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 8

(i) All Questions are compulsory
(ii) Question number 1 to 15 are of 2 marks each
(iii) Question number 16 to 25 are of 4 marks each
(iv) Question number 26 to 30 are of 6 marks each
 

Q1) A coin is tossed 12 times. What is the Probability of getting exactly 8 tails? (Marks 2)
Ans1)
When a coin is tossed, we have S = {H, T}
p = P (getting a tail) = 1/2
... q = 1 - p = 1/2
Here n = 12
Now the Probability Distribution
B(n, p) = B(12, 1/2)
i.e. (q + p)12 where q = p = 1/2
... P (X = 8) = 12C8 . q12 - 8 . p8  
= 12C4 . q4 p8 = 12x11x10x9 x (1/2)4 + 8  
                              4
= 498 = 495
    212   4096

Q2) Three coins are tossed simultaneously. List the sample space for the event. (Marks 2)
Ans2)
Sample space S, of tossing 3 coins simultaneously is given by:-
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Q3) Two cards are drawn without replacement from a well shuffled pack of 52 cards. What is the probability that one is red queen and the other is a king of black colour? (Marks 2)
Ans3)
We know that there are 2 red queens and two kings of black colour in a pack of 52 cards.
P (drawing a red queen) = 2/52 = 1/26
Since the card is not replaced, after the firs card, we have 51 cards left.
... P (drawing a black king) = 2/51
But these two draws are interchangeable.
... The reqd. probability = 2 x 1/26 x 2/51 = 2/(13 x 51) = 2/663

Q4) Find a unit vector perpendicular to both = 3 + - 2 and = 2 + 3 - (Marks 2)
Ans4) Here, = 3 + - 2 and = 2 + 3 -

= x      
3   1   -2
2   3   -1

= 5 - + 7
... = Unit vector  to  and
= ( x )/[ x ] =  (5 - + 7)/(52 + (-1)2 + 72)
= (5 - + 7)/(75) = 1/(53) (5 - + 7)

Q5) Find if = 4 - + and = - 2 + 2 are perpendicular to each other. (Marks 2)
Ans5) For the given vectors
= 4 - + and = - 2 + 2 to be  , we have
. = 0
= (4 - + ) ( - 2 + 2) = 0
= (4)() + (-1)(-2) + (1)(2) = 0
= 4 + 2 + 2 = 0 = = -1

Q6) Find the regression coefficients bxy and byx given that :-
n = 7, x = 24, y = 12, x2 = 374, y2 = 97 and xy = 157. (Marks 2)
Ans6) Given that n = 7, x = 24, y = 12, x2 = 374, y2 = 97 and xy = 157
now, bxy = (xy - 1/n x y)/(y2 - 1/n (y)2)
= 157 - (1/7) x 24 x 12
    97 - (1/7) x 12 x 12
= 1099 - 288 = 811  
    679 - 144     535
Also, byx = (xy - 1/n x y)/(x2 - 1/n (x)2)
= 157 - (1/7) x 24 x 12 = 1099 - 288
   374 - (1/7) x 24 x 24    2618 - 576
= 811  
  2042

Q7) Solve: (1 + x)(1 + y2)dx + (1 + y)(1 + x2)dy = 0. (Marks 2)
Ans7) The given D.E. is
(1 + x)(1 + y2)dx + (1 + y)(1 + x2)dy = 0
Separating the variables, we get:
1 + x dx + 1 + y dy = 0
1 + x2      1 + y2
on integration, we get,

1 + x dx +
1 + x2
1 + y dy = C
1 + y2

Now

1 + x dx = 
1 + x2
   1     dx +
1 + x2
   x     dx
1 + x2

= tan-1 x + 1/2 log (1 + x2)
Similarly,

1 + y dy = tan-1 y + 1/2 log (1 + y2)
1 + y2

Substituting these in (2), we get
tan-1 x + 1/2 log (1 + x2) + tan-1 y + 1/2 log (1 + y2) = c
= tan-1 x + tan-1y + 1/2 log(1 + x2)(1 + y2) = c
which is the reqd. solution of (1).

