
e =
a 2
Result :
Major Axis
(a) Centre C is the point of intersection of the axes of an ellipse. Also C is the mid point of AA.
(b) Another form of standard equation of ellipse
x 2 y2
a 2 b2
1 when a < b.
Directrix Minor Axis Directrix x = -a/e x = a/e
Let us assume that a2(1– e2 )= b2
The standard equation will be given by
x2 y2
a2 b2
2.1.1 Various parameter related with standard ellipse :
In this case major axis is BB= 2b which is along y- axis and minor axis is AA= 2a along x- axis. Focus S(0,be) and S(0,–be) and directrix are y = b/e and y = –b/e.
2.2 General equation of the ellipse
The general equation of an ellipse whose focus is (h,k) and the directrix is the line ax + by + c = 0 and the eccentricity will be e. Then let P(x1,y1) be any point on the ellipse which moves such that SP = ePM
Let the equation of the ellipse x
y2
a > b
(x –h)2 + (y –k)2 =
e 2 (ax1 by1 c) 2
a 2 b2
1 1 a 2 b2
(i) Vertices of an ellipse : The point of which ellipse cut the axis x-axis at (a,0) & (– a, 0) and y- axis at (0, b) & (0, – b) is called the vertices of an ellipse.
Hence the locus of (x1,y1) will be given by (a2 + b
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Objective:
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1. Total No. of questions in Parabola are-
InchapterExamples.......................................38
SolvedExamples...........................................24
TotalNo. ofquestions............................. ....62
PARABOLA
2. 1. DEFINITION
A parabola is the locus of a point which moves in
such a way that its distance from a fixed point is
equal to its perpendicular distance from a fixed
straight line.
1.1 Focus : The fixed point is called the focus of the
Parabola.
1.2 Directrix : The fixed line is called the directrix of the
Parabola.
2. TERMS RELATED TO PARABOLA
2.1 Eccentricity : If P be a point on the parabola and
PM and PS are the distances from the directrix and
focus S respectively then the ratio PS/PM is called
the eccentricity of the Parabola which is denoted by e.
Note: By the definition for the parabola e = 1.
If e > 1 Hyperbola, e = 0 circle, e < 1
ellipse
2.2 Axis : A straight line passes through the focus and
perpendicular to the directrix is called the axis of
parabola.
(focus)
2.3 Vertex : The point of intersection of a parabola and
its axis is called the vertex of the Parabola.
NOTE: The vertex is the middle point of the focus and
the point of intersection of axis and directrix
2.4 Focal Length (Focal distance) : The distance of any
point P (x, y) on the parabola from the focus is called
the focal length. i.e.
The focal distance of P = the perpendicular distance
of the point P from the directrix.
2.5 Double ordinate : The chord which is perpendicular
to the axis of Parabola or parallel to Directrix is called
double ordinate of the Parabola.
2.6 Focal chord : Any chord of the parabola passing
through the focus is called Focal chord.
2.7 Latus Rectum : If a double ordinate passes through
the focus of parabola then it is called as latus rectum.
2.7.1Length of latus rectum :
The length of the latus rectum = 2 x perpendicular
distance of focus from the directrix.
3. STANDARD FORM OF EQUATION OF
PARABOLA
If we take vertex as the origin, axis as x- axis and
distance between vertex and focus as 'a' then equation
of the parabola in the simplest form will be-
y2 = 4ax
y
M L(a,2a)
L’
S (a,0)
x
A
Z
Directrix x = a
Focus
axis
Focal chord
Double ordinate
Latus Rectum
Vertex
F
o
c
a
l
d
is
ta
n
c
e
y’
x+
a
=
0
P
(a,-2a)
L’
L
3. 3.1 Parameters of the Parabola y2 = 4ax
(i) Vertex A (0, 0)
(ii) Focus S (a, 0)
(iii) Directrix x + a = 0
(iv) Axis y = 0 or x– axis
(v) Equation of Latus Rectum x = a
(vi) Length of L.R. 4a
(vii) Ends of L.R. (a, 2a), (a, – 2a)
(viii) The focal distance sum of abscissa of
the point and distance between vertex and L.R.
(ix) If length of any double ordinate of parabola
y2 = 4ax is 2 then coordinates of end points
of this Double ordinate are
,
4a
2
and
,
4a
2
.
3.2 Other standard Parabola :
Examples
based on Standard form of an equation of Parabola
Ex.1 If focus of a parabola is (3,–4) and
directrix is x + y – 2 = 0, then its vertex is
(A) (4/15, – 4/13)
(B) (–13/4, –15/4)
(C) (15/2, – 13/2)
(D) (15/4, – 13/4)
Q1 P1
Y
S
X
M
Z S P1
Q1
X
M
A
Z
y2 = – 4 ax
x2 = 4ay x2 = – 4ay
Equationof Vertex Axis Focus Directrix Equation of Lengthof
Parabola Latus rectum Latus rectum
y2 = 4ax (0,0) y= 0 (a,0) x = –a x= a 4a
y2 = – 4ax (0,0) y= 0 (–a, 0) x = a x= –a 4a
x2 = 4ay (0,0) x=0 (0,a) y = a y = a 4a
x2 = – 4ay (0,0) x=0 (0, –a) y = a y = –a 4a
4. Sol. First we find the equation of axis of parabola,
which is perpendicular to directrix. So its equation
is x – y + k = 0. It passes through focus
S (3, –4)
3 – (–4) + k = 0 k = –7
Let Z is the point of intersection of axis and
directrix.
Solving equation x + y –2 = 0 and x –y–7=0
gives Z (9/2, –5/2)
Vertex A is the mid point of Z and S
A
2
2
5
4
,
2
9
3
2
= A
A
4
13
,
4
15
Ans.[D]
Ex.2 The equation of the parabola whose vertex is
(–3, 0) and directrix is x + 5 = 0 is-
(A) y2 = 8(x + 3)
(B) y2 = – 8(x + 3)
(C) x2 = 8(y + 3)
(D) y2 = 8(x + 5)
Sol. A line passing through the vertex (–3, 0) and
perpendicular to directrix x + 5 = 0 is x-axis
which is the axis of the parabola by definition.
