SlideShare a Scribd company logo

General Quiz (1- ) 12th ABCD.pdf

Q.1 A circular arch having width 24m and height 9m is to be constructed. What is the radius of the circle of which the arch is an arc? (A) 10m (B*) 12.5 m (C) 13.5m (D) 14m Q.2 The sum of the solutions on the interval (0, 2] to the equation: 0 = – (2 sin2 x + 2 cos2 x)  , is (cos x  0 )  cos x  (A*) 3 (B) 4 (C) 5 (D) 6 Q.3 Let XOY be a right triangle with XOY = 90°. Let M and N be the midpoints of legs OX and OY respectively. If XN = 19 and YM = 22, then XY equals (A*) 26 (B) 13 (C) 32.5 (D) 41 [Sol. (XY)2 = 4(a2 + b2) now, 4a2 + b2 = (19)2 a2 + 4b2 = (22)2  a2 + b2 = 169  4(a2 + b2) = (XY)2 = 4 × 169  XY = 2 × 13 = 26 Ans.] Q.4 Given parallelogram ABCD, with AB = 10, AD = 12 and BD = 2 . The length of (AC) is (A) 10 (B) 2 (C) 12 (D*) 2 [Hint: 2(a2 + b2) = d2 + d2 ] Q.5 ABC is a right triangle with hypotenuse AB and AC = 15 cm. Altitude CH divides AB into segments AH and HB, with HB = 16cm. The area of ABC, is (A) 120 cm2 (B) 144cm2 (C*) 150cm2 (D) 216cm2 [Sol. Equating the two values of p 15 sin  = 16 cot   15 sin2 = 16 cos  solving cos  = 3/5 (cos  = – 5/3 rejected)  sin  = 4/5 2512 x = 15 cos  = 9; p = 15 sin  = 12; Area = 2 = 150cm2] Q.6 Consider the eight digit number N = 11115556. Which of the following statements are true? I. N is divisible by 11 II. N – 9 is a prime III N is a perfect square (A) I (B) II (C*) III (D) I, II and III [Hint: I SE = 12; So = 13  not divisible by 11 II N – 9 = 11115556 – 9 = 11115547 = 3331 × 3337 Note : N – 9 = (3334)2 – 32 = (3337) × (3331)  N – 9 can not be prime III N = 11115556 is a perfect square of 3334 ] Q.7 The value of the series 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... – 99 – 100, is (A) –100 (B) 0 (C) 1 (D*) 100 [Sol. 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... – 99 – 100 = (1 – 3) + (2 – 4) + (5 – 7) + (6 – 8) + ... = (–2) × 50 = – 100 ] One or more than one is/are correct. Q.110/mod If x = t3 + t + 5 & y = sin t then (3t 2 + 1) sin t + 6 t cos t d2y dx2 = (3t 2 + 1) sin t + 6 t cos t (A*)  (3t 2 + 1)3 (B) (3t 2 + 1)2 (C)  (3t 2 + 1) sin t + 6 t cos t 2 (D) cost 2 Q.212/modIf y = (3t 2 + 1) 1 then 2x2 + 3x + 1 d2y dx2 at x =  2 is : 3t + 1 (A*) 38 27 1 (B)  38 27 2 (C) 27 38 L(2x + 2)  (2x + 1) (D) none [Hint : y = (2x + 1)(x + 1) = (2x + 1) (2x + 2) = 2 MN(2x + 1) (2x + 2) ] Q.3 If y2 = P(x), is a polynomial of degree 3, then 2  d  dx  y . d2y 2  equals :  dx  (A) P  (x) + P  (x) (B) P  (x) . P  (x) (C*) P (x) . P  (x) (D) a constant d [Sol: 2 (y3 y ) = 2 (y3.y + 3 y2 y y ). Now differentiate y2 = P(x) thrice)] dx 2 3 1 2 Q.425/modIf f (x) = x  2 & g (x) = f ( f (x)) then for x > 20, g  (x) = (A*) 1 (B)  1 (C) 0 (D) none [Hint: for x > 20, f (x) = x – 2 ; g () = f (x) – 2 = x – 4 ] Q.537/modPeople living at Mars, instead of the usual definition of derivative D f(x), define a new kind of derivative, D*f(x) by the form

1 of 5
Download to read offline
Q.1 Acirculararchhaving width24m andheight 9m isto be constructed. Whatis the radius ofthecircleof
which the archis an arc?
(A)10m (B*) 12.5 m
(C) 13.5m (D)14m
Q.2 The sum ofthe solutions on theinterval (0, 2] to the equation:
0 = –   






 

x
cos
x
sin
x
sin
x
cos
2
x
sin
2
3
2
2 , is (cos x  0 )
(A*) 3 (B) 4 (C) 5 (D) 6
Q.3 Let XOY be a right triangle with XOY = 90°. Let M and N be the midpoints of legs OX and OY
respectively. If XN = 19 andYM = 22, then XYequals
(A*) 26 (B) 5
13 (C) 32.5 (D) 41
[Sol. (XY)2 = 4(a2 + b2)
now, 4a2 + b2 = (19)2
a2 + 4b2 = (22)2
 a2 + b2 = 169
 4(a2 + b2) = (XY)2 = 4 × 169
 XY = 2 × 13 = 26 Ans.]
Q.4 Given parallelogramABCD, withAB = 10,AD = 12 and BD = 13
2 .The length of (AC) is
(A) 10 (B) 13
2 (C) 3
12 (D*) 109
2
[Hint: 2(a2 + b2) = 2
1
d + 2
2
d ]
Q.5 ABC is arighttrianglewithhypotenuseABandAC =15cm.Altitude
CH dividesABinto segmentsAH and HB, with HB =16cm.The area
of ABC, is
(A) 120 cm2 (B) 144cm2 (C*) 150cm2 (D) 216cm2
[Sol. Equatingthe twovalues ofp
15 sin  = 16 cot   15 sin2 = 16 cos 
solving cos  = 3/5 (cos  = – 5/3 rejected)
 sin  = 4/5
x = 15 cos  = 9; p = 15 sin  = 12; Area =
2
12
25
= 150cm2]
Q.6 Consider the eightdigit numberN =11115556.Which of the followingstatements are true?
I. N isdivisible by11
II. N – 9 is a prime
III N is a perfect square
(A) I (B) II (C*) III (D) I, II and III
[Hint: I SE = 12; So = 13  not divisibleby11
II N – 9 = 11115556 – 9 = 11115547 = 3331 × 3337
Note : N – 9 = (3334)2 – 32 = (3337) × (3331)  N – 9 can not be prime
III N = 11115556 is a perfect square of 3334 ]
Q.7 The valueoftheseries
1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... – 99 – 100, is
(A) –100 (B) 0 (C) 1 (D*) 100
[Sol. 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... – 99 – 100
= (1 – 3) + (2 – 4) + (5 – 7) + (6 – 8) + ...
= (–2) × 50 = – 100 ]
One or more than one is/are correct.
Q.110/mod If x = t3 + t + 5 & y = sin t then
d y
dx
2
2 =
(A*) 
 
 
3 1 6
3 1
2
2
3
t t t t
t
 

sin cos
(B)
 
 
3 1 6
3 1
2
2
2
t t t t
t
 

sin cos
(C) 
 