Q8) Evaluate:- 

(x2 - 4x + 2) dx   (Marks 2)

Ans8) Let I =

(x2 - 4x + 2) dx
((x - 2)2 - 2) dx
((x - 2)2 - (2)2) dx
= (x - 2)/2 ((x - 2)2 - (2)2) - (2)2/2 log |(x - 2) + ((x - 2)2 - (2)2)
= (x - 2)/2 (x2 - 4x + 2) - log |(x - 2) + (x2 - 4x + 2) + c

Q9) Evaluate:- 

(2x + 4) (x2 + 4x + 3) dx   (Marks 2)

Ans9) Let I =

(2x + 4) (x2 + 4x + 3) dx
putting  x2 + 4x + 3 = t
((2x + 4) dx = dt
(t) dt = t / (3/2) + c
= 2/3 (x2 +4x +3)3/2 + c

Q10) Evaluate:- 

     dx    
50 + 2x2    (Marks 2)

Ans10) Let I =

I = 1/2     dx   
25 + x2 
= 1/2     dx   
52 + x2 
= 1/2 [1/5 tan-1 (x/5)] + c
= 1/10 tan-1(x/5) + c

Q11) If y = tan-1 x, Show that (1 + x2)d2y/dx2 + 2x dy/dx = 0  (Marks 2)
Ans11) y = tan-1 x
Differentiating w.r.t. 'x', we get,
dy/dx = 1/(1 + x2)
= (1 + x2)dy/dx = 1
Again Differentiating w.r.t 'x', we get,
(1 + x2)d2y/dx2 + 2x dy/dx = 0

Q12) Discuss the applicability of Rolle's Theorem for the function f(x) = x2/3 on (-1, 1). (Marks 2)
Ans12) Here, f(x) = x2/3 on (-1, 1)
f '(x) = 2/3 x-1/3 = 2 / 3x1/3 which doesn't exist at x = 0
where 0 ]-1, 1[
... The conditions of Rolle's Theorem are not satisfied.
= Rolle's Theorem is not applicable to the given function.

Q13) Evaluate: lim Sin 3x + 7x    (Marks 2)
                     x-0 4x + Sin 2x

Ans13) lim Sin 3x + 7x  
          x-0 4x + Sin 2x
= lim (Sin 3x)/x + 7 
  x-0 4 + (Sin 2x)/x
= lim 3(Sin 3x)/3x + 7 
  x-0 4 + lim 2(Sin 2x)/2x
              x-0
= 3 x 1 + 7 = 10 = 5 
   4 + 2 x 1     6     3

Q14) If

A = 1  -3   2
2   0   2
B = 2  -1  -1
1   0  -1 

find the matrix C such that A + B + C is a zero matrix. (Marks 2)
Ans14) Here

A = 1  -3   2
2   0   2
B = 2  -1  -1
1   0  -1 

we want to find C such that A + B + C = 0
<= C = 0 - A - B = -A - B

= - 1  -3   2
2   0   2
 - 2  -1  -1
1   0  -1 
= -1   3  -2
-2   0  -2
 + -2   1  1
-1   0  1
= -3   4  -1
-3   0  -1
Q15) Construct a 2 x 3 matrix whose elements in the ith row and the jth coloumn are given by:- aij = (3i - j)/2 (Marks 2)
Ans15)
We want a 2 x 3 matrix
= 1 i 2 and 1 j 3
where aij = (3i - j)/2
... a11 = (3 - 1)/2 = 1; a12 = (3 - 2)/2 = 1/2; a13 = (3 - 3)/2 = 0;
a21 = (6 - 1)/2 = 5/2; a22 = (6 - 2)/2 = 2; a23 = (6 - 3)/2 = 3
... The reqd matrix is :-
a11   a12   a13
a21   a22   a23
 =  1     1/2   0
5/2   2    3/2

Q16) Show that the points with position vectors 6 - 7, 16 - 19 - 4, 3 - 6 and 2 - 5 + 10 are coplanar. (Marks 4)
Ans16)
Given A(6 - 7), B(16 - 19 - 4), C(3 - 6) and D(2 - 5 + 10)
Now A, B, C, D are coplanar.
<= , , are coplanar
= P.V. of B - P.V. of A
= (16 - 19 - 4) - (6 - 7) = 10 - 12 - 4
= (3 - 6) - (6 - 7) = -6 + 10 - 6
= (2 - 5 + 10) - (6 - 7) = -4 + 2 + 10
we know that vectors , , are coplanar
iff [ ] = 0

Now [, , ] =  10  -12  -4
-6   10   -6
-4    2   10

= 10(100 + 12) + 12(-60 - 24) - 4(-12 + 40)
= 1120 + (-1008) - 112
=1120 - 1120 = 0
... The points A, B, C, D are coplanar.
[As the scalar triple product of , , is 0]