Let focus of the parabola is (a, 0). Since
vertex, is the middle point of Z(–5, 0) and
focus S, therefore
1
–
a
2
)
5
a
(
3
Focus = (–1,0)
Thus the equation to the parabola is
(x + 1)2 + y2 =(x + 5)2
y2 = 8(x +3) Ans. [A]
Ex.3 The equation of the directrix of the parabola
y2 = 12x is-
(A) x + 3 = 0 (B) y + 3 = 0
(C) x – 3 = 0 (D) y – 3 = 0
Sol. Here a = 3, so the equation of the directrix
is given by x = – a x= – 3
x + 3 = 0
Ans. [A]
Ex.4 The equation of the latus rectum of the
parabola x2 = – 12y is-
(A) y = 3 (B) x = 3
(C) y = – 3 (D) x = –3
Sol. Here a = 3 so the equation of the L.R. is
given by y = – a y = – 3 Ans. [C]
4. REDUCTION TO STANDARD EQUATION
If the equation of a parabola is not in standard form
and if it contains second degree term either in y or
in x (but not in both) then it can be reduced into
standard form. For this we change the given equation
into the following forms-
(y – k)2 = 4a (x – h) or (x – p)2 = 4b (y – q)
And then we compare from the following table for the
results related to parabola.
Examples
based on Reduction to Standard eqn. of a Parabola
Ex.5 The axis of the parabola
9y2 –16x –12y –57 = 0 is
(A) 3y = 2 (B) 2x = 3
(C) x +3y = 3 (D) y = 3
Equation of Vertex Axis Focus Directrix Equation of L.R. Length of L.R.
Parabola
(y– k)2 = 4a (x–h) (h,k) y = k (h + a, k) x + a – h = 0 x = a + h 4a
(x–p)2 = 4b ( y–q) (p, q) x = p (p,b+ q) y + b – q = 0 y = b + q 4b
5. Sol. 9y2 –12y = 16x + 57
(3y – 2)2 = 16x + 57
16
61
x
9
16
3
2
y
2
Which shows that equation of the axis is
y –2/3 = 0 or 3y = 2 Ans. [A]
Ex.6 Vertex, focus, latus rectum, length of the
latus rectum and equation of directrix of the
parabola y2 = 4x +4y are
(A) (1, 2), (0, 2), y = 0, 4,x = – 2
(B) (–1, 2), (0, 2), x = 0, 4, x = –2
(C) (–1, 2), (1, 2), x = 0, 4, x = 2
(D) (–1, 2) (0, 2), y = 0, 2 , y = – 2
Sol. Given parabola y2 = 4x + 4y
or (y – 2)2 =4(x +1)
or Y2 = 4X
Here X = x +1, Y= y – 2
vertex = (X = 0, Y = 0)
or (x + 1 = 0, y – 2 = 0) (–1, 2)
Focus (X = 1, Y = 0)
or (x + 1 = 1, y – 2 = 0) (0, 2)
Latus rectum X = 1 x = 0
Length of Latus rectum = 4
equation of the directrix is
X = –1 x +1 = –1 x = –2
Ans. [B]
Ex.7 The vertex of the parabola x2 – 8y – x + 19 = 0 is-
(A)
2
1
,
32
75
(B)
32
75
,
2
1
(C)
32
75
,
2
1
(D) None of these
Sol. The given equation of Parabola can be written
as
2
2
1
x
– 8y + 19 –
4
1
= 0
2
2
1
x
= 8y –
4
1
76
2
2
1
x
= 8
32
75
y
vertex =
32
75
,
2
1
Ans.[B]
5. GENERAL EQUATION OF A PARABOLA
If (h,k) be the locus of a parabola and the equation
of directrix is ax + by + c = 0, then its equation is
given by
2
2
)
k
y
(
)
h
x
(
= 2
2
b
a
c
by
ax
which
gives (bx– ay)2 + 2gx + 2fy + d = 0 where g, f, d are the
constants.
Note:
(i) It is a second degree equation in x and y and the
terms of second degree forms a perfect square and it
contains at least one linear term.
(ii) The general equation of second degree
ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represents a
parabola, if
(a) h2 = ab
(b) = abc + 2fgh – af2 – bg2 – ch2 0
6. EQUATION OF PARABOLA WHEN ITS VERTEX
AND FOCUS ARE GIVEN
6.1 If both lie on either of the coordinate axis :
In this case first find distance 'a' between these
points and taking vertex as the origin suppose the
equation as y2 = 4ax or x2 = 4ay. Then shift the origin
to the vertex.
6.2 When both do not lie on any coordinate axes
In this case first find the coordinates of Z and
equation of the directrix, then write the equation of
the parabola by the definition.
Examples
based on
Equation of Parabola when its vertex and
focus are given
Ex.8 If (0,4) and (0,2) are vertex and focus of
parabola respectively, then its equation is-
(A) y2 + 8x = 32 (B) y2 – 8x = 32
(C) x2 + 8y = 32 (D) x2 – 8y = 32
Sol. Since vertex and focus are on y-axis, so
y-axis is the axis of the parabola.
Distance between vertex and focus = 'a' = 2.
So on taking vertex as origin, the equation of
the parabola is x2
= – 4ay
(negative because vertex lies above the
focus)
or x2
= – 8y.
Now shifting the origin to its original position,
the required equation will be
x2
= –8(y – 4)
x2
+ 8y = 32
Ans. [C]
6. 7. PARAMETRIC EQUATION OF PARABOLA
The parametric equation of Parabola y2 = 4ax are
x = at2, y = 2at
Hence any point on this parabola is (at2, 2 at) which
is called as 't' point.