 
3 1 6
3 1
2
2
2
t t t t
t
 

sin cos
(D)
cost
t
3 1
2

Q.212/modIf y =
1
2 3 1
2
x x
 
then
d y
dx
2
2 at x =  2 is :
(A*)
38
27
(B) 
38
27
(C)
27
38
(D) none
[Hint: y =
1
2 1 1
( )( )
x x
 
=
2
2 1 2 2
( )( )
x x
  = 2
( ) ( )
( )( )
2 2 2 1
2 1 2 2
x x
x x
  
 
L
N
M O
Q
P ]
Q.323/modIf y2 = P(x), is a polynomial of degree 3, then 2
d
dx





 y
d y
dx
3
2
2
.





 equals :
(A) P  (x) + P  (x) (B) P  (x) . P  (x) (C*) P (x) . P  (x) (D) a constant
[Sol: 2
d
dx
(y3 y2) = 2 (y3.y3 + 3 y2 y1y2). Now differentiate y2 = P(x) thrice)]
Q.425/modIf f (x) = x  2 & g (x) = f ( f (x)) then for x > 20, g  (x) =
(A*) 1 (B)  1 (C) 0 (D) none
[Hint: for x > 20, f (x) = x – 2 ; g () = f (x) – 2 = x – 4 ]
Q.537/modPeoplelivingatMars,insteadoftheusualdefinition ofderivativeDf(x),defineanewkindofderivative,
D*f(x)bytheformula
D*f(x) = Limit
h
f x h f x
h

 
0
2 2
( ) ( )
where f(x) means [f(x)]2. If f(x) = x lnx then
D f x x e
* ( )  has the value
(A) e (B) 2e (C*) 4e (D) none
[Hint: D*f(x) = 2f(x).f (x)
D*(x lnx) = 2x lnx (1 + lnx)]
Q.6503/mod If f(x) = x . x, then its derivative is :
(A) 2x (B) 2x (C*) 2x (D*) 2x sgn x
Q.1127/mod If y =
1
x
3
x
1
x
x
2
2
4




and
dx
dy
= ax + b then the value of a + b is equal to
(A) cot
8
5
(B*) cot
12
5
(C) tan
12
5
(D) tan
8
5
[Sol. y =
1
x
3
x
x
3
)
1
x
(
2
2
2
2




=
x
3
1
x
)
x
3
1
x
)(
x
3
1
x
(
2
2
2






dx
dy
= 2x – 3  a = 2 & b = – 3
a + b = 2 – 3 = tan
12

= cot
12
5
Ans. ]
Q.2 Valueofsin 705° is
(A)
4
2
6 
(B*)
4
6
2 
(C)
4
2
3 
(D)
2
3
1
Q.3 Findallrealfunctionsofonerealvariablethatsatisfythefollowingfunctionalequation,
f (x) + 2 f 





x
1
= x
[Sol. f (x) + 2 f 





x
1
= x ....(1)
Replacingxby1/x and multiplyby2 yields
2 f 





x
1
+ 4 f (x) =
x
2
....(2)
Andso,subtractingequation(1) from(2) gives
3 f (x) =
x
2
– x  f (x) = 





 x
x
2
3
1
]
Q.4 In the figure, each region T represents an equilateral triangle and each
region S a semicircle. The complete figure is a semicircle of radius 6
withitscentreO.Thethreesmallersemicirclestouchthelargesemicircle
at pointsA, B and C.What is the radius of a semicircle S?
[Sol. If we cut out part of the diagram and label appropriately
we find that DL = r and DO = 3
r .
This gives OB = r(1 + 3 ) = 6
or r =
3
1
6

= 3( 1
3  ) Ans. ]
Q.5 Solve 3
3
28
x
6
28
x
6 

 = 2. [Ans. x = 6]
[Sol. let a = 3
28
x
6  and b = 3
28
x
6 
a – b = 2
(a – b)3 = 8
a3 – b3 – 3ab(a – b) = 8 (using a – b = 2)
56 – 6 · 3
28
x
6  · 3
28
x
6  = 8  48 = 6 · 3
28
x
6  · 3
28
x
6 
 83 = (6x + 28)(6x – 28)  36x2 – 784 = 512  36x2 = 1296
 x2 = 36  x = ± 6 ]
Q.1130/mod If f (x) = x5 + 2x3 + 2x and g is the inverse of f then g'(–5) is equal to
(A*)
13
1
(B) 1 (C)
7
1
(D) none
[Sol. y = x5 + 2x3 + 2x
dx
dy
= 5x4 + 6x2 + 2 
dy
dx
=
2
x
6
x
5
1
2
4


= g ' (y)
when y = – 5, x5 + 2x3 + 2x + 5 = 0
(x + 1)(x4 – x3 + 3x2 – 3x + 5) = 0  x = – 1

1
x
5
y
dy
dx




=
2
6
5
1


=
13
1
]
Q.2126/mod Suppose the function f (x) – f (2x) has the derivative 5 at x = 1 and derivative 7 at x = 2. The
derivative of the function f (x) – f (4x) at x = 1, has the value equal to
(A*) 19 (B) 9 (C) 17 (D) 14
[Sol. y = f (x) – f (2x)
y' = f ' (x) – 2 f ' (2x)
y'(1) = f ' (1) – 2 f ' (2) = 5 ....(1)
and y'(2) = f ' (2) – 2 f ' (4) = 7 ....(2)
now let y = f (x) – f (4x)
y' = f ' (x) – 4 f ' (4x)
y ' (1) = f ' (1) – 4 f ' (4) ....(3)
substituting the value of f ' (2) = 7 + 2 f ' (4) in (1)
f ' (1) – 2 [7 + 2 f ' (4)] = 5
f ' (1) – 4 f ' (4) = 19  (A) ]
Q.3 Findthenumberoftriangles withsides of integerlength whoseperimeteris 10.
(A) 4 (B) 3 (C*) 2 (D) 0
[Sol. Let the sides are x, y, 10 – (x + y), x, y  N
Now x + y > 10 – (x + y)
x + y > 5 ....(1)
also 10 – x > x  0 < x < 5 ....(2)
|||ly 0 < y < 5 ....(3)
If x = 1, y can be 5, 6, 7..... not possible
x = 2, y = 4, z = 4; x = 3, y = 3, z = 4  (2, 4, 4) & (3, 3, 4) ]
Q.4 The value of the parameter m for whichthe equation
1
x
3
2
x


=
m
x
6
1
x
2


has no solution for x, is
(A) 12 (B) 18 (C) 14 (D*) 13
[Sol. x =
m
13
1
m
2


, for no solution m = 13 ]
Q.5 A circle of area 20 is centered at the point indicated bya solid dot in the
accompanying figure. Suppose that ABC is inscribed in that circle and
has area 8. The central angles ,  and  are as shown. What is the value
of sin  + sin  + sin ?
(A*)
5
4
(B)
4
3
(C)
3
2
(D)
2

