1. Studymate Solutions to CBSE Board Examination 2012-2013
Series : SKS/1 Code No. 65/1/1
Candidates must write the Code on
Roll No. the title page of the answer-book.
 Please check that this question paper contains 8 printed pages.
 Code number given on the right hand side of the question paper should be written on the title page of
the answer-book by the candidate.
 Please check that this question paper contains 29 questions.
 Please write down the Serial Number of the questions before attempting it.
 15 minutes time has been allotted to read this question paper. The question paper will be distributed at
10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will not
write any answer on the answer script during this period.
MATHEMATICS
[Time allowed : 3 hours] [Maximum marks : 100]
General Instructuions:
(i) All questions are compulsory.
(ii) Questions numbered 1 to 10 are very short-answer questions and carry 1 mark each.
(iii) Questions numbered 11 to 22 are short-answer questions and carry 4 marks each.
(iv) Questions numbered 23 to 29 are also short-answer questions and carry 6 marks each.
(v) Use of calculators is not allowed.
-(1)-

2. STUDYmate
SECTION-A
Question numbers 1 to 10 carry 1 mark each.
Write the principal value of tan1 1  cos 1   
1
1.  
 2
 1
 
–1 –1
Ans. tan (1) + cos
 2
   
= tan1  tan   cos 1   cos 
 4  3
    
=  cos 1 cos     
4   3 
 2
= 
4 3
3  8  11
= =
12 12
2. Write the value of tan  2 tan 1 1 
 
 5
 1 1 
Ans. tan  2 tan 
 5
  1 
 2
1 
tan  tan  5  
2
=
  1 

 1   5   
   
 2  5  
= tan  tan1  
  24  

  5 
= tan  tan1   
 12  
5
=
12
 a  b 2a  c   1 5 
3. Find the value of a if 2a  b 3c  d    0 13 
   
Ans. a – b = –1
2a – b = 0
–a = –1
a=1
x 1 x 1 4 1
4. If  , then write the value of x.
x 3 x 2 1 3
Ans. (x + 1) (x + 2) – (x – 3) (x – 1) = 12 + 1
2 2
(x + 3x + 2) – (x – 4x + 3) = 13
7x – 1 = 13
7x = 14
x=2
-(2)-

3. STUDYmate
 9 1 4  1 2 1
5. If  2 1 3   A  0 4 9  , then find the martix A.
   
 9 1 4  1 2 1
Ans. A =  2 1 3   0 4 9 
   
 8 3 5 
A=  
 2 3 6 
2 4
3 d y 
2
 dy 
6. Write the degree of the differential equation x  2   x   0
 dx   dx 
Ans. Degree = 2
 
7. If a  xi  2 ˆ  zk and b  3i  yj  k are two equal vectors, then write the value of x + y + z.
ˆ j ˆ ˆ ˆ ˆ

Ans. a  xi  2 ˆ  zk  3i  yj  k
ˆ j ˆ ˆ ˆ ˆ
 x=3 y = –2 z = –1
 x+y+z=0
  
8. If a unit vector a makes angles with iˆ, with ˆ and an acute angle  with k , then find
j ˆ
3 4
the value of .
2  
Ans. cos  cos 2  cos 2   1
3 4
1 1
  cos 2  = 1
4 2
2 1
cos  =
4
1
cos  = , as  is an acute angle.
2

=
3
9. Find the Cartesian equation of the line which passes through the point (–2, 4, –5) and is
x 3 4y z 8
parallel to the line   .
3 5 6
Ans. The required equation of the line
x 2 y 4 z 5
 
3 5 6
10. The amount of pollution content added in air in a city due to x-diesel vehicles is given by P(x)
3 2
= 0.005x + 0.02x + 30x. Find the marginal increase in pollution content when 3 diesel
vehicles are added and write which value is indicated in the above question.
3 2
Ans. P(x) = 0.005x + 0.02x + 30x
2
P'(x) = 3 × 0.005x + 2(0.02)x + 30
P'(3) = 3 × 0.005 × 9 + 2(0.02)(3) + 30 = 30.255
-(3)-