Note:
(i) Parametric equation of the Parabola x2 = 4ay is
x = 2at, y = at2
(ii) Any point on Parabola y2 = 4ax may also be written
as (a/t2, 2 a/t)
(iii) The ends of a double ordinate of a parabola can be
taken as (at2, 2 at) and (at2, – 2at)
(iv) Parametric equations of the parabola
(y – h)2 = 4a (x – k)2 is x – k = at2 and y – h= 2 at
Examples
based on
Parametric equation of Parabola
Ex.9 The parametric equation of the curve y2 = 8x are-
(A) x = 2t, y = 4t2 (B) x = 2t2, y = 4t
(C) x = t2, y = 2t (D) None of these
Sol. Here a = 2; y = 2at y = 2.2.t = 4t
x = at2 x = 2t2 Ans. [B]
Ex.10 The parametric equation of the curve
(y– 2)2 = 12 (x – 4) are-
(A) 6t, 3t2 (B) 2 + 3t, 4 + t2
(C) 4 + 3t2,2 + 6t (D) None of these
Sol. Here a = 3
y – 2 = 2at y = 2 + 2.3t = 2 + 6t
x – 4 = at2 x = 4 + 3.t2 = 4 + 3t2
Ans. [C]
Ex.11 Parameter t of a point (4,–6) of the parabola
y2 = 9x is-
(A) 4/3 (B) – 4/3 (C) – 3/4 (D) – 4/5
Sol. Parametric coordinates of any point on
parabola y2 = 4 ax are (at2, 2 at)
Here 4 a = 9 a = 9/4
y coordinate 2 at = – 6
2 (9/4) t= – 6 t = – 4/3 Ans. [B]
8. CHORD
8.1 Equation of chord joining any two points of a parabola
Let the points are (at1
2, 2 at1) and (at2
2, 2 at2) then
equation of chord is-
(y– 2at1) = 2
1
2
2
1
2
at
at
at
2
at
2
(x – at1
2)
y – 2at1 =
2
1 t
t
2
(x – at1
2)
(t1 + t2) y = 2x + 2at1 t2
Note :
(i) If 't1' and 't2' are the Parameters of the ends
of a focal chord of the Parabola y2 = 4ax, then
t1t2 = – 1
(ii) If one end of focal chord of parabola is
(at2, 2at) , then other end will be (a/t2, – 2a/t)
and length of focal chord = a (t + 1/t)2.
(iii) The length of the chord joining two points 't1
'
and 't2' on the parabola y2 =4ax is
a (t1 – t2) 4
)
t
t
( 2
2
1
8.2 Length of intercept = )
mc
a
)(
m
1
(
a
m
4 2
2
Examples
based on
Chord
Ex.12 If one extremity of the focal chord of the
parabola y2 = 16x is (1,–4) then find other
extremity.
Sol. Here 2at = – 4 and a = 4
t =
4
.
2
4
= –
2
1
other extremity is
2
/
1
4
.
2
,
4
/
1
4
= (16,16) Ans.
Ex.13 Find the length of a chord of a parabola
y2 = 4x which passes through vertex and
makes an angle of 45º with x- axis.
Sol. Let the equation of chord is y = mx + c
Here m = tan 45º = 1
as it passes through vertex (0,0) c = 0
Hence the equation is y = x and a = 1
= 4 )
0
1
)(
1
1
(
1
= 4 2 Ans.
Ex.14 Length of intercept by the line 4y = 3x – 48
on the parabola y2 = 64x is-
7. (A)
1600
9
(B)
9
1600
(C)
9
160
(D) None of these
Sol. 4y = 3x – 48 m = 3/4, c = –12
y2 = 64x a = 16
Length of intercept = )
mc
a
(
)
m
1
(
a
m
4 2
2
=
4
3
12
16
16
9
1
16
16
9
4
=
9
1600
Ans. [B]
9. POSITION OF A POINT AND A LINE WITH
RESPECT TO A PARABOLA
9.1 Position of a point with respect to a parabola
A point (x1, y1) lies inside, on or outside of the region
of the parabola y2 = 4ax according as y1
2 – 4ax1 <
= or > 0
9.2 Line and Parabola :
The line y = mx + c will intersect a parabola
y2 = 4ax in two real and different, coincident or
imaginary point, according as a – mc >, = < 0
Ex.15 For the parabola y2 = 8x point (2, 5) is
(A) Inside the parabola
(B) Focus
(C) Outside the parabola
(D) On the parabola
Sol. (y2 – 8x)x = 2, y = 5 = (5)2 – 8 × 2 = 9 > 0
Point (2, 5) is outside parabola y2 = 8x
Ans. [C]
10. TANGENT TO THE PARABOLA
10.1 Condition of Tangency : If the line y = mx + c touches
a parabola y2 = 4ax then c = a/m
Note:
(i) The line y = mx + c touches parabola x2 = 4ay
if c = – am2
(ii) The line x cos + y sin = p touches the parabola
y2 = 4ax if a sin2 + p cos = 0
(iii) If the equation of parabola is not in standard form,
then for condition of tangency, first eliminate one
variable quantity (x or y) between equations of
straight line and parabola and then apply the condition
B2 = 4AC for the quadratic equation so obtained.
Examples
based on
Condition of tangency
Ex.16 The line y = mx + 1 is a tangent to the
parabola y2 = 4x if
(A) m = 1 (B) m = 4
(C) m = 2 (D) m = 3
Sol. Here a = 1, c = 1, m = m. Now applying
condition of tangency c = a/m, we have
1 = 1/m m = 1
Ans. [A]
Ex.17 If the line 2x – 3y = k touches the parabola
y2 = 6x, then the value of k is
(A) 27/4 (B) –81/4
(C) – 7 (D) – 27/4
Sol. Given
2
k
y
3
x
...(1)
and y2 = 6x ...(2)
2
k
y
3
6
y2
y2 = 3(3y + k)
y2 – 9y – 3k = 0 ...(3)
If line (1) touches parabola (2) then roots of
quadratic equation (3) is equal
(– 9)2 = 4 × 1 × (– 3k)
4
27
k
Ans. [D]
10.2Equation of Tangent
10.2.1 Point Form : The equation of tangent to the
parabola y2 = 4ax at the point (x1, y1) is
yy1 = 2a(x + x1) or T = 0
10.2.2 Parametric Form : The equation of the
tangent to the parabola at t i.e. (at2, 2at) is
ty = x + at2
10.2.3 Slope Form : The equation of the tangent of the
parabola y2 = 4ax is
y = mx +
m
a
(at ,2at)
2
Note :
(i) y = mx + a/m is a tangent to the parabola y2 = 4ax
for all value of m and its point of contact is (a/m2, 2a/m).