[Hint:
2
1
r2(sin  + sin  + sin ) = 8 and r2 = 20 ]
Q.195/mod Which of the following could be the sketch graph of y=
d
dx
(x ln x)?
(A) (B) (C*) (D)
Q.278/mod If y = (A + Bx) emx + (m  1)2 ex then
d y
dx
2
2  2m
dy
dx
+ m2y is equal to :
(A*) ex (B) emx (C) emx (D) e(1  m) x
[Hint: multiplygivenequationby emx &then differentiatetwice]
Q.3 Number ofsides in aregularpolygonifthesumofits interioranglesis 2520°, is
(A) 14 (B*) 16 (C) 18 (D) 20
[Hint (n – 2) = 2520° external angle = n 




 


n
2
= 2520°
 n = 16 n – 2 = 2520°  (n – 2)180 = 2520  n – 2 = 14 n = 6]
Q.4 The value of x such that 4(1 + y)x2 – 4x + 1 – y = 0 is true for all real values of y, is
(A) – 1/2 (B) – 2/3 (C) 2/3 (D*) 1/2
[Sol. x =
)
y
1
(
8
)
y
1
(
4


or
)
y
1
(
8
)
y
1
(
4


; hence x =
2
1
Ans. ]
Q.5 The lengths of the sides of triangleABC are 60, 80 and 100
with A=90°.The lineAD divides triangleABC into two
triangles ofequal perimeter. Calculatethelength ofAD.
[Ans. 24 5 ]
[Sol. 60 + x = 100 – x + 80
2x = 120  x = 60
now usingcosinelaw inACD
y2 = (60)2 + x2 – 2x · 60 cos C (cosC = 3/5)
= 3600 + 3600 – 2 · 60 · 60 ·
5
3
= 7200 – 24 · 60 · 3 = 2880
y = 2880 = 24 5 Ans.
Alternatively: sin
2
C
= 60
·
2
y
=
120
y
y = 120 sin
2
C
....(1)
now cos C =
5
3
1 – 2 sin2
2
C
=
5
3
 2 sin2
2
C
=
5
2
 sin
2
C
=
5
1
.
Hence from (1), y =
5
120
= 5
24 Ans. ]

Recommended

Method of differentiation.pdf
Method of differentiation.pdfMethod of differentiation.pdf
Method of differentiation.pdfSTUDY INNOVATIONS
 
MATHS- (12th & 13th) Paper-1.pdf
MATHS- (12th & 13th) Paper-1.pdfMATHS- (12th & 13th) Paper-1.pdf
MATHS- (12th & 13th) Paper-1.pdfSTUDY INNOVATIONS
 
Application of Derivative.pdf
Application of Derivative.pdfApplication of Derivative.pdf
Application of Derivative.pdfSTUDY INNOVATIONS
 
Definite & Indefinite Integration Q.B..pdf
Definite & Indefinite Integration Q.B..pdfDefinite & Indefinite Integration Q.B..pdf
Definite & Indefinite Integration Q.B..pdfSTUDY INNOVATIONS
 
General Quiz WA (1- ) 12th ABCD.pdf
General Quiz WA (1- ) 12th ABCD.pdfGeneral Quiz WA (1- ) 12th ABCD.pdf
General Quiz WA (1- ) 12th ABCD.pdfSTUDY INNOVATIONS
 
Probability Quiz Only Jpr WA.pdf
Probability Quiz Only Jpr WA.pdfProbability Quiz Only Jpr WA.pdf
Probability Quiz Only Jpr WA.pdfSTUDY INNOVATIONS
 

More Related Content

Similar to General Quiz (1- ) 12th ABCD.pdf

Probability Quiz Only Jpr WA_watermark.pdf
Probability Quiz Only Jpr WA_watermark.pdfProbability Quiz Only Jpr WA_watermark.pdf
Probability Quiz Only Jpr WA_watermark.pdfSTUDY INNOVATIONS
 
Dpp (31-35) 11th J-Batch Maths.pdf
Dpp (31-35) 11th J-Batch Maths.pdfDpp (31-35) 11th J-Batch Maths.pdf
Dpp (31-35) 11th J-Batch Maths.pdfSTUDY INNOVATIONS
 
Definite & Indefinite Integration.pdf
Definite & Indefinite Integration.pdfDefinite & Indefinite Integration.pdf
Definite & Indefinite Integration.pdfSTUDY INNOVATIONS
 
Dpp (25-30) 11th J-Batch Maths.pdf
Dpp (25-30) 11th J-Batch Maths.pdfDpp (25-30) 11th J-Batch Maths.pdf
Dpp (25-30) 11th J-Batch Maths.pdfSTUDY INNOVATIONS
 
DIFFERENTIAL EQUATION & AREA UNDER CURVE
DIFFERENTIAL EQUATION & AREA UNDER CURVEDIFFERENTIAL EQUATION & AREA UNDER CURVE
DIFFERENTIAL EQUATION & AREA UNDER CURVESTUDY INNOVATIONS
 
Conic Section (Para_Ellipse_Hyperbola).pdf
Conic Section (Para_Ellipse_Hyperbola).pdfConic Section (Para_Ellipse_Hyperbola).pdf
Conic Section (Para_Ellipse_Hyperbola).pdfSTUDY INNOVATIONS
 
IPT-1-MATHS-SOLUTION XII& XIIIpdf
IPT-1-MATHS-SOLUTION XII& XIIIpdfIPT-1-MATHS-SOLUTION XII& XIIIpdf
IPT-1-MATHS-SOLUTION XII& XIIIpdfSTUDY INNOVATIONS
 
Dpp (13-15) 11th J-Batch Maths.pdf
Dpp (13-15) 11th J-Batch Maths.pdfDpp (13-15) 11th J-Batch Maths.pdf
Dpp (13-15) 11th J-Batch Maths.pdfSTUDY INNOVATIONS
 
MATHS- 11th (PQRS & J) Partial Marking.pdf
MATHS- 11th (PQRS & J) Partial Marking.pdfMATHS- 11th (PQRS & J) Partial Marking.pdf
MATHS- 11th (PQRS & J) Partial Marking.pdfSTUDY INNOVATIONS
 
Dpp (10-12) 11th J-Batch Maths.pdf
Dpp (10-12) 11th J-Batch Maths.pdfDpp (10-12) 11th J-Batch Maths.pdf
Dpp (10-12) 11th J-Batch Maths.pdfSTUDY INNOVATIONS
 
Definite integration_watermark.pdf
Definite integration_watermark.pdfDefinite integration_watermark.pdf
Definite integration_watermark.pdfSTUDY INNOVATIONS
 
MATHS 13th Paper-1 TEST-3.pdf
MATHS  13th Paper-1 TEST-3.pdfMATHS  13th Paper-1 TEST-3.pdf
MATHS 13th Paper-1 TEST-3.pdfSTUDY INNOVATIONS
 

Similar to General Quiz (1- ) 12th ABCD.pdf (20)

Probability Quiz Only Jpr WA_watermark.pdf
Probability Quiz Only Jpr WA_watermark.pdfProbability Quiz Only Jpr WA_watermark.pdf
Probability Quiz Only Jpr WA_watermark.pdf
 
Dpp (31-35) 11th J-Batch Maths.pdf
Dpp (31-35) 11th J-Batch Maths.pdfDpp (31-35) 11th J-Batch Maths.pdf
Dpp (31-35) 11th J-Batch Maths.pdf
 