4. STUDYmate
SECTION-B
Question numbers 11 to 22 carry 4 marks each.
11. Show that the function f in A =   
2
3
defined as f(x) =
4x  3
6x  4
is one-one and onto. Hence
–1
find f .
4x  3
Ans. f (x ) 
6x  4
2
Let x1, x 2  A      ; x1  x 2
3 
Consider, f (x1) = f (x2)
4x1  3 4x 2  3
i.e. 
6x1  4 6x 2  4
i.e. (4x1 + 3) (6x2 – 4) = (4x2 + 3) (6x1 – 4)
i.e. 24x1x2 – 16x1 + 18x2 – 12 = 24 x1x2 – 16x2 + 18 x1 – 12
i.e. –34x1 = –34x2
i.e. x1 = x2
 f is one - one.
4x  3
Let y 
6x  4
6xy – 4y = 4x + 3
i.e. (6y – 4) x = 3+ 4y
3  4y
x
6y  4
4 2
 y i .e.
6 3
2
 y  
3 
 f is onto
Since f is one-one and onto
3  4y
 xf 1
y  
6y  4
i.e. y (6x – 4) = 4x + 3
1  3  4x
 f x 
6x  4
12. Find the value of the following:
1  1 2x 1  y2 
tan  sin  cos 1  , |x| < 1, y > 0 and xy < 1.
2 1  x2 1  y2 
OR
1  1  1  1  1  1  
Prove that: tan    tan    tan   
2 5 8 4
1
  2x  1  1  y 2 

Ans. tan  sin1  2 
 cos 1  2 
2
 1  x  2  1  y  
-(4)-

5. STUDYmate
1 1 1 1 
= tan  (2 tan x )  (2tan y )
2 2 
= tan (tan–1 x + tan–1 y)
x y
= 1  xy
OR
1 1
   1  x y 
Ans. tan  2 5   tan1
1 [using tan1 x  tan1 y  tan1   , xy  1]
1  1  8  1  xy 
 
 10 
7 1
= tan1    tan1  
 9 8
7 1
  
= tan  9 8 
1
1  7 
 
 72 
 56  9 
= tan1  
 72  7 
 65 
= tan1  
 65 
= tan–1 (1)

=
4
13. Using properties of determinants, prove the following:
1 x x2
x  1  x 3 
2
x2 1
x x2 1
Ans. Operating C1  C1 + C2 + C3, we obtain
1 x  x2 x x2
1 x  x2 1 x
L.H.S. =
1 x  x2 x2 1
Take out 1 + x + x2 from C1
1 x x2
1 1 x
= (1 + x + x2)
1 x2 1
Operate R2  R2 – R1 : R3  R3 – R1
1 x x2
0 1 x x (1  x )
= (1 + x + x2)
0 x (x  1) 1  x 2
Expand with C1
1 x x (1  x )
= (1 + x + x2) ,
 x (1  x ) (1  x 2 )
Take out 1 – x from C1 and same from C2
-(5)-

6. STUDYmate
1 x
= (1 + x + x2) (1 – x)2  x 1  x
= (1 + x + x2) (1 – x)2 (1 + x + x2)
= {(1 + x + x2) (1 – x)}2
= (1 – x3)2 = R.H.S.
14. Differentiate the following function with respect to x:
x log x
(log x) + x
Ans. Let y = (log x)x + xlog x,
dy d
Then  {(log x )x  x log x }
dx dx
d d log x
= (log x )x  (x )
dx dx
d d  d v d 
= (log x )x {(x log(log x )}  x log x (log x log x )  (u )  uv (v log u )
dx dx  dx dx 
  1 1
 
  1  d d 
  log(log x )  x  (log x log x )  ((log x )2 ) 
log x
= (log x )x x   2(log x ) 
  log x  x
 
  x  dx dx 
 1   log x  log x
= (log x )x   log(log x )  2  x
 log x   x 
2 2 d 2y dy
15. If y  log[x  x  a ], show that (x 2  a 2 ) 2
x  0.
dx dx
Ans. y  log [x  x 2  a 2 ]
dy 1  2x 
 1  
dx x  x 2  a 2  2 x2  a2 
 
1   x 
x 2  a 2 ____
y1   

x a 2 
 x 2  ____
 x2  a2  
 y1( x 2  a 2 )  1
1.2 x .y1
 y2 ( x 2  a 2 )  0
2 x 2  a2
2 2
 y2 (x + a ) + xy1 = 0
16. Show that the function f (x ) | x  3|, x  , is continuous but not differentiable at x = 3.
OR
d 2y
If x = a sin t and y = a (cot t + log tan t/2), find .
dx 2
Ans. f(x) = |x – 3|; x = 3
– f  3  h   f  3
LHD = f '(3 ) = lim
h 0 h
h 0
|3  h  3|  |3  3|
= lim
h 0 h
h 0
-(6)-