(ii) y = mx – am2 is a tangent to the parabola x2 = 4ay
for all value of m and its point of contact is
(2am, am2)
8. (iii) Point of intersection of tangents at points t1 and t2
of parabola is [at1t2, a(t1+t2)]
(iv) Two perpendicular tangents of a parabola meet on its
directrix. So the director circle of a parabola is its
directrix or tangents drawn from any point on the
directrix are always perpendicular.
(v) The tangents drawn at the end points of a focal
chord of a parabola are perpendicular and they meet
at the directrix.
Examples
based on
Equation of tangent
Ex.18 The equation of the tangent to the parabola
y2 = 4ax at the points (3, 2) is
(A) 3y + x +3 = 0 (B) 3x + y +3 = 0
(C) 3x = y +3 (D) 3y = x +3
Sol. Since (3, 2) lies on the parabola y2 = 4ax, so
4 = 12a a = 1/3
Now using T = 0, the equation of the tangent
at (3, 2) is
y(2) = 2a(x + 3)
y =
3
1
(x + 3)
3y = x + 3
Ans. [D]
Ex.19 The equation of the tangent to the parabola
y2 = 12x drawn at the upper end of its latus
rectum is
(A) y = x + 3 (B) x + y = 3
(C) y = x – 3 (D) None of these
Sol. Here a = 3, so the upper end of LR = (3, 6).
Hence the equation of the tangent at this
point is y(6) = 6(x + 3)
y = x + 3 Ans. [A]
Ex.20 The slope of tangent lines drawn from (3, 8)
to the parabola y2 = –12x are
(A) –3, –1/3 (B) 3, 1/3
(C) 3, –1/3 (D) –3, 1/3
Sol. Since 82 + 12 × 3 0, therefore the point
(3, 8) is not on the parabola
Now equation of any tangent to the parabola
y2 = –12x is written as
y = mx –
m
3
Since this line passes through (3, 8)
8 = 3m –
m
3
3m2 – 8m – 3 = 0
Solving this equation, we get m = 3, –
3
1
Ans. [C]
Ex.21 If a tangent to the parabola y2 = ax makes
an angle of 45º with x- axis, then its point of
contact is-
(A)
2
a
,
4
a
(B)
2
a
,
2
a
(C)
4
a
,
2
a
(D)
2
a
,
4
a
Sol. The slope of the tangent = tan 45º = 1
m = 1 and a a/4
Point of contact =
1
4
/
a
.
2
,
1
4
/
a
2 =
2
a
,
4
a
Ans. [A]
Ex.22 Find the common tangent of the parabola
x2 =4ay and y2 = 4ax(m > 0).
Sol. Equation of tangent for x2 = 4ay is y = mx – am2
as this line also touches y2 = 4ax
– am2 = a/m (c = a/m)
m3 = – 1
m = – 1
equation of tangent x + y + a = 0
Ans.
11. NORMAL TO THE PARABOLA
11.1 Equation of Normal :
11.1.1 Point Form : The equation to the normal at the
point (x1, y1) of the parabola y2 = 4ax is given by
y – y1 =
a
2
y1
(x – x1)
11.1.2 Parametric Form : The equation to the normal
at the point (at2,2at) is y + tx = 2at + at3
11.1.3 Slope Form : Equation of normal in terms of
slope m is y = mx – 2am – am3.
Note :
(i) the foot of the normal is (am2, –2am)
(ii) Condition for normal : The line y = mx + c
is a normal to the parabola y2 = 4ax if
c = –2am – am3 and x2 = 4ay
if c = 2a + 2
m
a
(iii) Normal Chord : If the normal at 't1' meets
the parabola y2 = 4ax again at the point ' t2'
then this is called as normal chord. Again for
normal chord t2 = – t1 –
1
t
2
9. (iv) If two normal drawn at point ' t1' and ' t2'
meet on the parabola then t1t2 =2
Examples
based on
Equation and properties of Normal
Ex.23 The equation of normal at the point (a/m2, 2a/m)
on the parabola y2 = 4ax is-
(A) m3y = m2x – 2am2 – a
(B) y = mx – 2am – am3
(C) m3y = 2am2 – m2x + a
(D) None of these
Sol. The equation of the normal at (at2, 2at) of the
parabola y2 = 4ax is written as
y + tx = 2at + at3
For the given point t = 1/m, therefore equation
of required normal is written as
3
m
1
a
m
1
a
2
x
m
1
y
m3y = 2am2 – m2x + a
Ans. [C]
Ex.24 Find the equation of a normal at the parabola
y2 = 4x which is parallel to the line y = 3x + 4.
Sol. Let the equation is y = 3x + c
This is normal to the parabola y2 = 4x then
c = – 2 am – am3
= – 2.3 – 33 ( a = 1)
= – 33
Hence equation is y = 3x – 33
Ans.
Ex.25 If the line x + y = 1 is a normal to the parabola
y2 = kx, then the value of k is
Sol. As line y = –x + 1 is the normal to the
parabola y2 = kx then
1 = – 2(a)(– 1) – a(– 1)3
a = 1/3
4a = k k = 4/3
Ans.
Ex.26 Find the equation of a normal at the parabola
y2 = 4x which passes through (3,0).
Sol. Equation of Normal y = mx – 2 am – am3
Here a = 1 and it passes through (3,0)
0 = 3m – 2m – m3
m3 – m = 0
m = 0, ± 1
for m = 0 y = 0
m = 1 y = x – 3
m = –1 y = – x + 3
Ans.
Ex.27 If the normal of the parabola y2 = 4ax drawn
at (a, 2a) meets the parabola again at the
point (at2, 2at) then t is equal to-
(A) 3 (B) 1 (C) –1 (D) –3
Sol. If t' be the parameter of the given point, then
2at' = 2a t' = 1
Now t= – t' –
t
2
t = –1 –
2
1
= – 3
Ans. [D]
Ex.28 Find the point where the normal at (4,4)
meets the parabola y2 = 4x
Sol. Here a = 1 and
at1
2 = 4 t1 = 2
t2 = – 2 –
2
2
= – 3
Point is (at2
2, 2 at2) = {(–3)2,2.1.(–3)}= (9,– 6)
Ans.