Q17) Find the foot of the perpendicular from (0, 2, 7) on the line (x + 2)/-1 = (y - 1)/3 = (z - 3)/-2  (Marks 4)
Ans17)
Any point on the line (x + 2)/-1 = (y - 1)/3 = (z - 3)/-2 = K (say)
i.e. P(-K - 2, 3K + 1, -2K + 3)
Let it be the foot of  from Q(0, 2, 7) on (1)
Then d.r.s. of PQ are :-
-k - 2 - 0 , 3k + 1 - 2, -2k + 3 - 7
i.e. -(k + 2), 3k - 1, -(2k + 4)
As PQ is  to line (1),so
-(k + 2) x (-1) + (3k - 1) x 3 - (2k + 4) x (-2) = 0
= k + 2 + 9k - 3 + 4k + 8 = 0
= k = -1/2
... The foot of the  is given by:-
P(1/2 - 2, -3/2 + 1, 1 + 3) = (-3/2, -1/2, 4)

Q18) Three balls are drawn without replacement from a bag containing 5 white and 4 red balls. Find the probability distribution of the number of red balls drawn. (Marks 4)
Ans18)
  5   4
           W   R
Let A be the event of getting a red ball, here n = 3
... following cases arise:-
(i) 0 red and 3 white
 or
(ii) 1red and 2 white
 or
(iii) 2 red and 1 white
 or
(iv) 3 red and 0 white
... the reqd. probability distribution is:-

X 0 1 2 3
P(X) (5C3)/(9C3) (4C1) x (5C2)/(9C3) (4C2) x (5C1)/(9C3) (4C3) x (5C0)/(9C3)
 
X 0 1 2 3
P(X) 5/42 20/42 15/42 2/42
is the reqd. probability distribution.

Q19) Evaluate

      x2       dx   (Marks 4)
(x2 - 4x + 3)

Ans19) Let 

I=       x2       dx
(x2 - 4x + 3)
1 +     4x - 3     dx
      (x2 - 4x + 3)
1 dx  +          4x - 3       dx
(x2 - 4x + 3)
= x +      4x - 3    dx
(x - 1)(x - 3)
= x + 1/2    -1      9    dx
(x - 1)    (x - 3)
(Resolving into Partial fractions)
= x - 1/2   1  dx + 9
x - 1       2
  1    dx
x - 3

                 = x - 1/2 log (x - 1) + 9/2 log (x - 3) + c

Q20) Evaluate

x cos-1 x dx  (Marks 4)

Ans20) Let 

I= x cos-1 x dx
= x2/2 cos-1 x - +  x2   -1    dx  
2    (1 - x2
(Integrating by Parts, taking x as the II nd function)
= 1/2 x2 cos-1 x - 1/2 -x2/(1 - x2) dx
= 1/2 x2 cos-1 x - 1/2 ((1 - x2) - 1)/(1 - x2) dx
= 1/2 x2 cos-1 x - 1/2 (1 - x2) dx + 1/2 dx/(1 - x2)

                 = 1/2 x2 cos-1 x -1/2 [x(1 - x2)/2 + 1/2 sin-1 x] + 1/2 sin-1 x + c
                 = 1/2 [x2 cos-1 x - (x(1 - x2))/2 + 1/2 sin-1 x] + c
                 = 1/4 [2x2 cos-1 x - x(1 - x2) + sin-1 x] + c

Q21) Sketch the region common to the circle x2 + y2 = 16 and the parabola x2 = 6y. Also, find the area of the region using Integration. (Marks 4)
Ans21) Here the curves are
x2 + y2 = 16   ...(1)
x2 = 6y          ...(2)
Solving (1) and (2)
6y + y2 = 16
y2 + 6y - 16 = 0
y2 + 8y - 2y - 16 = 0
y(y + 8) - 2(y + 8) = 0
y = 2, y = -8
... (1) and (2) meet at A(23, 2) and B(-23, 2)
Also, (1) meets x-axis at C(4, 0) and C'(-4, 0)

The reqd. area = QABDO
= 2 Area OAD (because of symmetry)
= 2[area of OEAD - area OEA]

= 2 23
 (16 - x2) dx  -
0
23
 x2/6 dx
0
=2 23
 (42 - x2) - 1/6 x2
0
dx

= 2 [x/2 (16 - x2) + 16/2 sin-1 x/4 - x3/18]o23
= 2[1/2 (23) x 2 + 8 sin-1 23/4 - 243/18]
= [43 + 16 sin-1 (3/2) - (83)/3]
= 43 - (83)/3 + 16 . /3
= ((43)/3 + 16/3) sq. units