Definite & Indefinite Integration.pdf
Definite & Indefinite Integration.pdfDefinite & Indefinite Integration.pdf
Definite & Indefinite Integration.pdf
 
Dpp (25-30) 11th J-Batch Maths.pdf
Dpp (25-30) 11th J-Batch Maths.pdfDpp (25-30) 11th J-Batch Maths.pdf
Dpp (25-30) 11th J-Batch Maths.pdf
 
DIFFERENTIAL EQUATION & AREA UNDER CURVE
DIFFERENTIAL EQUATION & AREA UNDER CURVEDIFFERENTIAL EQUATION & AREA UNDER CURVE
DIFFERENTIAL EQUATION & AREA UNDER CURVE
 
Conic Section (Para_Ellipse_Hyperbola).pdf
Conic Section (Para_Ellipse_Hyperbola).pdfConic Section (Para_Ellipse_Hyperbola).pdf
Conic Section (Para_Ellipse_Hyperbola).pdf
 
DPP - 34-36-Answer
DPP - 34-36-AnswerDPP - 34-36-Answer
DPP - 34-36-Answer
 
IPT-1-MATHS-SOLUTION XII& XIIIpdf
IPT-1-MATHS-SOLUTION XII& XIIIpdfIPT-1-MATHS-SOLUTION XII& XIIIpdf
IPT-1-MATHS-SOLUTION XII& XIIIpdf
 
DPP-42-44-Answer
DPP-42-44-AnswerDPP-42-44-Answer
DPP-42-44-Answer
 
Dpp (13-15) 11th J-Batch Maths.pdf
Dpp (13-15) 11th J-Batch Maths.pdfDpp (13-15) 11th J-Batch Maths.pdf
Dpp (13-15) 11th J-Batch Maths.pdf
 
MATHS- 11th (PQRS & J) Partial Marking.pdf
MATHS- 11th (PQRS & J) Partial Marking.pdfMATHS- 11th (PQRS & J) Partial Marking.pdf
MATHS- 11th (PQRS & J) Partial Marking.pdf
 
Dpp 12th Maths WA.pdf
Dpp 12th Maths WA.pdfDpp 12th Maths WA.pdf
Dpp 12th Maths WA.pdf
 
Dpp (10-12) 11th J-Batch Maths.pdf
Dpp (10-12) 11th J-Batch Maths.pdfDpp (10-12) 11th J-Batch Maths.pdf
Dpp (10-12) 11th J-Batch Maths.pdf
 
Definite integration_watermark.pdf
Definite integration_watermark.pdfDefinite integration_watermark.pdf
Definite integration_watermark.pdf
 
Definite integration
Definite integrationDefinite integration
Definite integration
 
Ph-1,2,3 & Binomial(F).pdf
Ph-1,2,3 & Binomial(F).pdfPh-1,2,3 & Binomial(F).pdf
Ph-1,2,3 & Binomial(F).pdf
 
FLCD-MOD-ITF.pdf
FLCD-MOD-ITF.pdfFLCD-MOD-ITF.pdf
FLCD-MOD-ITF.pdf
 
FLCD-MOD-ITF.pdf
FLCD-MOD-ITF.pdfFLCD-MOD-ITF.pdf
FLCD-MOD-ITF.pdf
 
MATHS 13th Paper-1 TEST-3.pdf
MATHS  13th Paper-1 TEST-3.pdfMATHS  13th Paper-1 TEST-3.pdf
MATHS 13th Paper-1 TEST-3.pdf
 
DPP-52-54-Answer
DPP-52-54-AnswerDPP-52-54-Answer
DPP-52-54-Answer
 

More from STUDY INNOVATIONS

CHALCOGENS & XENON FLUORIDE_watermark.pptx
CHALCOGENS & XENON FLUORIDE_watermark.pptxCHALCOGENS & XENON FLUORIDE_watermark.pptx
CHALCOGENS & XENON FLUORIDE_watermark.pptxSTUDY INNOVATIONS
 
Physics_dpp2_Rotational Motion_watermark.pdf
Physics_dpp2_Rotational Motion_watermark.pdfPhysics_dpp2_Rotational Motion_watermark.pdf
Physics_dpp2_Rotational Motion_watermark.pdfSTUDY INNOVATIONS
 
Physics_dpp2_Properties of Matter_watermark.pdf
Physics_dpp2_Properties of Matter_watermark.pdfPhysics_dpp2_Properties of Matter_watermark.pdf
Physics_dpp2_Properties of Matter_watermark.pdfSTUDY INNOVATIONS
 
Physics_dpp4_Law of Motion_watermark.pdf
Physics_dpp4_Law of Motion_watermark.pdfPhysics_dpp4_Law of Motion_watermark.pdf
Physics_dpp4_Law of Motion_watermark.pdfSTUDY INNOVATIONS
 
Physics_dpp3_law of motion_watermark.pdf
Physics_dpp3_law of motion_watermark.pdfPhysics_dpp3_law of motion_watermark.pdf
Physics_dpp3_law of motion_watermark.pdfSTUDY INNOVATIONS
 
04-CIRCLE SYSTEM (ASSIGNMENT) PART-I_watermark.pdf
04-CIRCLE SYSTEM (ASSIGNMENT) PART-I_watermark.pdf04-CIRCLE SYSTEM (ASSIGNMENT) PART-I_watermark.pdf
04-CIRCLE SYSTEM (ASSIGNMENT) PART-I_watermark.pdfSTUDY INNOVATIONS
 
02-DETERMINANT-(E)- THEORY_watermark.pdf
02-DETERMINANT-(E)- THEORY_watermark.pdf02-DETERMINANT-(E)- THEORY_watermark.pdf
02-DETERMINANT-(E)- THEORY_watermark.pdfSTUDY INNOVATIONS
 
02-QUADRATIC EQUATION-(E)-THEORY_watermark.pdf
02-QUADRATIC EQUATION-(E)-THEORY_watermark.pdf02-QUADRATIC EQUATION-(E)-THEORY_watermark.pdf
02-QUADRATIC EQUATION-(E)-THEORY_watermark.pdfSTUDY INNOVATIONS
 
DIFFERENTIAL EQUATION-3 maths_watermark.pdf
DIFFERENTIAL EQUATION-3 maths_watermark.pdfDIFFERENTIAL EQUATION-3 maths_watermark.pdf
DIFFERENTIAL EQUATION-3 maths_watermark.pdfSTUDY INNOVATIONS
 
INDEFINITE-INTEGRAL-1 maths_watermark.pdf
INDEFINITE-INTEGRAL-1 maths_watermark.pdfINDEFINITE-INTEGRAL-1 maths_watermark.pdf
INDEFINITE-INTEGRAL-1 maths_watermark.pdfSTUDY INNOVATIONS
 
Physics -27 -08 -06 _ watermark .pdf.
Physics  -27 -08 -06  _ watermark  .pdf.Physics  -27 -08 -06  _ watermark  .pdf.
Physics -27 -08 -06 _ watermark .pdf.STUDY INNOVATIONS
 
Physics -27 -08- 06 _ watermark. pdf.
Physics   -27 -08- 06 _ watermark.  pdf.Physics   -27 -08- 06 _ watermark.  pdf.
Physics -27 -08- 06 _ watermark. pdf.STUDY INNOVATIONS
 