7. STUDYmate
| h |
= lim
h 0 h
h 0
h
= lim
h 0 h
h 0
= –1
f  3  h   f  3
R H D = f '(3+) = lim
h 0 h
h 0
|3  h  3| |3  3|
= lim
h 0 h
h 0
|h |
= lim
h 0 h
h 0
h
= lim
h 0 h
h 0
= 1
As LHD  RHD  f is not differentiable.
Again, LHL = lim | x  3| (at x = 3)
x 3
= lim|3  h  3|
h 0
= lim| h | 0

h 0
RHL = lim | x  3| (at x = 3)
x 3
= lim|3  h  3|
h 0
= 
lim|h | 0
h 0
Since, LHL = RHL
 f is continuous at x = 3.
OR
 t
Ans. y = a  cos t  log tan  ; x = a sin t
 2
dy  1  t  1
= a   sin t   sec2    
dt
 t   2  2
tan  

 2 

 t  
 cot   
1 2  1
a   sin t   
=  2 t  2  t 
sin   cos  

 2  2
 1 
= a   sin t 
 sin t 

1  sin2 t 
= a 
 sin t 
cos2 t
= a
sin t
dx
= a cos t
dt
-(7)-

8. STUDYmate
 a cos2 t 
dy  sin t  = cot t
  
dx a cos t
d 2y d  dy 
 
dx 2 dx  dx 
 
=  cosec 2t  
dt 
 
 dx 
1 cosec2t .sec t
=   cosec t  .
2

a cos t a
cosec2t
=
a cos t
sin(x  a )
17. Evaluate :  sin(x  a dx
OR
5x  2
Evaluate :  1  2x  3x 2
dx
sin(x  a )
Ans. I   dx
sin(x  a )
sin(t  2a )
I dx
sin t
sin t cos 2a cos t sin2a
 dt   dt
sin t sin t
= t cos 2a – sin 2a .  cot dt
= t cos 2a – sin 2a ln | sin t | + C
= (x + a) cos 2a – sin 2a ln | sin (x + a) | + C
OR
5x  2
Ans. I   1  2x  3x 2
dx
d
5x  2  A (1  2x  3x 2 )  B
dx
 5x – 2 = A (2 + 6x) + B
Comparing the co-efficients we get
5 = 6A
5
 A=
6
– 2 = 2A + B
 B = – 2A – 2
5
=  2
3
11
= 
3
-(8)-

9. STUDYmate
5 11
(2  6x )
 I 6 dx   3 dx
1  2x  3x 2 1  2x  3x 2
I1 I2
5 2  6x
6  1  2x  3x 2
I1  dx
Let 1 + 2x + 3x2 = t
 (2 + 6x)dx = dt
5 dt 5
6 t
 I1   ln t
6
5
= ln|1  2x  3x 2 | C1
6
11 dx
I2 
3  3x 2  2x  1
11 dx
9  x2  2 x  1  1  1

3 3 9 9
11 dx

9   1 2 2
x   
 3 9
1
x
11 1 3 C
 . tan1 2
9 2 2
3 3
= I = I 1 – I2
5 11  3x  1 
= ln|1  2x  3x 2 |  tan1  C
6 3 2  2 
x2
18. Evaluate :  (x 2
 4) (x 2  9)
dx
1 2x 2  4  9  4  9
2   x 2  4  x 2  9
Ans. I = dx
1 x2  4 1 x2  9 1 x2  4 1 13dx
=
2   x 2  4 x 2  9 dx  2   x 2  4 x 2  9 dx  2   x 2  4 x 2  9dx  2   x 2  4  x 2  9dx
1 1 1  x  1 1 1  x  1 3  1 1 
= 2 . 3 tan  3   2 . 2 tan  2   2 . 5   2
     dx
 x  4  x  9 
2
1 x 1  x  13  x  13 x
= tan1    tan1    tan1    tan1  C
6 3 4  2  20  2  30 3
1  x   1 13  1  x   1 13 
= tan       tan     C
 3   6 30   2   4 20 

3 x 2 x
= tan1    tan1    C
5 3 5 2
-(9)-

10. STUDYmate
19. Evaluate :
 (|x |  | x  2| | x  4|)dx
4
Ans. | x |  | x  2|  | x  4|dx  I
0
0<x <4  |x| = + x
0<x <2  |x – 2| = – (x – 2)
0<x <4  |x – 2| = + (x – 2)
0<x <4  |x – 4| = – (x – 4)
4 4 4 4
 I = 
0

0

0

x  2  x  (x  2)  4  x
0
4 2 4 4
 x2   x2   x2   x2 
    2x     2x   4x  
 2 0 
   2 0  2
 2 
  2 0
= (8) + [(4 – 2) – 0] + [(8 – 8) – (2 – 4)] + [16 – 16/2]
= 8 + 2 + 2 + 16/2
= 20
      