Ex.29 If two of the normal of the parabola y2
= 4x,
that pass through (15, 12) are 4x + y = 72,
and 3x – y = 33, then the third normal is
(A) y = x – 3 (B) x + y = 3
(C) y = x + 3 (D) None of these
Sol. Here, If m1
, m2
, m3
are slopes of normal, then
m1
+m2
+m3
=0 and m1
m2
m3
=
a
y1
a = 1 here m1
= –4, m2
= 3
– 4 + 3 + m3
= 0 m3
= 1
Also (–4) (3) (1) = –
1
12
is satisfied.
But (15, 12) satisfies y = x – 3
Hence (A) is correct answer
Ans. [A]
Note: As seen value of m3
is not sufficient to locate
the correct answer. The answer should be
confirmed before it is reported.
10. Ex.30 If two normal drawn from any point to the
parabola y2
= 4ax makes angle and with the
axis such that tan .tan = 2, then locus of this
point is-
(A) y2
= 4ax (B) x2
= 4ay
(C) y2
= – 4ax (D) x2
= –4ay
Sol. Let the point is (h, k). The equation of any
normal to the parabola y2
= 4ax is
y = mx – 2am – am3
passes through (h, k)
k = mk – 2am – am3
am3
+ m(2a – h) + k = 0 ...(i)
m1
, m2
, m3
are roots of the equation
then m1
.m2
.m3
= –
a
k
but m1
m2
= 2, m3
= –
a
2
k
m3
is root of (i)
a
3
a
2
k
–
a
2
k
(2a – h) + k = 0 k2
= 4ah
Thus locus is y2
= 4ax
Ans. [A]
12. PAIR OF TANGENTS
If the point (x1, y1 ) is outside the parabola, then two
tangents can be drawn. The equation of pair of
tangents drawn to the parabola y2 = 4ax is given by
SS1 = T2
i.e. (y2 – 4ax) (y1
2 – 4ax1) = [yy1 – 2a(x + x1)]2
12.1Locus of point of Intersection of tangents
The locus of point of intersection of tangent to the
parabola y2 = 4ax which are having an angle between
them is given by y2 – 4ax = (a+x)2 tan2
Note :
(i) If = 0º or then locus is (y2 –4 ax) = 0
which is the given parabola.
(ii) If = 90º, then locus is x + a = 0 which is
the directrix of the parabola.
Ex.31 The equation of pair of tangents drawn from
(1,4) to the parabola y2 =12x is-
(A) 3x2 + y2 – 10 x + 4y – 3 = 0
(B) 3x2 + y2 – 10 x + 4xy + 4y – 3 = 0
(C) 3x2 + y2 + 10 x – 4xy – 4y + 3 = 0
(D) x2 + 3y2 + 10 x + 4xy – 4y – 3 = 0
Sol. From formula SS1 = T2
(y2 – 12x). 4 = (4y – 6x – 6)2
3x2 +y2 +10x– 4xy –4y+3 = 0
Ans. [C]
Ex.32 Find the locus of perpendicular tangents of
the parabola x2 – 4x + 2y + 3 = 0
Sol. The given equation of parabola can be
written as
x2 – 4x + 4– 4 + 2y + 3 = 0
(x–2)2 +2y – 1 = 0
(x–2)2 = – 2 (y –1/2) 4a = 2
a =1/2
Directrix y – 1/2 = 1/2 y = 1
is the equation of locus.
Ans.
13. CHORD OF CONTACT
The equation of chord of contact of tangents drawn
from a point (x1, y1) to the parabola y2 = 4ax is
yy1 = 2a (x + x1)
This equation is same as equation of the tangents
at the point (x1, y1).
Note :
(i) The chord of contact joining the point of contact of
two perpendicular tangents always passes through
focus.
(ii) Lengths of the chord of contact is
a
1
)
a
4
y
)(
ax
4
y
( 2
2
1
1
2
1
(iii) Area of triangle formed by tangents drawn from
(x1, y1) and their chord of contact is
a
2
1 2
/
3
1
2
1 )
ax
4
y
(
11. Examples
based on
Chord of contact
Ex.33 The chord of contact drawn from (2,4) to the
parabola y2 = 4x is-
(A) 2y = x – 2 (B) y = 2x + 2
(C) y = 2x – 1 (D) 2y = x + 2
Sol. Here a = 1 equation of chord of contact is
yy1= 2a (x + x1)
4y = 2 (x + 2) 2y = x + 2 Ans.[D]
Ex.34 The area of triangle made by the chord of
contact and tangents drawn from point (4,6)
to the parabola y2 = 8x is-
(A)
2
1
(B) 2
(C)
2
1
3 (D) None of these
Sol. Area =
a
2
1
(y1
2 – 4ax1)3/2
Here a = 2, (x1, y1) = (4,6)
Area of triangle =
2
.
2
1
(36– 32)3/2
=
( ) /
4
4
3 2
= 2
Ans. [B]
Ex.35 The length of chord of contact of tangents
drawn from point (–2, –1) to the parabola
y2 = 4x is-
(A) 2 2 (B) 3 5
(C) 8 (D) None of these
Sol. Here a = 1, (x1, y1) (–2, –1)
Hence length of chord of contact
=
1
a
(y 4ax )(y 4a )
1
2
1 1
2 2
= (1 4)(1 4.1.( 2))
= 5 9
. = 3 5
Ans. [B]
14. EQUATION OFTHE CHORD WITH GIVEN
MIDPOINT
The equation of the chord at the parabola
y2 = 4ax bisected at the point (x1, y1) is given by
T = S1 i.e. yy1 – 2a(x + x1) = y1
2 – 4ax1
Ex.36 Find the equation of a chord of the parabola
y2 = 6x if mid point of the chord is (– 1,1).
Sol. Here a = 3/2
y = – 2
3
2
(x –1) = 1 – 6 (–1)
y – 3x + 3 = 7 y – 3 x = 4
Ans.
15. DIAMETER OF THE PARABOLA
The locus of the mid points of a system of parallel
chords of a parabola is called a diameter of the
parabola.
The equation of a system of parallel chord
y = mx + c with respect to the parabola y2 = 4ax is
given by y
a
m
2
Ex.37 The equation of system of parallel chords of the
parabola y2 =
3
2
x is y + 2x + 1 = 0 then find its
diameter.