P-24-09-2006 13th Paper-2_watermark.pdf
P-24-09-2006 13th  Paper-2_watermark.pdfP-24-09-2006 13th  Paper-2_watermark.pdf
P-24-09-2006 13th Paper-2_watermark.pdfSTUDY INNOVATIONS
 
Modern Physics physics _watermark. pdf
Modern  Physics  physics _watermark. pdfModern  Physics  physics _watermark. pdf
Modern Physics physics _watermark. pdfSTUDY INNOVATIONS
 
Particle dynamics physics _watermark.pdf
Particle dynamics physics _watermark.pdfParticle dynamics physics _watermark.pdf
Particle dynamics physics _watermark.pdfSTUDY INNOVATIONS
 
Mechanical Waves physics _watermark.pdf
Mechanical Waves physics  _watermark.pdfMechanical Waves physics  _watermark.pdf
Mechanical Waves physics _watermark.pdfSTUDY INNOVATIONS
 
Geometrical Optics physics _watermark.pdf
Geometrical Optics physics _watermark.pdfGeometrical Optics physics _watermark.pdf
Geometrical Optics physics _watermark.pdfSTUDY INNOVATIONS
 
Fluid mechanics physics _watermark.pdf
Fluid mechanics physics   _watermark.pdfFluid mechanics physics   _watermark.pdf
Fluid mechanics physics _watermark.pdfSTUDY INNOVATIONS
 
Calorimetry & Heat Transfer_watermark.pdf
Calorimetry & Heat Transfer_watermark.pdfCalorimetry & Heat Transfer_watermark.pdf
Calorimetry & Heat Transfer_watermark.pdfSTUDY INNOVATIONS
 
Modern Physics _watermark .pdf physics
Modern Physics _watermark  .pdf  physicsModern Physics _watermark  .pdf  physics
Modern Physics _watermark .pdf physicsSTUDY INNOVATIONS
 

More from STUDY INNOVATIONS (20)

CHALCOGENS & XENON FLUORIDE_watermark.pptx
CHALCOGENS & XENON FLUORIDE_watermark.pptxCHALCOGENS & XENON FLUORIDE_watermark.pptx
CHALCOGENS & XENON FLUORIDE_watermark.pptx
 
Physics_dpp2_Rotational Motion_watermark.pdf
Physics_dpp2_Rotational Motion_watermark.pdfPhysics_dpp2_Rotational Motion_watermark.pdf
Physics_dpp2_Rotational Motion_watermark.pdf
 
Physics_dpp2_Properties of Matter_watermark.pdf
Physics_dpp2_Properties of Matter_watermark.pdfPhysics_dpp2_Properties of Matter_watermark.pdf
Physics_dpp2_Properties of Matter_watermark.pdf
 
Physics_dpp4_Law of Motion_watermark.pdf
Physics_dpp4_Law of Motion_watermark.pdfPhysics_dpp4_Law of Motion_watermark.pdf
Physics_dpp4_Law of Motion_watermark.pdf
 
Physics_dpp3_law of motion_watermark.pdf
Physics_dpp3_law of motion_watermark.pdfPhysics_dpp3_law of motion_watermark.pdf
Physics_dpp3_law of motion_watermark.pdf
 
04-CIRCLE SYSTEM (ASSIGNMENT) PART-I_watermark.pdf
04-CIRCLE SYSTEM (ASSIGNMENT) PART-I_watermark.pdf04-CIRCLE SYSTEM (ASSIGNMENT) PART-I_watermark.pdf
04-CIRCLE SYSTEM (ASSIGNMENT) PART-I_watermark.pdf
 
02-DETERMINANT-(E)- THEORY_watermark.pdf
02-DETERMINANT-(E)- THEORY_watermark.pdf02-DETERMINANT-(E)- THEORY_watermark.pdf
02-DETERMINANT-(E)- THEORY_watermark.pdf
 
02-QUADRATIC EQUATION-(E)-THEORY_watermark.pdf
02-QUADRATIC EQUATION-(E)-THEORY_watermark.pdf02-QUADRATIC EQUATION-(E)-THEORY_watermark.pdf
02-QUADRATIC EQUATION-(E)-THEORY_watermark.pdf
 
DIFFERENTIAL EQUATION-3 maths_watermark.pdf
DIFFERENTIAL EQUATION-3 maths_watermark.pdfDIFFERENTIAL EQUATION-3 maths_watermark.pdf
DIFFERENTIAL EQUATION-3 maths_watermark.pdf
 
INDEFINITE-INTEGRAL-1 maths_watermark.pdf
INDEFINITE-INTEGRAL-1 maths_watermark.pdfINDEFINITE-INTEGRAL-1 maths_watermark.pdf
INDEFINITE-INTEGRAL-1 maths_watermark.pdf
 
Physics -27 -08 -06 _ watermark .pdf.
Physics  -27 -08 -06  _ watermark  .pdf.Physics  -27 -08 -06  _ watermark  .pdf.
Physics -27 -08 -06 _ watermark .pdf.
 
Physics -27 -08- 06 _ watermark. pdf.
Physics   -27 -08- 06 _ watermark.  pdf.Physics   -27 -08- 06 _ watermark.  pdf.
Physics -27 -08- 06 _ watermark. pdf.
 
P-24-09-2006 13th Paper-2_watermark.pdf
P-24-09-2006 13th  Paper-2_watermark.pdfP-24-09-2006 13th  Paper-2_watermark.pdf
P-24-09-2006 13th Paper-2_watermark.pdf
 
Modern Physics physics _watermark. pdf
Modern  Physics  physics _watermark. pdfModern  Physics  physics _watermark. pdf
Modern Physics physics _watermark. pdf
 
Particle dynamics physics _watermark.pdf
Particle dynamics physics _watermark.pdfParticle dynamics physics _watermark.pdf
Particle dynamics physics _watermark.pdf
 
Mechanical Waves physics _watermark.pdf
Mechanical Waves physics  _watermark.pdfMechanical Waves physics  _watermark.pdf
Mechanical Waves physics _watermark.pdf
 
Geometrical Optics physics _watermark.pdf
Geometrical Optics physics _watermark.pdfGeometrical Optics physics _watermark.pdf
Geometrical Optics physics _watermark.pdf
 
Fluid mechanics physics _watermark.pdf
Fluid mechanics physics   _watermark.pdfFluid mechanics physics   _watermark.pdf
Fluid mechanics physics _watermark.pdf
 
Calorimetry & Heat Transfer_watermark.pdf
Calorimetry & Heat Transfer_watermark.pdfCalorimetry & Heat Transfer_watermark.pdf
Calorimetry & Heat Transfer_watermark.pdf
 
Modern Physics _watermark .pdf physics
Modern Physics _watermark  .pdf  physicsModern Physics _watermark  .pdf  physics
Modern Physics _watermark .pdf physics
 

Recently uploaded

BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (GLOB...
BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (GLOB...BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (GLOB...
BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (GLOB...Nguyen Thanh Tu Collection
 
A TEXTBOOK OF INTELLECTUAL ROPERTY RIGHTS
A TEXTBOOK OF INTELLECTUAL ROPERTY RIGHTSA TEXTBOOK OF INTELLECTUAL ROPERTY RIGHTS
A TEXTBOOK OF INTELLECTUAL ROPERTY RIGHTSDr.M.Geethavani
 