Q22) Evaluate o3f(x) dx, when f(x) = |x| + |x - 1| + |x - 2| (Marks 4)
Ans22) Here, f(x) = |x| + |x - 1| + |x - 2|
Now o3f(x)dx = o1f(x)dx + 12f(x)dx + 23f(x)dx 
= o1[x - (x - 1) - (x - 2)]dx + 12[x + (x - 1) - (x - 2)]dx + 23[x + (x - 1) + (x - 2)]dx
= o1(-x + 3)dx + 12(x + 1)dx + 23(3x - 3)dx
= [-x2/2 + 3x]o1 + [x2/2 + x]12 + [3x2/2 - 3x]23
= [(-1/2) + 3] + [2 + 2 - (1/2) - 1] + [(27/2) - 9 - 6 + 6]
= 5/2 + 5/2 + 9/2 = 19/2

Q23) Prove, using the properties of determinants (Marks 4)

b+c   c+a   a+b
c+a   a+b   b+c
a+b   b+c   c+a
 = 2  a   b   c
b   c   a
c   a   b

Ans23) Here

= b+c   c+a   a+b
c+a   a+b   b+c
a+b   b+c   c+a
          applying  C1 C1 + C2 + C3
= 2(a+b+c)   c+a   a+b
2(a+b+c)   a+b   b+c
2(a+b+c)   b+c   c+a
= 2 (a+b+c)   c+a   a+b
(a+b+c)   a+b   b+c
(a+b+c)   b+c   c+a
No we operate: C2 C2 - C1, C3 C3 - C1
= 2 (a+b+c)   -b   -c
(a+b+c)   -c   -a
(a+b+c)   -a   -b
operate C1 C1 + C2 + C3
= 2

a   -b   -c
b   -c   -a
c   -a   -b

= 2(-1)(-1) a   b   c
b   c   a
c   a   b
= 2

a   b   c
b   c   a
c   a   b

= RHS

Q24) A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as x-coordinate. (Marks 4)
Ans24) Given dy/dt = 8 dx/dt    --- (i)
also the equation of curve is
6y = x3 + 2
differentiating w.r.t. 't' 6dy/dt = 3x2dx/dt   --- (ii)
(ii)/(i) = 6 = 3x2/8
= x2 = 16 = x = 4
when x = +4:-
6y = 64 + 2  = y = 11
... one point is (4, 11)
putting x = -4
6y = -64 + 2 = -62
= y = -62/6 = -31/3
... other point is (-4, -31/3)
... two points are:- (4, 11) and (-4, -31/3).

25) Find the derivative of cos (3x + 2) w.r.t. x from the first principal. (Marks 4)
Ans25) Let y = cos (3x + 2)   --- (1)
Let x be a small increment in the value of x and y be the corresponding increment in the value of y.
... y + y = cos [3(x + x) + 2]   ---(2)
subtracting (2) from (1), we get
y = cos [3(x + x) + 2] - cos (3x + 2) 
= -2 sin [(3(x + x) + 2 + 3x + 2)/2] sin [(3 + (x + x) + 2 - 3x - 2)/2]
= 2 sin [3x + 2 + (3/2) x] sin [(3/2) x]
Dividing both sides by x, we get:-
y/x = -2 sin(3x + 2 + (3/2) x) sin (3/2 x)/x
= -sin (3x + 2 + (3/2) x). 3 sin (3/2 x)/[(3/2)x]
Proceeding to limits as x0, we get
dy/dx = -3 sin (3x + 2)

Q26) Given the following pairs of values of variables x & y:

x: 3 5 7 12 20 22 24
y: 30 25 24 16 11 9 5
(i) Find the karl Pearson's coefficient of Correlation.
(ii) Interpret the result
(iii) Confirm your interpretation with the help of a scatter diagram. 
(Marks 6)

Ans26) (i) Here n = 7, Assume A = 13 and B = 17
x y x - A = dx y - B = dy dxdy dx2 dy2
3
5
7
12
20
22
24
30
25
24
16
11
9
5
-10
-8
-6
-1
7
9
11
13
8
7
-1
-6
-8
-12
-130
-64
-42
1
-42
-72
-132
100
64
36
1
49
81
121
169
64
49
1
36
64
144
  2 1 -481 452 527
... P = Karl Pearson coefficient of Correlation :-
= [ndxdy - dx dy] / [ndx2 - (dx)2 . (ndy2 - (dy)2)]
= [7 x (-481) - 2 x 1] / [(7 x 452 - 4) (7 x 527 - 1)]
= (-3369) / (3160 x 3688) = (-3369)/3413.81 = -0.987
Scatter Diagram:-

The scatter diagram shows that y is decreasing in the proportion in which x is increasing. Therefore it is a very high degree negative correlation between the variables x and y, which confirms with the coefficient of correlation.
 