20. If a and b are two vectors such that |a  b |  |a |, then prove that vector 2a  b is

perpendicular to vector b.
  2  2
Ans. |a  b |  |a |
 2    2  2
 |a |  2a .b  |b | |a |
   
2a .b  b .b  0
  
(2a  b ). b  0
  
 2a  b  b
x  2 y 1 z  2
21. Find the coordinates of the point, where the line   intersects the plane
3 4 2
x – y + z – 5 = 0. Also, find the angle between the line and the plane.
OR
Find the vector equation of the plane which contains the line of intersection of the planes
 ˆ 
r . (i  2 ˆ  3k )  4  0 and r . (2i  ˆ  k )  5  0 and which is perpendicular to the plane
j ˆ ˆ j ˆ

r . (5i  3 ˆ  6k )  8  0.
ˆ j ˆ
x  2 y 1 z  2
Ans.    [x = 2 + 2, y = 4 – 1; z = 2 + 2]
3 4 2
This point lies on the plane x – y + z – 5 = 0
(3 + 2) – (4 – 1) + (2 + 2) – 5 = 0
3 + 2 – 4 + 1 + 2 + 2 – 5 = 0
=0
-(10)-

11. STUDYmate
 Point of Intersection = [2, –1, 2]
 3  1  4  1  2 1
sin  =
9  16  4 1  1  1
342
sin  =
29 3
1
sin  =
87
1 1
 = sin
87
OR
Ans. Equation of plane through 1 & 2
(x + 2y + 3z – 4) + k(2x + y – z + 5) = 0
x (1 + 2k) + y(2 + k) + z (3 – k) – 4 + 5k = 0
This plane is perpendicular to 5x + 3y – 6z + 8 = 0
 5 (1 + 2k) + 3 (2 + k) – 6 (3 – k) = 0
5 + 10 k + 6 + 3k – 18 + 6k = 0
19 k – 7 = 0
7
k
19
 Required plane is will be
 14   7   7 
x 1    y 2    z 3  
 19   19   19 
85
4  0
19
33x + 45y + 50z + 9 = 0
22. A speak truth in 60% of the cases, while B in 90% of the cases. In what percent of cases are
they likely to contradict each other in stating the same fact? In the cases of contradiction do
you think, the statement of B will carry more weight as he speaks truth in more number of
cases than A?
60 40
Ans. P (AT) = P (AC) =
100 100
90 10
P (BT) = P (BC) =
100 100
60 10 400 90
P (contradiction)    
100 100 100 100
600  3600 4200
 
10000 10000
% P (contradiction) = 42 %
In general parlace we understand who speaks truth, will have a better say and accepted
more.
-(11)-

12. STUDYmate
SECTION-C
Question numbers 23 to 29 carry 6 marks each.
23. A school wants to award its students for the values of Honesty, Regularity and Hard work with
a total cash award of ` 6,000. Three times the award money for Hardwork added to that given
for honesty amounts to ` 11,000. The award money given for Honesty and Hard work together
is double the one given for Regularity. Represent the above situation algebraically and find
the award money for each value, using matrix method. Apart from these values, namely,
Honesty, Regularity and Hard work, suggest one more value which the school must include
for award.
Ans. Let Honesty award = ` x
Regularity award = ` y
Hard work award = ` z
According to question;
x + y + z = 6000
3z + x = 11000
x + z = 2y
x + y + z = 6000
x + 0y + 3z = 11000
x – 2y + z = 0
We can represent these equations using Matrices as:
1 1 1   x   6000 
1 0 3 y   11000 
    