Managing Choice, Coherence and Specialisation in Upper Secondary Education - ...
Managing Choice, Coherence and Specialisation in Upper Secondary Education - ...Managing Choice, Coherence and Specialisation in Upper Secondary Education - ...
Managing Choice, Coherence and Specialisation in Upper Secondary Education - ...EduSkills OECD
 
Database Security Methods, DAC, MAC,View
Database Security Methods, DAC, MAC,ViewDatabase Security Methods, DAC, MAC,View
Database Security Methods, DAC, MAC,ViewDr-Dipali Meher
 
2024-02-24_Session 1 - PMLE_UPDATED.pptx
2024-02-24_Session 1 - PMLE_UPDATED.pptx2024-02-24_Session 1 - PMLE_UPDATED.pptx
2024-02-24_Session 1 - PMLE_UPDATED.pptxgdgsurrey
 
A Free eBook ~ Mental Exercise ...Puzzles to Analyze.pdf
A Free eBook ~ Mental Exercise ...Puzzles to Analyze.pdfA Free eBook ~ Mental Exercise ...Puzzles to Analyze.pdf
A Free eBook ~ Mental Exercise ...Puzzles to Analyze.pdfOH TEIK BIN
 
Healthy Habits for Happy School Staff - presentation
Healthy Habits for Happy School Staff - presentationHealthy Habits for Happy School Staff - presentation
Healthy Habits for Happy School Staff - presentationPooky Knightsmith
 
A LABORATORY MANUAL FOR ORGANIC CHEMISTRY.pdf
A LABORATORY MANUAL FOR ORGANIC CHEMISTRY.pdfA LABORATORY MANUAL FOR ORGANIC CHEMISTRY.pdf
A LABORATORY MANUAL FOR ORGANIC CHEMISTRY.pdfDr.M.Geethavani
 
EDL 290F Week 2 - Good Company (2024).pdf
EDL 290F Week 2  - Good Company (2024).pdfEDL 290F Week 2  - Good Company (2024).pdf
EDL 290F Week 2 - Good Company (2024).pdfElizabeth Walsh
 
Peninsula Channel Commanders Rules Handbook
Peninsula Channel Commanders Rules HandbookPeninsula Channel Commanders Rules Handbook
Peninsula Channel Commanders Rules Handbookpccwebmasterhmb
 
The basics of sentences session 6pptx.pptx
The basics of sentences session 6pptx.pptxThe basics of sentences session 6pptx.pptx
The basics of sentences session 6pptx.pptxdeputymitchell2
 
Successful projects and failed programmes – the cost of not designing the who...
Successful projects and failed programmes – the cost of not designing the who...Successful projects and failed programmes – the cost of not designing the who...
Successful projects and failed programmes – the cost of not designing the who...Association for Project Management
 
Discussing the new Competence Framework for project managers in the built env...
Discussing the new Competence Framework for project managers in the built env...Discussing the new Competence Framework for project managers in the built env...
Discussing the new Competence Framework for project managers in the built env...Association for Project Management
 
How To Create Record Rules in the Odoo 17
How To Create Record Rules in the Odoo 17How To Create Record Rules in the Odoo 17
How To Create Record Rules in the Odoo 17Celine George
 
Can Brain Science Actually Help Make Your Training & Teaching "Stick"?
Can Brain Science Actually Help Make Your Training & Teaching "Stick"?Can Brain Science Actually Help Make Your Training & Teaching "Stick"?
Can Brain Science Actually Help Make Your Training & Teaching "Stick"?Aggregage
 
Dr.M.Florence Dayana-Cloud Computing-Unit - 1.pdf
Dr.M.Florence Dayana-Cloud Computing-Unit - 1.pdfDr.M.Florence Dayana-Cloud Computing-Unit - 1.pdf
Dr.M.Florence Dayana-Cloud Computing-Unit - 1.pdfDr.Florence Dayana
 
EmpTech Lesson 7 - Online Creation Tools, Platforms, and Applications for ICT...
EmpTech Lesson 7 - Online Creation Tools, Platforms, and Applications for ICT...EmpTech Lesson 7 - Online Creation Tools, Platforms, and Applications for ICT...
EmpTech Lesson 7 - Online Creation Tools, Platforms, and Applications for ICT...liera silvan
 
Mycobacteriology update 2024 Margie Morgan.ppt
Mycobacteriology update 2024 Margie Morgan.pptMycobacteriology update 2024 Margie Morgan.ppt
Mycobacteriology update 2024 Margie Morgan.pptMargie Morgan
 
Andreas Schleicher_ Strengthening Upper Secondary Education in Lithuania
Andreas Schleicher_ Strengthening Upper Secondary  Education in LithuaniaAndreas Schleicher_ Strengthening Upper Secondary  Education in Lithuania
Andreas Schleicher_ Strengthening Upper Secondary Education in LithuaniaEduSkills OECD
 

Recently uploaded (20)

BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (GLOB...
BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (GLOB...BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (GLOB...
BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (GLOB...
 
A TEXTBOOK OF INTELLECTUAL ROPERTY RIGHTS
A TEXTBOOK OF INTELLECTUAL ROPERTY RIGHTSA TEXTBOOK OF INTELLECTUAL ROPERTY RIGHTS
A TEXTBOOK OF INTELLECTUAL ROPERTY RIGHTS
 
Managing Choice, Coherence and Specialisation in Upper Secondary Education - ...
Managing Choice, Coherence and Specialisation in Upper Secondary Education - ...Managing Choice, Coherence and Specialisation in Upper Secondary Education - ...
Managing Choice, Coherence and Specialisation in Upper Secondary Education - ...
 
CLUBE PERLINGUAS .
CLUBE PERLINGUAS                        .CLUBE PERLINGUAS                        .
CLUBE PERLINGUAS .
 
Database Security Methods, DAC, MAC,View
Database Security Methods, DAC, MAC,ViewDatabase Security Methods, DAC, MAC,View
Database Security Methods, DAC, MAC,View
 
2024-02-24_Session 1 - PMLE_UPDATED.pptx
2024-02-24_Session 1 - PMLE_UPDATED.pptx2024-02-24_Session 1 - PMLE_UPDATED.pptx
2024-02-24_Session 1 - PMLE_UPDATED.pptx
 
A Free eBook ~ Mental Exercise ...Puzzles to Analyze.pdf
A Free eBook ~ Mental Exercise ...Puzzles to Analyze.pdfA Free eBook ~ Mental Exercise ...Puzzles to Analyze.pdf
A Free eBook ~ Mental Exercise ...Puzzles to Analyze.pdf
 
Healthy Habits for Happy School Staff - presentation
Healthy Habits for Happy School Staff - presentationHealthy Habits for Happy School Staff - presentation
Healthy Habits for Happy School Staff - presentation
 
A LABORATORY MANUAL FOR ORGANIC CHEMISTRY.pdf
A LABORATORY MANUAL FOR ORGANIC CHEMISTRY.pdfA LABORATORY MANUAL FOR ORGANIC CHEMISTRY.pdf
A LABORATORY MANUAL FOR ORGANIC CHEMISTRY.pdf
 
EDL 290F Week 2 - Good Company (2024).pdf
EDL 290F Week 2  - Good Company (2024).pdfEDL 290F Week 2  - Good Company (2024).pdf
EDL 290F Week 2 - Good Company (2024).pdf
 
Peninsula Channel Commanders Rules Handbook
Peninsula Channel Commanders Rules HandbookPeninsula Channel Commanders Rules Handbook
Peninsula Channel Commanders Rules Handbook
 
The basics of sentences session 6pptx.pptx
The basics of sentences session 6pptx.pptxThe basics of sentences session 6pptx.pptx
The basics of sentences session 6pptx.pptx
 
Successful projects and failed programmes – the cost of not designing the who...
Successful projects and failed programmes – the cost of not designing the who...Successful projects and failed programmes – the cost of not designing the who...
Successful projects and failed programmes – the cost of not designing the who...
 