Sol. Here 4a =
3
2
a =
6
1
and m = – 2
Diameter is y
a
m
2
y =
2
6
1
.
2
y = –
6
1
This is the required equation of diameter. Ans.
15.1 Properties of Diameter :
(i) Every diameter of a parabola is parallel to its
axis.
(ii) The tangents at the end point of a diameter
is parallel to corresponding system of parallel
chords.
(iii) The tangents at the ends of any chord meet
on the diameter which bisects the chord.
16.GEOMETRICAL PROPERTIES OF THE
PARABOLA
(i) The sub tangent at any point on the parabola
is twice the abscissa or proportional to
square of the ordinate of the point.
(ii) The sub normal is constant for all points on
the parabola and is equal to its semi latus
rectum 2a.
12. (iii) The semi latus rectum of a parabola is the
H.M. between the segments of any focal
chord of a parabola i.e. if PQR is a focal
chord, then
QR
PQ
QR
.
PQ
2
a
2
(iv) The tangents at the extremities of any focal
chord of a parabola intersect at right angles and
their point of intersection lies on directrix i.e. the
locus of the point of intersection of perpendicular
tangents is directrix.
(v) If the tangent and normal at any point P of
parabola meet the axes in T and G respectively,
then
(a) ST = SG = SP
(b) PSK is a right angle, where K is the point
where the tangent at P meets the directrix.
(c) The tangent at P is equally inclined to the
axis and the focal distance.
(vi) The locus of the point of intersection of the
tangent at P and perpendicular from the focus on
this tangent is the tangent at the vertex of the
parabola.
(vii) If a circle intersect a parabola in four points,
then the sum of their ordinates is zero.
(viii) The area of triangle formed inside the parabola
y2 = 4ax is
a
8
1
(y1 – y2)(y2 – y3) (y3 – y1) where y1,y2, y3
are ordinate of vertices of the triangle.
(ix) The abscissa of point of intersection R of tangents
at P(x1, y1) and Q(x2, y2) on the parabola is G.M. of
abscissa of P and Q and ordinate of R is A.M. of
ordinate of P and Q thus R
2
x
x
,
x
x 2
1
2
1
(x) If vertex and focus of a parabola are on the
x-axis and at distance a and a' from origin respectively
then equation of parabola
y2 = 4(a' – a) (x – a)
(xi) The area of triangle formed by three points on a
parabola is twice the area of the triangle formed by
the tangents at these points.
Examples
based on
Geometrical properties of a parabola
Ex.38 The latus rectum of a parabola whose focal
chord is PSQ such that SP=3 and SQ= 2 is
given by-
(A) 24/5 (B) 6/5
(C) 12/5 (D) None of these
Sol. Since the semi-latus rectum of a parabola is
the HM of segments of a focal chord.
Semi-latus rectum =
SQ
SP
SQ
.
SP
2
=
5
12
Latus rectum =
5
24
Ans. [A]
13. Review Chart for Standard Parabolas
y2 = 4ax x2 = 4ay
Diagram
Vertex (A) (0, 0) (0, 0)
Focus (S) (a, 0) (0,a)
Axis y = 0 x = 0
Directrix x + a = 0 y + a = 0
Equation of LR x – a = 0 y – a = 0
Length of LR 4a 4a
Extremities of LR(L1, L2) (a, 2a); (a, –2a) (2a, a) ; (–2a, a)
Focal distance of (x, y) x + a y + a
Parametric equations x = at2, y = 2at x = 2at, y = at2
Parametric points (at2, 2at) (2at, at2)
Condition of tangency c = a/m c = –am2
(for y = mx + c)
Tangent at (x1, y1) yy1 = 2a (x + x1) xx1 = 2a (y + y1)
Tangent in slope form y = mx + a/m y = mx – am2
point of contact of above (a/m2, 2a/m) (2am, am2)
Tangent at ' t' point ty = x + at2 tx = y + at2
Slope of tangent at ' t ' 1/t t
Normal at (x1, y1) y – y1 =–
2a
y1
(x – x1) y – y1 =
1
x
a
2
(x – x1)
Normal in slope form y = mx – 2am – am3 y = mx + 2a + a/m2
Foot of above normal (am2, –2am) (–2a/m, a/m2)
Normal at ' t ' point y + tx = 2at + at3 ty + x = 2at + at3
Slope of normal at 't' –t – 1/t
Condition of normal c = – 2am – am3 c = 2a + a/m2
(for y = mx + c)
Director circle x + a = 0 y + a = 0
Diameter w.r.t. (y = mx + ) y = 2a/m x = 2am
14. Ex.1 The vertex of the parabola y2
+6x – 2y +13 =0 is
(A) (1 , –1) (B) (–2, 1)
(C) (3/2, 1) (D) (– 7/2,1)
Sol. We have : y2
+ 6x – 2y + 13 = 0
y2
– 2y = –6x – 13
(y – 1)2
= – 6(x + 2)
Clearly, the vertex of this parabola is (–2, 1)
Ans. [B]
Ex.2 If vertex of parabola is (2, 0) and directrix is y-
axis, then its focus is-
(A) (2, 0) (B) (–2, 0)
(C) (–4, 0) (D) (4, 0)
Sol. Since the axis of the parabola is the line which
passes through vertex and perpendicular to the
directrix, therefore x-axis is the axis of the
parabola.
Obviously Z (0, 0).
Let focus of the parabola is S (a, 0). Since vertex
(2,0) is mid point of ZS, therefore
2
0
a
= 2 a = 4.
Focus is (4, 0)
Ans. [D]
Ex.3 If the focus of a parabola is (1, 0) and its directrix
is x + y = 5, then its vertex is-
(A) (0, 1) (B) (0, –1)
(C) (2, 1) (D) (3, 2)
Sol. Since axis is a line perpendicular to directrix, so
it will be x – y = k. It also passes from focus,
therefore k = 1.
So equation of axis is x – y = 1.
Solving it with x + y = 5, we get
Z (3, 2).
If vertex is (a, b), then a = 2, b = 1.
Hence vertex is (2, 1).
Ans. [C]
Ex.4 The directrix and axis of the parabola
4y2
– 6x – 4y = 5 are respectively.
