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

Maths Class - XII  (CBSE)
You are on Set no 1 Answer 1 to 8

Q1) Find from first principle the derivative of (x² + 1)/x w.r.t. x
Ans1) Let y = (x² + 1)/x = x + 1/x   - (i)
Let x be a small incremental in the value of x and y be the corresponding incremental in the value of y, then
y + y = x + x + 1/(x + x)   - (ii)
Now (ii) - (i)
= y = x + 1/(x + x) - 1/x
y = x - x/(x(x + x)) = y/x = 1 - 1/(x(x + x))

Q2) Evaluate : 

1 - cos 5x 1 - cos 6x

Ans2) 

	1 - cos 5x    =  1 - cos 6x			2sin2 (5x/2) 2sin2 (3x)
	(5/2)2[sin (5x/2)/(5x/2)]2      32 [sin 3x/3x]2		= 25/36

Q3) Using differentials, find the approximate value of 29
Ans3) To find 29
Let x = 27 and x = 2 then x + x = 27 + 2 = 29
Taking y = x^1/3, we get dy/dx = 1/3 x^-2/3
Also y + y = (x + x)^1/3 = 29^1/3
where y = x^1/3 = 27^1/3 = 3
.^.. 3 + y = 29^1/3
Again y = dydx x = 1/3 x^-2/3 . 2 = 2/3(27)^-2/3 = 2/27
29^1/3 = 3 + y = 3 + (2/27) = 83/27

Q4) Solve the differential equation:
x (dy/dx) = x + y
Ans4) x (dy/dx) = x + y
= dy/dx = (x + y)/x   - (i)
Put y = vx
= dy/dx = v + x dv/dx   - (ii)
Now (i) and (ii) = v + x dv/dx = (x + vx)/x = 1 + v
= x dv/dx = 1
= dv = dx/x
Integrating, we get
v = log x + log c = log cx
= y/x = log cx = y = xlog cx is the reqd. sol.

Q5)  Solve the differential equation dy/dx = ysin 2x given that y(0) = 1
Ans5) dy/dx = ysin 2x
= dy/y = sin 2x . dx + c
Integrating, we get
log y = -1/2 . cos 2x + c   - (i)
By hypothesis, y(0) = 1 i.e. x = 0 = y = 1
log 1 = -1/2 cos 0 + c = c = 1/2
Thus (i) = log y = 1/2 -1/2 cos 2x = 1/2 (1 - cos 2x)
= 1/2 . 2sin² x = sin² x = y = e^{sin2 x} is the reqd. sol.

Q6) Using determinants, find that the area of the triangle with vertices (-3, 5), (3, -6), (7, 2).
Ans6) The area of the given triangle

= 1/2

-3  5  1 3  -6  1 7  2  1

operates : R₁ R₁ - R₂; R₂ R₂ - R₃

= 1/2

-6  11  0 -4  -8  0 7   2   1

Expanding along C3 = 1/2 . 1

-6  11 -4  -8

= 1/2 (48 + 44) = 46 sq units

Q7) By using elementary row transformations, find the inverse of the matrix

A =

5  2 2  1

Ans7) Now IA = A

1  0 0  1

A =

5  2 2  1

= R₁ R₁ - 2R₂ 

1  -2 0  1

A =

1  0 2  1

=  R₂ R₂ - 2R₁ 

1  -2 -2  5

A =

1  0 0  1

A-1 =

1  -2 -2  5

Pls verify : A.A^-1 = I

Q8) If = - 2 + 3 and = 2 + 3 - 5 then find x . Verify that and x are perpendicular to each other. 
Ans8) Given  = - 2 + 3 and = 2 + 3 - 5

... x =

1  -2  3 2  3  -5

= + 11 + 7

Now . ( x ) = ( - 2 + 3) . ( + 11 + 7)
= 1 . 1 + (-2) . 11 + 3 . 7 = 0 =   |  x

Q9) Using vector show that the line segment joining the mid-points of two side of a triangle is parallel to the third side.
Ans9) 

Let , , be position vectors of vertices A, B, and C. Let D and E be the mid points of AB and AC.
Thus position vector of D = ( + )/2 = ( + )/2
also position vector of E = ( + )/2 = ( + )/2
Now = - = (( + )/2) - (( + )/2)
= 1/2 [ - ]
= 1/2 BC
.^.. = 1/2
= || and || = 1/2|| i.e. DE = 1/2 BC.