Discussing the new Competence Framework for project managers in the built env...
Discussing the new Competence Framework for project managers in the built env...Discussing the new Competence Framework for project managers in the built env...
Discussing the new Competence Framework for project managers in the built env...
 
How To Create Record Rules in the Odoo 17
How To Create Record Rules in the Odoo 17How To Create Record Rules in the Odoo 17
How To Create Record Rules in the Odoo 17
 
Can Brain Science Actually Help Make Your Training & Teaching "Stick"?
Can Brain Science Actually Help Make Your Training & Teaching "Stick"?Can Brain Science Actually Help Make Your Training & Teaching "Stick"?
Can Brain Science Actually Help Make Your Training & Teaching "Stick"?
 
Dr.M.Florence Dayana-Cloud Computing-Unit - 1.pdf
Dr.M.Florence Dayana-Cloud Computing-Unit - 1.pdfDr.M.Florence Dayana-Cloud Computing-Unit - 1.pdf
Dr.M.Florence Dayana-Cloud Computing-Unit - 1.pdf
 
EmpTech Lesson 7 - Online Creation Tools, Platforms, and Applications for ICT...
EmpTech Lesson 7 - Online Creation Tools, Platforms, and Applications for ICT...EmpTech Lesson 7 - Online Creation Tools, Platforms, and Applications for ICT...
EmpTech Lesson 7 - Online Creation Tools, Platforms, and Applications for ICT...
 
Mycobacteriology update 2024 Margie Morgan.ppt
Mycobacteriology update 2024 Margie Morgan.pptMycobacteriology update 2024 Margie Morgan.ppt
Mycobacteriology update 2024 Margie Morgan.ppt
 
Andreas Schleicher_ Strengthening Upper Secondary Education in Lithuania
Andreas Schleicher_ Strengthening Upper Secondary  Education in LithuaniaAndreas Schleicher_ Strengthening Upper Secondary  Education in Lithuania
Andreas Schleicher_ Strengthening Upper Secondary Education in Lithuania
 