(A) 8x + 11 = 0; y – 1 = 0
(B) 8x – 11 = 0, 2y – 1 = 0
(C) 8x + 11 = 0; 2y – 1 = 0
(D) None of these
Sol. Here 4y2
– 4y = 6x + 5
4
2
2
1
y
= 6 (x + 1)
Put y –
2
1
= Y, x + 1 = X
The equation in standard form Y2
=
2
3
X
4a =
2
3
a =
8
3
Directrix, X + a = 0
x +1+
8
3
= 0 8x + 11 = 0
Axis is Y = 0 y –
2
1
= 0 2y – 1 = 0
Ans. [C]
Ex.5 The angle subtended by double ordinate of
length 8a at the vertex of the parabola
y2
= 4ax is-
(A) 45º (B) 90º (C) 60º (D) 30º
Sol. Let (x1
, y1
) be any point on the parabola
y2
= 4ax, then length of double ordinate
2y1
= 8a y1
= 4a
y1
2
= 4ax1
x1
= 4a
vertices of double ordinate are
P(4a, 4a); Q(4a, –4a)
If A is the vertex(0, 0), then
Slope of AP = 1 = m1
Slope of AQ = –1 = m2
m1
m2
= –1 PAQ = 90º Ans. [B]
Ex.6 The length of latus rectum of a parabola, whose
focus is (2, 3) and directrix is the line
x – 4y + 3 = 0 is-
(A)
17
7
(B)
21
14
(C)
21
7
(D)
17
14
SOLVED EXAMPLES
(3/8, 0) X
O
Y
(–3/8, 0)
15. Sol. The length of latus rectum
= 2 × perp. from focus to the directrix
= 2 ×
2
2
)
4
(
)
1
(
3
)
3
(
4
2
=
17
14
The numerical length =
17
14
Ans. [D]
Note:- The negative sign of the latus rectum may only
be ignored if its length is asked. For other
calculations it should be used.
Ex.7 The coordinates of an endpoint of the latus
rectum of the parabola (y – 1)2
= 4(x + 1) are
(A) (0,–3)(B) (0,–1)
(C) (0,1) (D) (1, 3)
Sol. Shifting the origin at (–1, 1) we have
1
Y
y
1
X
x
...(i)
Using (i), the given parabola becomes.
Y2
= 4X
The coordinates of the endpoints of latus rectum
are
(X = 1, Y = 2) and (X =1, Y= –2)
Using (i), the coordinates of the end point of the
latus rectum are (0,3) and (0, –1)
Ans. [B]
Ex.8 The locus of the middle points of all focal chords
of the parabola y2
= 4ax is-
(A) y2
= 2a(x + a) (B) y2
= 2ax ( C )
y2
= 2a (x – a) (D) None of these
Sol. Let the middle point be (h, k), then equation of
the chord is given by
k2
– 4ah = yk – 2a(x + h)
Since it passes through (a, 0),
k2
– 4ah = – 2a(h + a)
k2
= 2a(h – a)
Hence required locus is
y2
= 2a(x – a) which is parabola.
Ans. [C]
Ex.9 The length of the chord of parabola x2
= 4ay
passing through the vertex and having slope tan
is-
(A) 4a cosec cot (B) 4a tan sec ( C )
4a cos cot (D) 4a sin tan
Sol. Let A be the vertex and AP be a chord of
x2
= 4ay such that slope of AP is tan . Let the
coordinates of P be(2at, at2
) Then,
Slope of AP =
at
2
at2
=
2
t
tan =
2
t
t 2tan
Now, AP =
2
2
2
)
0
at
(
)
0
at
2
(
= at 2
t
4
= 2a tan a
tan
4
4 2
= 4a tan sec
Ans. [B]
Ex.10 The point on y2
= 4ax nearest to the focus has
its abscissa equal to
(A) –a (B) a (C) a/2 (D) 0
Sol. Let P(at2
, 2at) be a point on the parabola
y2
= 4ax and S be the focus of the parabola.
Then, SP = a + at2
[ focal distance = x + a]
Clearly, SP is least for t = 0.
Hence, the abscissa of P is at2
= a × 0 = 0
Ans. [D]
Ex.11 Tangents are drawn to the points of intersection
of the line 7y – 4x = 10 and parabola y2
= 4x,.then
the point of intersection of these tangents is
(A)
2
5
,
5
7
(B)
2
7
,
2
5
(C)
2
7
,
2
5
(D)
2
5
,
2
7
Sol. Here let the point be (x1
, y1
)
chord of contact yy1
= 2 (x + x1
)
compare with the given line
10
x
2 1
=
7
y1
=
4
2
x1
= 5/2, y1
= 7/2
Point reqd.
2
7
,
2
5
Ans. [B]
Ex.12 If the line x + my + am2
= 0 touches the parabola
y2
= 4ax, then the point of contact is
(A) (am2
, – 2am) (B)
m
a
2
,
m
a
2
(C) (–am2
, – 2am)
(D) The line does not touch.
Sol. Here x = –my – am2
16. Putting in y2
= 4ax, we get
y2
= 4a (–my – am2
)
or y2
+ 4amy + 4a2
m2
= 0
or (y + 2am)2
= 0
Which is a perfect square, hence the line is a
tangent y = –2am, also x = am2
the point reqd. is (am2
, –2am) Ans. [A]
Note:- For condition of tangency, solve the equation of
line and curve and put B2
–4AC = 0 for the
quadratic equation obtained by the two,
(the line and curve equation)
Ex.13 The common tangent of the parabola
y2
= 8ax and the circle x2
+ y2
= 2a2
is
(A) y = x + a (B) y = x – a
(C) y = x –2a (D) y = x + 2a
Sol. Any tangent to parabola is
y = mx +
m
a
2
Solving with the circle
x2
+ (mx +
m
a
2
)2
= 2a2
B2
– 4AC = 0 gives m = ± 1
Otherwise
Perp. from (0, 0) = radius a 2
2
m
1
m
a
2
= a 2 m = ± 1
Tangent y = ± x ± 2a
y = x + 2a is correct option.