Q10) Evaluate

1     ex     dx 0 1 + e2x

Ans10) Let I =

1     ex     dx 0 1 + e2x

Putting ex = t = ex dx = dt
Also x = 0 = t = 1 and x = 1 = t = e

e     dt     = [tan-1 t]1e dx 1 1 + t2

= tan^-1 e - tan^-1 1 = tan^-1 [(e - 1)/(1 + e)]

Q11) Evaluate

1         dx  (2 + 2x - x2)

Ans11) Let I = 

1         dx  (2 + 2x - x2)

1         dx  = sin-1 [(x - 1)/3] + c  ((3)2 - (x - 1)2)

Q12) In a group there are 3 men and 2 women 3 person are selected at random from this group. Find the probability that 1 man and 2 women or 2 men and 1 woman are selected.
Ans12) Given Men (3) and women (2)
Total no. of cases = ⁵C₃
P(2 M and 1 W or 1 M and 2 W) = P(2 M and 1 W) + P(1 M and 2 W)
= (³C₂ . ²C₁)/⁵C₃ + (³C₁ . ²C₂)/⁵C₃ = (3 . 2)/10 + (3 . 1)/10 = 9/10

Q13) One card is drawn from a well shuffled pack of 52 cards. If E is the event the card drawn is a king or queen and F is the event the card drawn is a queen or an ace, then find the probability of the conditional event E/F.
Ans13) Event E : The card is King or Queen
Event : F The card is Queen or Ace
= E F : The card is Queen
P(E) = P(F) = 8/52 = 2/13
P(E F) = 4/52 = 1/13
Now P(E/F) = P(E  F)/P(F)
= (1/13)/(2/13) = 1/2

Q14) A die is thrown 7 times, if getting an "even number" is success, find the probability of getting at least 6 successes.
Ans14) Let p = Prob. of getting an even number
= 3/6 = 1/2
= q = 1 - p = 1/2
Also n = 7
.^.. Binomial distribution is
B(n, p) i.e. (q, p)ⁿ = (1/2 + 1/2)⁷
Now Prob. of getting atleast 6 successes
= p(X 6) = p(X = 6) + p(X = 7)
= ⁷C₆qp⁶ + ⁷C₇p⁷ = 7(1/2)(1/2)⁶ + 1(1/2)⁷
= (7 + 1)/2⁷ = 1/16

Q15) The two lines of regression for a bivariate distribution (x, y) are 3x + 2y = 7 and x + 4y = 9. Find the regression coefficient b_yx and b_xy
Ans15) Let the line of regression of x upon y be : x = -2/3 y + 7/3
.^.. b_xy = -2/3
and the line of regression of y on x be : y = -1/4 x + 9/4,
.^.. b_yx = -1/4
Now b_yx . b_xy = 1/6 < 1
.^.. our supposition is correct

Q16) Find the radius of the circular section of the sphere x² + y² + z² = 49 cut by the plane 2x + 3y - z - 514 = 0
Ans16)  

2x + 3y - z = 514
Given Eq of sphere is x² + y² + z² = 49   - (i)
.^.. Its centre O (0, 0, 0) and radius, r = 7
Also eq. of the plane cutting (i) is
2x + 3y - z - 514 = 0   - (ii)
Now p =  |  from O on (ii)

=

2 . 0 + 3 . 0 - 0 - 514 (22 + 32 + (-1)2)