General Quiz (1- ) 12th ABCD.pdf

  • 1. Q.1 Acirculararchhaving width24m andheight 9m isto be constructed. Whatis the radius ofthecircleof which the archis an arc? (A)10m (B*) 12.5 m (C) 13.5m (D)14m Q.2 The sum ofthe solutions on theinterval (0, 2] to the equation: 0 = –             x cos x sin x sin x cos 2 x sin 2 3 2 2 , is (cos x  0 ) (A*) 3 (B) 4 (C) 5 (D) 6 Q.3 Let XOY be a right triangle with XOY = 90°. Let M and N be the midpoints of legs OX and OY respectively. If XN = 19 andYM = 22, then XYequals (A*) 26 (B) 5 13 (C) 32.5 (D) 41 [Sol. (XY)2 = 4(a2 + b2) now, 4a2 + b2 = (19)2 a2 + 4b2 = (22)2  a2 + b2 = 169  4(a2 + b2) = (XY)2 = 4 × 169  XY = 2 × 13 = 26 Ans.] Q.4 Given parallelogramABCD, withAB = 10,AD = 12 and BD = 13 2 .The length of (AC) is (A) 10 (B) 13 2 (C) 3 12 (D*) 109 2 [Hint: 2(a2 + b2) = 2 1 d + 2 2 d ] Q.5 ABC is arighttrianglewithhypotenuseABandAC =15cm.Altitude CH dividesABinto segmentsAH and HB, with HB =16cm.The area of ABC, is (A) 120 cm2 (B) 144cm2 (C*) 150cm2 (D) 216cm2 [Sol. Equatingthe twovalues ofp 15 sin  = 16 cot   15 sin2 = 16 cos  solving cos  = 3/5 (cos  = – 5/3 rejected)  sin  = 4/5 x = 15 cos  = 9; p = 15 sin  = 12; Area = 2 12 25 = 150cm2] Q.6 Consider the eightdigit numberN =11115556.Which of the followingstatements are true? I. N isdivisible by11 II. N – 9 is a prime III N is a perfect square (A) I (B) II (C*) III (D) I, II and III [Hint: I SE = 12; So = 13  not divisibleby11 II N – 9 = 11115556 – 9 = 11115547 = 3331 × 3337 Note : N – 9 = (3334)2 – 32 = (3337) × (3331)  N – 9 can not be prime III N = 11115556 is a perfect square of 3334 ] Q.7 The valueoftheseries 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... – 99 – 100, is (A) –100 (B) 0 (C) 1 (D*) 100 [Sol. 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... – 99 – 100 = (1 – 3) + (2 – 4) + (5 – 7) + (6 – 8) + ... = (–2) × 50 = – 100 ]
  • 2. One or more than one is/are correct. Q.110/mod If x = t3 + t + 5 & y = sin t then d y dx 2 2 = (A*)      3 1 6 3 1 2 2 3 t t t t t    sin cos (B)     3 1 6 3 1 2 2 2 t t t t t    sin cos (C)      3 1 6 3 1 2 2 2 t t t t t    sin cos (D) cost t 3 1 2  Q.212/modIf y = 1 2 3 1 2 x x   then d y dx 2 2 at x =  2 is : (A*) 38 27 (B)  38 27 (C) 27 38 (D) none [Hint: y = 1 2 1 1 ( )( ) x x   = 2 2 1 2 2 ( )( ) x x   = 2 ( ) ( ) ( )( ) 2 2 2 1 2 1 2 2 x x x x      L N M O Q P ] Q.323/modIf y2 = P(x), is a polynomial of degree 3, then 2 d dx       y d y dx 3 2 2 .       equals : (A) P  (x) + P  (x) (B) P  (x) . P  (x) (C*) P (x) . P  (x) (D) a constant [Sol: 2 d dx (y3 y2) = 2 (y3.y3 + 3 y2 y1y2). Now differentiate y2 = P(x) thrice)] Q.425/modIf f (x) = x  2 & g (x) = f ( f (x)) then for x > 20, g  (x) = (A*) 1 (B)  1 (C) 0 (D) none [Hint: for x > 20, f (x) = x – 2 ; g () = f (x) – 2 = x – 4 ] Q.537/modPeoplelivingatMars,insteadoftheusualdefinition ofderivativeDf(x),defineanewkindofderivative, D*f(x)bytheformula D*f(x) = Limit h f x h f x h    0 2 2 ( ) ( ) where f(x) means [f(x)]2. If f(x) = x lnx then D f x x e * ( )  has the value (A) e (B) 2e (C*) 4e (D) none [Hint: D*f(x) = 2f(x).f (x) D*(x lnx) = 2x lnx (1 + lnx)] Q.6503/mod If f(x) = x . x, then its derivative is : (A) 2x (B) 2x (C*) 2x (D*) 2x sgn x
  • 3. Q.1127/mod If y = 1 x 3 x 1 x x 2 2 4     and dx dy = ax + b then the value of a + b is equal to (A) cot 8 5 (B*) cot 12 5 (C) tan 12 5 (D) tan 8 5 [Sol. y = 1 x 3 x x 3 ) 1 x ( 2 2 2 2     = x 3 1 x ) x 3 1 x )( x 3 1 x ( 2 2 2       dx dy = 2x – 3  a = 2 & b = – 3 a + b = 2 – 3 = tan 12  = cot 12 5 Ans. ] Q.2 Valueofsin 705° is (A) 4 2 6  (B*) 4 6 2  (C) 4 2 3  (D) 2 3 1 Q.3 Findallrealfunctionsofonerealvariablethatsatisfythefollowingfunctionalequation, f (x) + 2 f       x 1 = x [Sol. f (x) + 2 f       x 1 = x ....(1) Replacingxby1/x and multiplyby2 yields 2 f       x 1 + 4 f (x) = x 2 ....(2) Andso,subtractingequation(1) from(2) gives 3 f (x) = x 2 – x  f (x) =        x x 2 3 1 ] Q.4 In the figure, each region T represents an equilateral triangle and each region S a semicircle. The complete figure is a semicircle of radius 6 withitscentreO.Thethreesmallersemicirclestouchthelargesemicircle at pointsA, B and C.What is the radius of a semicircle S? [Sol. If we cut out part of the diagram and label appropriately we find that DL = r and DO = 3 r . This gives OB = r(1 + 3 ) = 6 or r = 3 1 6  = 3( 1 3  ) Ans. ] Q.5 Solve 3 3 28 x 6 28 x 6    = 2. [Ans. x = 6] [Sol. let a = 3 28 x 6  and b = 3 28 x 6  a – b = 2 (a – b)3 = 8 a3 – b3 – 3ab(a – b) = 8 (using a – b = 2) 56 – 6 · 3 28 x 6  · 3 28 x 6  = 8  48 = 6 · 3 28 x 6  · 3 28 x 6   83 = (6x + 28)(6x – 28)  36x2 – 784 = 512  36x2 = 1296  x2 = 36  x = ± 6 ]
  • 4. Q.1130/mod If f (x) = x5 + 2x3 + 2x and g is the inverse of f then g'(–5) is equal to (A*) 13 1 (B) 1 (C) 7 1 (D) none [Sol. y = x5 + 2x3 + 2x dx dy = 5x4 + 6x2 + 2  dy dx = 2 x 6 x 5 1 2 4   = g ' (y) when y = – 5, x5 + 2x3 + 2x + 5 = 0 (x + 1)(x4 – x3 + 3x2 – 3x + 5) = 0  x = – 1  1 x 5 y dy dx     = 2 6 5 1   = 13 1 ] Q.2126/mod Suppose the function f (x) – f (2x) has the derivative 5 at x = 1 and derivative 7 at x = 2. The derivative of the function f (x) – f (4x) at x = 1, has the value equal to (A*) 19 (B) 9 (C) 17 (D) 14 [Sol. y = f (x) – f (2x) y' = f ' (x) – 2 f ' (2x) y'(1) = f ' (1) – 2 f ' (2) = 5 ....(1) and y'(2) = f ' (2) – 2 f ' (4) = 7 ....(2) now let y = f (x) – f (4x) y' = f ' (x) – 4 f ' (4x) y ' (1) = f ' (1) – 4 f ' (4) ....(3) substituting the value of f ' (2) = 7 + 2 f ' (4) in (1) f ' (1) – 2 [7 + 2 f ' (4)] = 5 f ' (1) – 4 f ' (4) = 19  (A) ] Q.3 Findthenumberoftriangles withsides of integerlength whoseperimeteris 10. (A) 4 (B) 3 (C*) 2 (D) 0 [Sol. Let the sides are x, y, 10 – (x + y), x, y  N Now x + y > 10 – (x + y) x + y > 5 ....(1) also 10 – x > x  0 < x < 5 ....(2) |||ly 0 < y < 5 ....(3) If x = 1, y can be 5, 6, 7..... not possible x = 2, y = 4, z = 4; x = 3, y = 3, z = 4  (2, 4, 4) & (3, 3, 4) ] Q.4 The value of the parameter m for whichthe equation 1 x 3 2 x   = m x 6 1 x 2   has no solution for x, is (A) 12 (B) 18 (C) 14 (D*) 13 [Sol. x = m 13 1 m 2   , for no solution m = 13 ] Q.5 A circle of area 20 is centered at the point indicated bya solid dot in the accompanying figure. Suppose that ABC is inscribed in that circle and has area 8. The central angles ,  and  are as shown. What is the value of sin  + sin  + sin ? (A*) 5 4 (B) 4 3 (C) 3 2 (D) 2  [Hint: 2 1 r2(sin  + sin  + sin ) = 8 and r2 = 20 ]
  • 5. Q.195/mod Which of the following could be the sketch graph of y= d dx (x ln x)? (A) (B) (C*) (D) Q.278/mod If y = (A + Bx) emx + (m  1)2 ex then d y dx 2 2  2m dy dx + m2y is equal to : (A*) ex (B) emx (C) emx (D) e(1  m) x [Hint: multiplygivenequationby emx &then differentiatetwice] Q.3 Number ofsides in aregularpolygonifthesumofits interioranglesis 2520°, is (A) 14 (B*) 16 (C) 18 (D) 20 [Hint (n – 2) = 2520° external angle = n          n 2 = 2520°  n = 16 n – 2 = 2520°  (n – 2)180 = 2520  n – 2 = 14 n = 6] Q.4 The value of x such that 4(1 + y)x2 – 4x + 1 – y = 0 is true for all real values of y, is (A) – 1/2 (B) – 2/3 (C) 2/3 (D*) 1/2 [Sol. x = ) y 1 ( 8 ) y 1 ( 4   or ) y 1 ( 8 ) y 1 ( 4   ; hence x = 2 1 Ans. ] Q.5 The lengths of the sides of triangleABC are 60, 80 and 100 with A=90°.The lineAD divides triangleABC into two triangles ofequal perimeter. Calculatethelength ofAD. [Ans. 24 5 ] [Sol. 60 + x = 100 – x + 80 2x = 120  x = 60 now usingcosinelaw inACD y2 = (60)2 + x2 – 2x · 60 cos C (cosC = 3/5) = 3600 + 3600 – 2 · 60 · 60 · 5 3 = 7200 – 24 · 60 · 3 = 2880 y = 2880 = 24 5 Ans. Alternatively: sin 2 C = 60 · 2 y = 120 y y = 120 sin 2 C ....(1) now cos C = 5 3 1 – 2 sin2 2 C = 5 3  2 sin2 2 C = 5 2  sin 2 C = 5 1 . Hence from (1), y = 5 120 = 5 24 Ans. ]