Ans. [D]
Ex.14 If two parabolas y2
= 4x and x2
= 32y intersect in
the point (16, 8) at an angle , then is equals-
(A) tan–1
(3/5) (B)
(C) tan–1
(4/5) (D) /2
Sol. The slope of tangent at the point (16, 8) to the
parabola y2
= 4x is
2
1
a
y =
4
1
and slope of tangent
to the parabola x2
= 32y is 1.
tan = )
4
/
1
(
1
)
4
/
1
(
1
=
3
5
Ans. [A]
Ex.15 The slope of tangents drawn from a point
(4, 10) to the parabola y2
= 9x are-
(A)
4
3
,
4
1
(B)
4
9
,
4
1
(C)
3
1
,
4
1
(D) None of these
Sol. The equation of a tangent of slope m to the
parabola y2
= 9x is
y = mx +
m
4
9
If it passes through (4,10), then
10 = 4m +
m
4
9
16m2
– 40m + 9 = 0
(4m –1)(4m –9) = 0 m =
4
1
,
4
9
Ans. [B]
Ex.16 Tangents are drawn from the point (–2, –1) to the
parabola y2
= 4x. If is the angle between these
tangents then tan equals
(A) 3 (B) 2 (C) 1/3 (D) 1/2
Sol. Any tangent to y2
= 4x is
y = mx + 1/m
If it is drawn from (–2, –1), then
–1 = –2m + 1/m
2m2
– m – 1 = 0
If m = m1
, m2
then m1
+ m2
=1/2,
m1
m2
=–1/2
tan a =
2
1
2
1
m
m
1
m
m
=
2
1
2
1
2
2
1
m
m
1
m
m
4
)
m
m
(
2
/
1
1
2
4
/
1
= 3 Ans. [A]
Ex.17 If the straight line x + y = 1 is a normal to the
parabola x2
= ay, then the value of a is
(A) 4/3 (B) 1/2 (D) 3/4 (D) 1/4
Sol. We know that equation of normal to the parabola
x2
= ay is
y =mx +
4
a
2
+ 2
m
4
a
Given that x + y = 1 or y = –x + 1 is normal to
the parabola therefore
m = –1 and
2
a
+ 2
m
4
a
= 1
2
a
+
4
a
= 1
4
a
3
= 1 a =
3
4
Ans. [A]
17. Ex.18 If line y = 2x + k is normal to the parabola y2
=
4x at the point (t2
, 2t), then-
(A) k = –12, t = –2
(B) k = 12, t = –2
(C) k = 12, t = 2
(D) None of these
Sol. Since normal to the parabola y2
= 4x at (t2
, 2t) is y +
tx = 2t + t3
.
Comparing it with y = 2x + k, we get
t = –2, k = 2t + t3
= –12
Ans. [A]
Ex.19 Which of the following lines, is a normal to the
parabola y2
= 16x
(A) y = x – 11 cos – 3 cos 3
(B) y = x – 11 cos – cos 3
(C) y = (x – 11) cos + cos 3
(D) y = (x – 11) cos – cos 3
Sol. Here a = 4
condition of normality c = –2am – am3
(1) and (2) are not clearly the answer as
m = 1 for (3), (4) m = cos
c = –2(4) cos – 4 cos3
= – 8 cos – (3 cos + cos 3)
= –11 cos – cos 3
Hence (D) is correct
Ans. [D]
Ex.20 A normal is drawn to the parabola y2
= 4ax at the
point (2a, –2 2 a) then the length of the normal
chord, is
(A) 4 2 a (B) 6 2 a
(C) 4 3 a (D) 6 3 a
Sol. Here comparing (2a,–2 2 a) with
(am2
, –2am) we get m = 2
Now length of normal chord
= 2
m
a
4
(1 +m2
)3/2
=
2
a
4
(1 + 2)3/2
= 2a 3 3
= 6 3 a
Ans. [D]
Ex.21 If a normal chord of the parabola y2
=4ax subtend
a right angle at the vertex, its slope is-
(A) ± 1 (B) ± 2
(C) ± 3 (D) None of these
Sol. If P(at1
2
, 2at1
) be one end of the normal, the other
say Q(at2
2
, 2at2
)
then t2
= – t1
–
1
t
2
...(1)
Again slope of
OP = 2
1
1
at
at
2
=
1
t
2
Slope of OQ =
2
t
2
1
t
2
×
2
t
2
= –1
t1
t2
= –4 ...(2)
From (1) and (2)
–
1
t
4
= –t1
–
1
t
2
+
1
t
2
= t1
t1
2
= 2 t1
= ± 2 Ans. [B]
Note:- Slope of the normal at P is (–t1
)
Ex.22 If the tangents at P and Q on a parabola (whose
focus is S) meet in the point T, then SP, ST and
SQ are in-
(A) H.P. (B) G.P.
(C) A.P. (D) None of these
Sol. Let P (at1
2
, 2at1
) and Q (at2
2
, 2at2
) be any two
points on the parabola y2
= 4ax, then point of
intersection of tangents at P and Q will be
T [at1
t2
,a(t1
+ t2
)]
Now SP= a(t1
2
+ 1)
SQ = a(t2
2
+ 1)
ST =a )
1
t
)(
1
t
( 2
2
2
1
ST2
= SP.SQ
SP, ST and SQ are in G.P.
Ans. [B]
Ex.23 If the distance of 2 points P and Q from the
focus of a parabola y2
= 4ax are 4and 9
respectively, then the distance of the point of
intersection of tangents at P and Q from the
focus is
(A) 8 (B) 6 (C) 5 (D) 13
O
90°
Q
P
X
Y
18. Sol. If S is the focus of the parabola and T is the
point of intersection of tangents at P and Q, then
ST2
= SP × SQ ST2
= 4 × 9 ST = 6
Ans. [B]
Ex.24 From the point (–1, 2) tangents are drawn to
parabola y2
=4x, then the area of the triangle
formed by the chord of contact and the tangent
is-
(A) 2 2 (B) 3 2
(C) 4 2 (D) 8 2
Sol. Here chord of contact y(2) = 2(1) (x –1)
or y = x – 1
Length of the chord = 2
1
4
)
1
1
)(
1
1
(
1
= 8
Area required =
2
1
× 8 ×
(perpendicular from (–1, 2) to chord)
= 4 ×
1
1
1
1
2
= 8 2
Ans. [D]
Note:- Direct formula may be used.