= 514/14 = 5

Now R = Radius of the circular section
= (r² - p²) = (7² - 5²) = 24 = 26

Q17) Find dy/dx when y = x^{sin x - cos x} + (x² - 1)/(x² + 1)
Ans17) Here : y = x^{sin x - cos x} + (x² - 1)/(x² + 1)
= u + v   - (i)
where u = x^{sin x - cos x} and v = (x² - 1)/(x² + 1)
= log u = (sin x - cos x)log x and v = 1 - 2/(x² + 1)
= 1/u . du/dx = (cos x + sin x)log x + (sin x - cos x)/x
= du/dx = x^{sin x - cos x} [(cos x + sin x)log x + (sin x - cos x)/x]
and dv/dx = 0 - 2(-1)/(x² + 1)² 2x = 4x/(x² + 1)²   - (ii)
from (i), dy/dx = du/dx + dv/dx   - (iii)
= x^{sin x - cos x} [(cos x + sin x)log x + (sin x - cos x)/x] + 4x/(x² + 1)² 

Q18) For the function f(x) = 2x³ - 24x + 5, find
(a) the interval (s) where it is increasing;
(b) the interval (s) where it is decreasing
Ans18) f(x) = 2x³ - 24x + 5
.^.. f'(x) = 6(x² - 4)
= 6(x + 2)(x - 2)
putting f(x') = 0
= x = -2 , 2
which divide the number line in following intervals
(-, -2) , (-2, 2) and (2, )
Let us see the sign of f'(x) in these intervals

Interval	sign of  (x + 2)	sign of (x - 2)	sign of  f'(x) = 6(x + 2)(x - 2)	Nature of f'(x)
(-, -2)	-	-	+
(-2, 2)	+	-	-
(2, )	+	+	+

.^.. f(x) is on the (-, -2) U (2, ) and on (-2, 2)

Q19) A box containing 16 bulbs out of which 4 bulbs are defective, 3 bulbs are drawn one by one from the box without replacement. Find the probability distribution of the number of defective bulbs drawn.
Ans19)  

3 bulbs out of 16 can be drawn in ¹⁶C₃ ways
.^.. n(s) = ¹⁶C₃
considering getting defective bulbs to be a success. Let x denote no. of success, then prob. distribution of no. of successes is given by:

Pls note:-
P(0) + P(1) + P(2) + P(3) = 1

Q20) Prove that

1  x  x3 1  y  y3 1  z  z3

Ans20) 

=

1  x  x3 1  y  y3 1  z  z3

operates : R₁ R₁ - R₂; R₂ R₂ - R₃

=

0   x-y   x3-y3 0   y-z   y3-z3 1    z       z3

= = (x-y)(y-z)

0   1   x2+xy+y2 0   1   y2+yz+z2 1   z              z3

expanding by C1

= = (x-y)(y-z)

1   x2+xy+y2 1   y2+yz+z2

= (x - y)(y - z)[y² + yz + z² - (x² + xy +y²)]
= (x - y)(y - z)[y(z - x) + (z² - x²)]
= (x - y)(y - z)(z - x)(x + y + z)

Q21) Prove{( + ) x ( + )} . ( + ) = 2[ ]
Ans21) LHS = {( + ) x ( + )} . ( + )
= [ x + x + x + x ] . [ + ]
= [ x + x + x ] . [ + ]    ...(^..^. x = 0)
= x . + x . + x . + x . + x . + x .
= [, , ] + [, , ]    ...(^..^. [, , ] = [, , ] = [, , ] = [, , ] = 0)
= [, , ] + [, , ]
= 2[, , ]

Q22) Evaluate

I =

dx       0  5 + 2cos x

Ans22) Let  

I =

dx       0  5 + 2cos x

Putting tan x/2 = t
= sec² (x/2).(1/2)dx = dt
= dx = 2dt/sec² (x + 2) = 2dt/(1 + tan² x/2) = 2dt/(1 + t²)
and cos x = (1 - tan² x/2)/(1 + tan² x/2) = (1 - t²)/(1 + t²)
Also x = 0 = t = 0 and x =   = t =