1 2 1   z   0 
    
AX = B
–1
 X=A B
1 1 1 
Now |A| = 1 0 3  = 6 + 2 – 2 = 6  0
 
1 2 1 
 
–1
 A exists
 6 3 3 
 
adj. A =  2 0 2
 2 3 1
 
1
A = |A| adj. A 
–1

 6 3 3 
1
2 0 2
= 6 
 2 3 1
 
 6 3 3   6000 
1
2 0 2 11000 
X=A B= 6  
–1
 2 3 1  0 
  
 36000  33000  0 
1 
12000  0  0
= 6 
 12000  33000  0 
 
-(12)-

13. STUDYmate
 3000 
1
12000 
= 6 
21000 
 
 500 
2000 
=  
3500 
 
 Award for honesty = ` 500
Award for regularity = ` 2000
Award for hardwork = ` 3500
One more value can be punctuality.
24. Show that the height of the cylinder of maximum value, that can be inscribed in a sphere of
3R
radius R is . Also find the maximum volume.
3
OR
2
Find the equation of the normal at a point on the curve x = 4y which passes through the
point(1, 2). Also find the equation of the corresponding tangent.
Ans. Let x be the radius of the base and h be the height of the cylinder ABCD which is inscried in
a sphere of radius R. It is obvious that for maximum volumke the axis of the cylinder must be
along OA2 = OL2 + AL2
 AL = R2  x 2
M
Let V be the volume of the cylinder. Then, D C
V = (AL)2 × LM
 V =  (AL)2 × 2 (OL)
 V =  (R2 – x2) × 2x O
a
 V = 2 (R2x – x3)
dV d 2V 
  2 (R 2  3x 2 ) and  12 x
dx dx 2 A L B
For maximum or minimum values of V, we must have
dV
0
dx
 2(R2 – 3x2) = 0
R
 x=
3
 d 2V  R
 12  0
Clearly,  dx 2 x  R
  3
3
R 2R
Hence, V is maximum when x  and hence LM = 2x = .
3 3
OR
Differentiating x = 4y with respect to x, we get
2
dy x

dx 2
Let (h, k) be the coordinates of the point of contact of the normal to the curve x2 = 4y. Now, slope
of the tangent at (h, k) is given by
-(13)-

14. STUDYmate
dy  h
 
dx (h , k ) 2
2
Hence, slope of the normal at (h, k) =
h
Therefore, the equation of normal at (h, k) is
2
y k  (x  h ) … (i)
h
Since it passes through the point (1, 2), we have
2
2k  (1  h )
h
2
or k  2  (1  h ) … (ii)
h
Since (h, k) lies on the curve x2 = 4y, we have
h2 = 4k … (iii)
From (ii) and (iii), we have h = 2 and k = 1. Substituting the values of h and k in (i), we get the
required equation of normal as
2
y 1  (x  2)
2
or x+y=3
2
25. Using integration, find the area bounded by the curve x = 4y and the line x = 4y – 2.
OR
Using integration, find the area of the region enclosed between the two circles
2 2 2 2
x + y = 4 and (x – 2) + y = 4.
Ans. The equations of the given curves are
x2 = 4y … (i)
and x = 4y – 2 … (ii)
Eq. (i) represents a parabola with vertex at the origin
and axis along positive direction of y-axis. Eq. (ii)
represents a straight line which meets the
coordinate axes at (– 2, 0) and (2, 0) respectively.
To find the points of intersection of the given parabola
and the line, we solve (i) and (ii) simultaneously.
Solving the two equations simultaneously.
We obtain that the points of intersection of the given
parabola and the line are (2, 1) and (–1, 1/4).
The region whose area is to be found out is shaded in figure.
Let us slice the shaded region into vertical strips. We find that each vertical strip runs from
parabola (i) to the line (ii). So, the approximating rectangle shown in figure has
Width = x, Length = (y2 – y1), and the Area (y2 – y1) x.
Since the approximating rectangle can move from x = – 1 to x = 2.
 Required area A is given by
2
A  (y
1
2  y1 )dx
 P(x , y2 ) and Q(x , y1 ) lie on (ii) and (i) respec.
2
 x  2 x2   
A    dx
  y  x  2 and y  x
2
4 4  
1 
 2
 4
1
4 

-(14)-

15. STUDYmate
2
x2 1 x3 
 A  x 
8 2 12 
4 2 8  1 1 1  9
 A       sq. units
 8 2 12   8 2 12  8
OR
Ans. Equations of the given circles are
x2 + y2 = 4 ...(1)
and (x – 2)2 + y2 = 4 ...(2)
Equation (1) is a circle with centre O at the origin
and radius 2. Equation (2) is a circle with centre C
(2, 0) and radius 2. Solving equations (1) and (2), we
have
(x –2)2 + y2 = x2 + y2
or x2 – 4x + 4 + y2 = x2 + y2
or x = 1 which gives y =  3
Thus, the points of intersection of the given circles are A 1, 3  and A' 1,  3  as shown in
the figure.
Required area of the enclosed region OACA'O between circles
= 2 [area of the region ODCAO] (Why?)
= 2 [area of the region ODAO + area of the region DCAD]
1 2 
= 2   ydx   ydx 
0
 1 

1 2
2 
= 2   4   x  2 dx   4  x dx 
2
(Why?)
0
 1 

1 2
1 2 1 1  x  2   1 1 1 x 
= 2   x  2  4   x  2    4 sin 
2
   2  x 4  x   4 sin
 2 2  2 0 2 2 2 1

1 2
 2 1  x  2    1 x 
=  x  2  4   x  2   4 sin 
2
   x 4  x  4 sin
  2  0   2 1

