CBSE Set Qa1 Maths Sample Test Papers For Class 12th for students online

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Maths Class - XII (CBSE)
You are on Set no 1 Answer 1 to 4

Q1) Using elementary row transformations, find the inverse of the matrix

A = 

4  5
3  4

Ans1) Here

A = 

4  5
3  4

and I = 

1  0
0  1

Now A = IA

=

4  5
3  4

=

1  0
0  1

A

= Applying R1 (1/4)R1

1  5/4
3   4

=

1/4  0
0     1

A

= Applying R2 R2 - 3R1

1  5/4
0  1/4

=

1/4   0
-3/4 1

A

= Applying R2 4R2

1  5/4
0   1

=

1/4  0
-3    4

A

= Applying R1 R1 - (5/4)R2

1  0
0  1

=

4    -5
-3    4

A
A-1

4    -5
-3    4

Q2) For the matrices A and B, verify that (A, B)' = A'B' where 

A =

1  2
3  4

B = 

4  5
1  2

Ans2) Here

A =

1   2
3   4

B = 

4   5
1   2

=

6    9
16  23

=

6  16
9  23

R.H.S. = B'A' =

4  1
5  2

1  3
2  4

=

6  16
9  23

= L.H.S.

Q3) Obtain the equation of the sphere with the points (1, -1, 1) and (3, -3, 3) as the extremities of a diameter and find the co-ordinates of its centre.
Ans3) Eg of the sphere with diameter AB is (x - x1)(x - x2) + (y - y1)(y - y2) + (z - z1)(z - z2) = 0
= (x - 1)(x - 3) + (y + 1)(y + 3) + (z - 1)(z - 3) = 0
= x2 + y2 + z2 - 4x + 4y - 4z + 9 = 0
Here 2u = -4
u = -2
2v = 4, v = 2
2w = -4
w = -2
d = 9
co-ordinates of centre are (-u, -v, -w) = (2, -2, 2)

Q4) Find the work done by the force = 2 + + , acting on a particle, if the particle is displaced from the point with position vector 2 + 2 + 2 to the point with position vector 3 + 4 + 5.
Ans4) Here = 2 + +
A(2, 2, 2) and B(3, 4, 5)
... = - = (3 - 2) + (4 - 2) + (5 - 2)
= + 2 + 3
... work done = .
= (2 + + )( + 2 + 3)
= 2.I + I.2 + I.3 = 7 units

Q5)  Verify Rolle's theorem for the following f(x) in the interval |2, 3|:
f(x) = x2 - 5x + 6
Ans5) Here f(x) = x2 - 5x + 6 on the interval [2, 3] and derivable on (2, 3)
Also f(2) = 4 - 5.2 + 6 = 0
and f(3) = 9 - 5 . 3 + 6 = 0  i.e. f(2) = f(3)
... conditions of Rolle's theorem are satisfied
= f '(c) = 0 for some c (2, 3)
Now f '(x) = 2x - 5
Solving f '(x) = 0 we get x = 5/2
thus c = 5/2 which clearly lies in (2, 3)

Q6) Evaluate

cosec x - cot x
          x

Ans6)  

cosec x - cot x
          x
(1/sin x) - (cos x/sin x)
             x
1 - cos x
 x.sin x
      2sin2 (x/2)       
2sin (x/2)cos (x/2).x
sin (x/2)  .      1   
 2.(x/2)     cos x/2

= 1/2 . 1 . 1 = 1/2

Q7) Differentiate tan-1 [2x/(1 - x2)] w.r.t. tan-1 x
Ans7) Let u = tan-1 [2x/(1 - x2)] and v = tan-1 x
To find : du/dv
putting x = tan   in these , we get
u = tan-1 (2tan )/(1 - tan2 ) - tan-1 (tan 2) = 2
= 2 tan-1 x  ... du/dv = 2/(1 + x2)
and v = tan-1 (tan ) =
= tan-1 x
... dv/dx = 1/(1 + x2)
... du/dv = (du/dx)/(dv/dx) = 2

Q8) Evaluate

1/2   sin-1 xdx
0    (1 - x2)3/2

Ans8) Let I =

1/2 sin-1 xdx
0     (1 - x2)3/2

Putting x = sin
dx = cos d
when x = 0 , sin = 0
= = 0
when x = 1/2 sin = 1/2
= = /4

/4   . cos       
0      (1 - sin2 )3/2
/4
   . sec2 d   (integrating)      
0      

=

tan

/4
          -
0
1 1. tan d
0 
= /4 + 

log cos

/4
          =
0
/4 + log (1/2)