Q27) Find the cartesian as well as vector equations of the planes through the intersection of the planes . (2 + 6) + 12 = 0 and . (3 - + 4) = 0 which are at unit distance from origin.  (Marks 6)
Ans27)
The two planes are:-
2x + 6y + 12 = 0 and 3x - y + 4z = 0
... equation of family of planes through their line of intersection is:-
2x + 6y + 12 + k(3x - y + 4z) = 0
= (2 + 3k)x + (6 - k)y + 4zk + 12 = 0   --- (i)
Now, distance of (i) from origin is:-
= | 12 / [(2 + 3k)2 + (6 - k)2 + 16k2] | = 1(given)
= 144 = (2 + 3k)2 + (6 - k)2 + 16k2
= 144 = 4 + 9k2 + 12k + 36 + k2 - 12k + 16k2
= 104 = 26k2
= k2 = 4 = k = 2
... the 2 planes are:-
(2 + 3 . 2)x + (6 - 2)y + (4z . 2) + 12 = 0
or 8x + 4y + 8z + 12 = 0
or 2x + y + 2z + 3 = 0   --- (ii)
also putting k = -2, we get
(2 + 3 x -2)x + (6 + 2)y - 8z + 12 = 0
or x - 2y + 2z - 3= 0
or x - 2y + 2z - 3 = 0   --- (iii)
... (ii) and (iii) are reqd. cartesian equations and the Vector Equations are:-
. (2 + + 2) + 3 = 0 and . ( - 2 + 2) = 3

Q28) The Slope of the tangent at any point of a curve is times the slope of the straight line joining the point of contact to the origin, formulate the differential equation representing the problem and hence find the equation of the curve. (Marks 6)
Ans28)
Let P(x, y) be any point on curve.
Then the slope of the tangent at P is dy/dx.
if O(0, 0) be origin, then the slope of
OP = (y - 0)/(x - 0) = y/x
By hypothesis,
dy/dx = y/x
= dy/y = dx/x
Integrating we get
log y = log x + log c
= log x + log c
=log cx 
= y = cx  is the reqd. equation of the curve.

Q29) If
A = 

1  2  2
2  1  2
2  2  1

find A-1 and hence prove that A2 - 4A - 5I = 0. (Marks 6)
Ans29) 

A = 

1  2  2
2  1  2
2  2  1

= |A| = 1(1 - 4) - 2(2 - 4) + 2(4 - 2)
= -3 + 4 + 4 = 5 0
= A-1 exist and it is given by:-
A-1 = 1/|A| (adj A) = 1/5 (adj A)
Now finding the various co-factors we get

adj A =  -3  2  2
2  -3  2
2  2  -3
... A-1 = 1/5 -3  2  2
2  -3  2
2  2  -3
= 5A-1 = 1-4   2     2
 2   1-4    2
 2     2   1-4
= 5A-1 =

1  2  2
2  1  2
2  2  1

 -  4   2   2
2   4   2
2   2   4
= A - 4I

= 5A-1A = (A - 4I)A   (By Post Multiplying by A)
= 5I = A2 - 4A
or A2 - 4A - 5I = 03x3

Q30) An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when the depth of the tank is half of its width. (Marks 6)
Ans30)
Let 'a' be the side of the square base and 'h' be the depth of the given open tank. By hypothesis;
v = a2h (given) (i.e. v is a constant)
= h = v/a2 
Also surface area of the tank = a2 + 4ah sq units
if Rs. P per square unit be the cost of the material, then total cost is
c = (a2 + 4ah)P = (a2 + 4av/a2)P = (a2 + 4v/a)P (where P is a constant)
Now c is a function of 'a' and we want to find the least value of c. Differentiating c w.r.t x, we get
dc/da = P(2a - 4v/a2)
Now, dc/da = 0 = 2a = 4v/a2
= a3 = 2v
Now, d2c/da2 = P(2 + 8v/a3)
= P(2 + 8(a3/2)/a3) (when a3 = 2v)
= 6P 0 (...  P 0)
= c is minimum
Since h = v/a2 = a3/2/a2 = a/2
depth of the tank is equal to half of its width.