 1  1     1 1 
=   3  4sin     4 sin  1   4 sin 1  3  4 sin
1 1
   2    2
     
=   3  4    4     4   3  4  
 6 2  2 6
 2   2 
=  3   2    2  3  
 3   3 
8
= 2 3
3
x/y x/y
26. Show that the differential equation 2ye dx + (y – 2x e ) dy = 0 is homogeneous. Find the
particular solution of this differential equation, given that x = 0 when y = 1.
Ans. The given differential equation can be written as
x
dx 2xe y  y

dy x ...(i)
2ye y
-(15)-

16. STUDYmate
x
2xe y  y
Let F(x, y) = x
2ye y
 x 
 2xe y  y 
    0 F x , y 
Then F(x, y) =   
 x 
  2ye y 

Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential
equation is a homogeneous differential equation.
To solve it, we make the substitution
x = vy ...(ii)
Differentiating equation (ii) with respect to y, we get
dx dv
v y
dy dy
dx
Substituting the value of x and dy in equation (i), we get
dv 2vev  1
v y 
dy 2ev
dv 2vev  1
or y  v
dy 2ev
dv 1
or y  v
dy 2e
v
dy
or 2e dv = y
dy
 2e .dv   
v
or y
v
or 2e = –log |y| + C
x
and replacing v by y , we get
x
2e y  log|y |  C ...(iii)
Substituting x = 0 and y = 1 in equation (iii), we get
0
2e + log |1| = C = C = 2
Substituting the value of C in equation (iii), we get
x
y
2e  log|y |  2
which is particular solution of the given differential equation.
27. Find the vector equation of the plane passing thorugh the three points with position vectors
i  ˆ  2k , 2i  j  k and i  2 ˆ  k. Also find the coordinates of the point of intersection of this
ˆ j ˆ ˆ ˆ j ˆ

plane and the line r  3i  ˆ  k  (2i  2 ˆ  k ).
ˆ j ˆ ˆ j ˆ
   
Ans. (r  a ).(b  c )  0
ˆ ˆ k
i j ˆ
  
n  b  c  2 1 1  3iˆ  ˆ  5k
j ˆ
1 2 1
-(16)-

17. STUDYmate
 ˆ
 r  (i  ˆ  2k ) . (3iˆ  ˆ  5k )  0
j ˆ j ˆ
 
Equation of plane is

r .(3iˆ  ˆ  5k )  14
j ˆ
 ˆ
or r .(3i  ˆ  5k )  14
j ˆ

Now r  3i  ˆ  k  (2i  2 ˆ  k )
ˆ j ˆ ˆ j ˆ
This line intersects the plane
 [(3 + 2l) [(3  2 )i  (1  2 ) ˆ(1   )k ]. (3i  ˆ  5k )  14)
ˆ j ˆ ˆ j ˆ
 –=1
 =– 1
 Point of intersection is
i  ˆ  2k .
ˆ j ˆ
28. A cooperative society of farmers has 50 hectares of land to grow two crops A and B. The profits
from crops A and B per hectare are estimated as ` 10,500 and ` 9,000 respectively. To control
weeds, a liquid herbicide has to be used for crops A and B at the rate of 20 litres and 10 litres
per hectare, respectively. Further not more than 800 litres of herbicide should be used in
order to protect fish and wildlife using a pond which collects drainage from this land. Keeping
in mind that the protection of fish and other wildlife is more important than earning profit,
how much land should be allocated to each crop so as to maximize the total profit? Form an
LPP from the above and solve it graphically. Do you agree with the message that the protection
of wildlife is utmost necessary to preserve the balance in environment?
Ans. Let x hectares for crop A and y hectares for crop B be allocated.
A B
x y 50 hectares
10500 9000 Profit
20 L/H 10 L/H 800 litres atmost
Y
C
50
B(30, 20)
X
O A 50
40 x + y = 50
2x + y = 80
-(17)-