= /4 - 1/2 log 2

Q9) Evaluate

       dx     
x2 + 4x + 8

Ans9) I =

       dx     
x2 + 4x + 8
       dx     
(x + 2)2 + 22

= 1/2 tan-1 ((x + 2)/2) + c

Q10) Evaluate

xcot -1 xdx

Ans10)  

xcot -1 xdx    (integrate by parts)
= x2/2 cot-1 x - x2/2 . -1/(1 + x2) dx
= 1/2 x2 cot-1 x + 1/2 (x2 + 1)/(1 + x2) dx
= 1/2 x2 cot-1 x + 1/2 [1 - 1/(1 + x2)]dx

= 1/2 x2 cot-1 x + 1/2 [x - tan-1 x] + c

Q11) Evaluate

          dx        
 (x2 + 4x + 2)

Ans11)  

          dx        
 (x2 + 4x + 2)
          dx        
 ((x + 2)2 - 2)
          dx        
 ((x + 2)2 - (2)2

= log [(x + 2) + (x2 + 4x + 2)] + c

Q12) Evaluate

ex [tan x + log sec x]dx

Ans12) I = 

ex [tan x + log sec x]dx

= ex tan x + c  [as d/dx(log sec x) = (1/sec x) . sec x tan x = tan x and 

ex (f(x) + f '(x))dx = ex f(x) + c]

Q13) Five dice are thrown simultaneously, if the occurrence of 3, 4, or 5 in a single die is considered as a success, find the probability of at least  3 success.
Ans13) Let p be the prob of getting 3, 4 or 5 in a single throw then 
p = 3/6 = 1/2
... q = 1 - p = 1/2
Here n = 5 and Binomial dist is 
(q + p)n = (1/2 + 1/2)5
... Prob. of least 3 successes
= P(3) + P(4) + P(5)
= 5C3(1/2)2(1/2)3 + 5C4(1/2)(1/2)4 + 5C5(1/2)5
= (10 + 5 + 1)/25 = 16/25 = 1/2

Q14) Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is neither 9 or 11?
Ans14)
The no. of exhaustive cases when two dice are thrown = n(s) = 6 x 6 = 36
The favourable cases for getting a sum of 9 or 11 on the two faces are (4, 5), (5, 4), (6, 3), (3, 6), (5, 6), (6, 5)
i.e. the no. of favourable cases = 6
... The prob of getting a 9 or 11 is 6/36 = 1/6
= The reqd Prob = 1 - 1/6 = 5/6

Q15) If A and B are two independent events such that P(AUB) = 0.6 and P(A) = 0.2 find P(B).
Ans15) As A and B are independent events.
we have ; P(AB) = P(A) . P(B)
Also P(AUB) = 0.6 and P(A) = 0.2 given
Now P(AUB) = P(A) + P(B) - P(AB)
= P(A) + P(B) - P(A) . P(B)
= P(A) + P(B)[1 - P(A)]
= 0.6 = 0.2 + P(B) [1 - 0.2]
= 0.4 = P(B) . (0.8)
= P(B) = 4/8 = 0.5

Q16) Using properties of determinants prove that

 

x + 4    x       x
x       x + 4    x
 x       x     x + 4

= 16(3x + 4)

Ans16)  Let =

x + 4    x       x
x       x + 4    x
 x       x     x + 4

operate : C1 C1 + C2 + C3

3x + 4    x       x
3x + 4  x + 4    x
3x + 4    x  x + 4

= 3x + 4

1    x       x
1  x + 4    x
 1    x   x + 4

operate R2 R2 - R1, R3 R3 - R1

= 3x + 4

1    x     x
0    4     x
 0    0      4

expand by C1

= 3x + 4

4  0
0  4

= 16(3x + 4)

Q17)  

If A = 

3  2
2  1

verify that A2 - 4A - I = O, where

I = 

1  0
0  1

and O =

0  0
0  0

Hence, find A-1

Ans17)  

Given : A = 3  2
2  1
= A2 = A.A =

3  2
2  1

3  2
2  1

=

13  8
 8  5

... A2 - 4a - I =

13  8
 8  5

- 4

3  2
2  1

-

1  0
0  1

= 13  8
 8  5

-

12  8
 8  4

-

1  0
0  1

=

0  0
0  0 

= A2 - 4A - I = 0   - (i)
Now |A| = 3 x 1 - 2 x 2 = -1 0
= A-1 exists
= (A2 - 4A - I) . A-1 = 0 . A-1
= A - 4I = A-1  [as 0 . A-1 = 0 and A . A-1 = I]