18. STUDYmate
A.T.Q.
We need to maximise profit given by
Z = 10500x + 9000y subject to
x + y  50 (Area constraint)
20x + 10y  800 (Herbicide constraint)
x, y  0 (Non-negative constraint)
After plotting graph, the corner points are
Corner Point Value of Z
O(0, 0) Z= 0
A(40, 0) Z= 420000
B(30, 20) Z= 315000 + 180000 = 495000 Maximum
C(0, 50) Z= 450000
 30 hectares for crop A and 20 hectares for crop B.
29. Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditation
and yoga course reduces the risk of heart attack by 30% and prescription of certain drug
reduces its chance by 25%. At a time a patient can choose any one of the two options with
equal probabilities. It is given that after going through one of the two options, the patient
selected at random suffers a heart attack. Find the probability that the patient followed a
course of meditation and yoga. Interpret the result and state which of the above stated methods
if more beneficial for the patient.
Ans. Let E1: the patient follows meditation and yoga
E2: tha patient uses drug
1
then E1 and E2 are mutually exclusive and P(E1) = P(E2) =
2
Let E: the selected patient suffers a heart attack
40  30  28
then P(E/E1) = 1  
100  100  100
40  25  30
and P(E/E2) = 1  
100  100  100
Hence, P (patient who suffers heart attack follows meditation and yoga).
P  E/E1  P  E1 
= P(E1/E) = P E/E P E  P E/E P E (Using Bayes Theorem)
 1  1  2  2
28 1

100 2 28 14
=  
28 1 30 1 58 29
  
100 2 100 2
×·×·×·×·×
-(18)-

19. STUDYmate
Studymate Solutions to CBSE Board Examination 2012-2013
Series : SKS/1 Code No. 65/1/2
UNCOMMON QUESTIONS ONLY
SECTION-A
4
 dy  d 2y
9. Write the degree of the differential equation    3x  0.
 dx  dx 2
4
 dy   d 2y 
Ans.  dx   3x  2   0
   dx 
 
Degree = 1
16. P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of cases are
they likely to agree in stating the same fact?
Do you think, when they agree, means both are speaking truth?
70 30
Ans. P(PT) = P(PL) =
100 100
80 20
P(QT) = P(QL) =
100 100
P(agreement) = (PL & QT) or (PL & QL)
5600 600
= 
10000 10000
6200
=  62%
10000
No, agreement does not mean they are speaking truth.
 ˆ       
18. If a  i  ˆ  k and b  ˆ  k  k , find a vector c , such that a  c  b and a .c  3
j ˆ j ˆ ˆ
 ˆ ˆ ˆ
Ans. Let c  xi  yj  zk
i ˆ k
ˆ j ˆ
 
a c  1 1 1
x y z
i (z  y )  ˆ(z  x )  k (y  x )
ˆ j ˆ
   
Now, a  c  b (given) (b  ˆ  k ) j ˆ
 z – y = 0, x – z = 1, y – x = – 1
y = z and x – y = 1 ... (i)
 
Again, a .c  3 x + y + z = 3
x+y+y=3
x + 2y = 3 ... (ii)
-(19)-

20. STUDYmate
Solving (i) and (ii)
x  2y  3
x  y 1
  2
y z
3y  2 / 3 3
5
 x=1+y=
3
 5ˆ 2 2 ˆ
So, c  i  ˆ  k.
j
3 3 3
3
19. Evaluate :  | x  1|  | x  2|  | x  3| dx.
1
3
Ans. I =  | x  1|  | x  2|  | x  3| dy
1
2 3
  | x  1  2  x  3  x  dx   (x  1  x  2  3  x )dx
1 2
2 3

 (4  x ) dx  x dx
1

2
2 3
 x   x2 
2
  4x    

 2 
1  2  2
 
 1 9 
 6   4      2
 2 2 
7 5
6 
2 2
5 5
  5
2 2
x2 1
20. Evaluate :  (x 2
 4)(x 2  25)
dx
x2  1
Ans. I   (x 2
 4)(x 2  25)
dx
x2 1 A B
2 2
 2
 2
(x  4) (x  25) (x  4) x  25
x =0
A B 1
    25 A  4B  1 ... (i)
4 25 100
x=1
2 A B
  

5  26 5 26
A B 1
  
5 26 65
5
 13A + B=1
2
-(20)-

21. STUDYmate
 26A + 5B = 2 ... (ii)
1 8
 A ,B
7 7
1 dx 8 dx
 I=  2
7 x 4 7
 x 2
 25
1 1 x 8 1 x
 . tan1    . tan1    c
7 2 2 7 5 5
dy y  y 
28. Show that the differential equation x sin    x  y sin    0 is homogeneous. Find the
dx  x x

particular solution of this differential equation, given that x = 1 when y  .
2
dy y y
Ans. x sin  y sin  x
dx x x
dy y 1
 
dx x sin y
x
y 1
Let F (x, y) = 
x sin y
x
y 1
 F (x, y) = 
x  y 
sin  
 x 
 
y 1 
    
 x sin y 
 
 x
= ° F (x, y)
 Given d.e. is homogeneous
Let y = vx
dy dv
 v  x
dx dx
Now
 dv 
 x v  x sin v  x  vx  sin v  0
 dx 
dv
 vx sin v  x 2 sin v  x  vx sin v  0
dx
dx
  sin v dv   x
 – cos v = – ln |x| + c
y 
 – cos   = – ln |x| + c
x