= A-1 = 3  2
2  1
- 4 1  0
0  1
= -1  2
 2  -3

Q18) Find the equation of the Plane passing through point(1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5) find also the perpendicular distance of the origin from the plane.
Ans18) Let A = (1, 2, 1), B = (1, 4, 2), C = (2, 3, 5)
To find : the plane through A and  to the line BC
The e.g. of the plane through A (1, 2, 1) is
l(x - 1) + m(y - 2) + n(z - 1) = 0   - (i)
As line BC is to be  to (i)
= l, m, n are proportional to the direction ratios of BC i.e.  to 2 - 1, 3 - 4, 5 - 2
i.e. to 1, -1, 3
i.e. l/1 = m/-1 = n/3 = K (say)
= l = K, m = -K, n = 3K
substituting these in (i) and canceling K we obtain
(x - 1) - (y - 2) + 3(z - 1) = 0
= x - y + 3z - 2 = 0   - (ii)
which is the reqd. eq of the plane
Also  from (0, 0, 0) on (ii) is
= ( 0 - 0 + 0 - 2)/(12 + (-1)2 + 32)
= 2/11

Q19) Prove that
( x ) x ( x ) = [ ] - [ ]
Ans19)  Let = a1 + b1 + c1, = a2 + b2 + c2, = a3 + b3 + c3, = a4 + b4 + c4

... x        
a1  b1  c1
a2  b2  c2
= (b1c2 - c1b2) - (a1c2 - a2c1) + (a1b2 - a2b1)
... x        
a3  b3  c3
a4  b4  c4
= (b3c4 - c3b4) - (a3c4 - a4c3) + (a3b4 - a4b3)
... ( x ) x ( x ) =                                              
b1c2 - c1b2   a2c1 - a1c2      a1b2 - a2b1
b3c4 - c3b4   a4c3 - a3c4      a3b4 - a4b3
R.H.S. =  a1  b1  c1
a2  b2  c2
a4  b4  c4
(a3 + b3 + c3) -  a1  b1  c1
a2  b2  c2
a3  b3  c3
(a4 + b4 + c4)

Simplifying we get

R.H.S. =                                              
b1c2 - c1b2   a2c1 - a1c2      a1b2 - a2b1
b3c4 - c3b4   a4c3 - a3c4      a3b4 - a4b3

... L.H.S. = R.H.S.

Q20) P, Q, R and S are the mid points of the sides of a quadrilateral ABCD using vectors, show that PQRS is a parallelogram?
Ans20)
 Let ABCD be a quadrilateral. Let P, Q, R, A be mid points of their sides. Let , , , be position vectors of points A, B, C, D

Position vector of mid point P of AB = ( + )/2
Position vector of mid point Q of BC = ( + )/2
Position vector of mid point R of CD = ( + )/2
Position vector of mid point S of DA = ( + )/2
To show that PQRS is a ||gm
Now = -
= ( + )/2 - ( + )/2
= 1/2 ( - ) = 1/2
also = -
=  ( + )/2 - ( + )/2
= 1/2 ( - ) = 1/2
... =
... PQRS is parallelogram

Q21) There are 3 urns A, B, and C, urn A contains 4 red balls and 3 black balls. Urn B contains 5 red balls and 4 black balls. Urn C contains 4 red balls and 4 black balls. One ball is drawn from each of these urns, what is the probability that the 3 balls drawn consists of 2 red balls and a black ball?
Ans21) 

URNS A B C
Red 4 5 4
Black 3 4 4

From URN A
P1 = Prob (Red Ball) = 4/(4 + 3) = 4/7
P2 = Prob (Black Ball) = 3/7
From URN B
P3 = Prob (Red Ball) = 5/7
P4 = Prob (Black Ball) = 4/9
From URN C
P5 = Prob (Red Ball) = 4/8 = 1/2
P6 = Prob (Black Ball) = 1/2
The reqd Prob (two red and one black ball)
= P1P3P6 + P1P4P5 + P2P3P5
= 4/7 x 5/9 x 1/2 + 4/7 x 4/9 x 1/2 + 3/7 x 5/9 x 1/2
= (4 x 5 + 4 x 4 + 3 x 5)/(7 x 9 x 2) = 51/126 = 17/42

Q22) Find from first principle, the derivative of (sin x) w.r.t. x
Ans22) Let  y = (sin x)
... y = y = (sin (x + x))
... y = (sin (x + x)) - (sin x) 

... dy/dx =
 

y/x = [(sin (x + x)) - (sin x)]/x

=

[(sin (x + x)) - (sin x)]/x[(sin (x + x)) - (sin x)]

=

[2 cos (x + x + x)/2 . sin (x/2)] / 2[(sin (x + x)) - (sin x)]x/2

=

cos x/2(sin x) 

... d/dx (sin x) = cos x/2(sin x)

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