x  1, y 
2
 0=c
y
 cos  ln | x |
x
 x  e cos(y/x )
y
 cos  log x
x
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22. STUDYmate
29. Find the vector equation of the plane determined by the points A (3, –1, 2), B (5, 2, 4) and
C (–1, –1, 6). Also find the distance of Point P (6, 5, 9) from this plane.
x 3 y 1 z 2
Ans. 53 2 1 42  0
1  3 1  1 6  2
x  3 y 1 z  2
2 3 2 0
4 0 4
(x – 3) (12 – 0) – (y + 1) (8 + 8) + (z – 2) (0 + 12) = 0
12x – 36 – 16y – 16 + 12 z – 24 = 0
12x – 16y + 12z – 52 – 24 = 0
12x – 16y + 12z – 76 = 0
3x – 4y + 3z – 19 = 0
 ˆ
r .(3i  4 ˆ  3k )  19
j ˆ
Now, distance of plane from Points P (6, 5, 9)
Distance of plane from (6, 5, 9)
(3  6)  (4)(5)  (3)(9)  19
34
18  20  27  19 6

34 34
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23. STUDYmate
Studymate Solutions to CBSE Board Examination 2012-2013
Series : SKS/1 Code No. 65/1/3
UNCOMMON QUESTIONS ONLY
SECTION-A
 
2. Write a unit vector in the direction of the sum of vectors a  2i  ˆ  2k and b  i  ˆ  3k .
ˆ j ˆ ˆ j
  
Ans. c  a  b  i  0 ˆ  5k
ˆ j ˆ
  
c |a  b | 12  52  26


^
 ˆ
i 5
 a  b    ˆ
k
2b 2b
11. A speaks truth 75% of the cases, while B in 90 % of the cases. In what percent of cases are
they likely to contradict each other in stating the same fact? Do you think that statement of
B is true?
75 25
Ans. P(AT) = P(AL) =
100 100
90 10
P(BT) = P(BL) =
100 100
75 10 25 90
P (contradiction) =   
100 100 100 100
750  2250
=
10000
3000
=
10000
3
=
10
3
 % of P (Contradiction) = × 100 = 30%
10
5
14. Evaluate :  | x  2|  | x  3|  || x  5| dx.
2
Ans. 2 < x < 5  |x – 2|  + (x – 2)
2<x <3  |x – 3|  – (x – 3)
3<x <5  |x – 3|  x – 3
2<x <5  |x – 5|  5 – x
5 3 5 5
   
  (x  2)  (3  x )  (x  3)  (5  x )
2 2 3 2
5 3 5 5
x 2   x   x2 2   x2 
  2x    3x     3x    5x  
 2   x     2 
 2  2  x 3  2
 25    9   25  9    25  
   10    2  4     9    (6  2)    15     9     25    (10  2)
 2    2   2  2    2  
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24. STUDYmate
5  9   5 9   25 
   2    4         8
2  2   2 2  2 
 9   7  4 9 39
2   2   2  2  2
   
2x 2  1
15. Evaluate : x 2
(x 2  4)
dx
2x 2  1
Ans. I= x 2
(x 2  4)
dx
2
Put x = t for PFD
2t  1 A B
 
t (t  4) t t  4
 2t + 1 = A (t + 4) + Bt
1 7
Comp : t =2=A+B where A  B
4 4
Const 1=4A
1 dx 7 dx
I =
4 x 2

4 x 2
4
1 7 1 x 
   tan1    C
4x 4 2 2
1 7 x 
  tan1    C
4x 8 2
25. Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the
plane, passing through the point (2, 2, 1) (3, 0, 1) and (4, –1, 0).
Ans. Equation of line
x 3 y 4 z 5
 
1 1 6
General Point [x = –  + 3, y =  + 4, z = 6 – 5]
Equation of plane
x 2 y 2 z 1
32 02 1 1  0
4  2 1  2 0  1
x  2 y  2 z 1
1 2 0 0
2 3 1
(x – 2) {2 – 0} – (y – 2) {– 1 – 0} + {z – 1} {– 3 + 4} = 0
2x – 4 + y – 2 + z – 1 = 0
2x + y + z – 7 = 0 _______________ ,
General point of line is a point on ,
 2 ( –  + 3) + ( – 4) + (6 – 5) – 7 = 0
–2 + 6 +  – 4 + 6 – 5 – 7 = 0
5 – 10 = 0
=2
 Point of intersection of line and plane (1, –2, 7)
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