I =		2dt                    0   (1 + t2)(5 + 2 . (1 - t2)/(1 + t2)
= 2		dt       0   (7 + 3t2)
= 2/3		dt       0   t2 + (7/3)2

= 2/3 . 1/(7/3)

tan-1 (1/(7/3))

0

= 2/21 [/2 - 0] = /21

Q23) Evaluate :

(tan x) dx

Ans23) Let I =

(tan x) dx

Put : (tan x) = t
= tan x = t² = sec² xdx = 2tdt
= dx = 2tdt/(1 + t⁴)

I =		t 2tdt   = (1 + t4)		2t2   dt 1 + t4
=		(t2 - 1) + (t2 + 1) dt         1 + t4		t2 - 1 dt + t2 + 1 dt t4 + 1       t4 + 1
=		(1 - 1/t2) dt + t2 + 1/t2		1 + 1/t2 dt t2 + 1/t2
or I = I1 + I2   (i)
I1 =		1 - 1/t2 dt =  t2 + 1/t2		1 - 1/t2    dt =  (t + 1/t)2 - 2
putting t + 1/t = u = (1 - 1/t2)dt = du
I1 =		du      = 1/22 log u2 - (2)2		u - 2 u + 2

= 1/22 log

t + 1/t - 2 t + 1/t + 2

and I2 =		(1 + 1/t2) dt putting t - 1/t = v  = (1 + 1/t2)dt = dv (t2 + 1/t2)
... I2 =		dv       = 1/2 tan-1 (v/2) v2 + (2)2

... I = I1 + I2 = 1/22 log

tan x - (2tan x) + 1 tan x + (2tan x) + 1

+ 1/2 tan-1 (tan2 x - 1)  + c                  tan x

Q24) Find the value of 

3 (2x2 + 5x) dx 1

Ans24) Here f(x) = 2x² + 5x . a = 1, b = 3
.^.. nh = b - a = 2

By def

b   f(x)dx = lim h [f(a) + f)a + h) + ... + f(a + (n - 1)h)] a              h-0

where nh = b - a

= I =		[{2 . 12 + 5 . 1} + {2(1 + h)2 + 5 . (1 + h)} + ... + {2 . (1 + (n-1)h)2 + 5 . (1 + (n - 1)h)}]
=		[7n + ah(1 + 2 + 3 + ... + (n - 1)) + 2h2{12 + 22 + 32 + ... + (n - 1)2}]
=		[7nh + 9h2 . (n - 1)n/n + 2h3 (n - 1)n(2n - 1)/6]
where nh = 2
=		[7nh + 9/2 . (nh - h) . nh + 1/3 . (nh - h) . nh . (2nh - 0)]

= 7 . 2 + 9/2(2 - 0) . 2 + 1/3(2 - 0) . 2 . (2 . 2 - 0)
= 14 + 18 + 16/3 = 112/3

Q25) A population grows at the rate of 2% per year. How long does it take for the population to double itself.
Ans25) Rate of growth of population = 2/100
= dP/dt = 2/100 P, where P is population at any time t
= 50 dP/P = dt
integrating, we get
50 log P = t + c
Let P =P_o when t = 0, where Po is initial Population
.^.. 50 log P_o = C
= 50 log P = t + 50 log P_o
Let t = t₁ when P = 2P_o
= 50 log 2P_o = t₁ + 50 log P_o
= t₁ = 50(log 2P_o - log P_o)
= 50 log 2 yrs

Q26) Define the line of shortest distance between two skew lines. Find the shortest distance and the vector equation of the line of shortest distance between the lines given by 
= (3 + 8 + 3) + (3 - + )
and r = (-3 - 7 + 6) + (-3 + 2 + 4)
Ans26) Given lines
= (3 + 8 + 3) + (3 - + )
= (-3 - 7 + 6) + (-3 + 2 + 4)
can be written in cartesian form
i.e. (x - 3)/3 = (y - 8)/-1 = (z - 3)/1 and (x + 3)/-3 = (y + 7)/2 = (z - 6)/4
Let (x - 3)/3 = (y - 8)/-1 = (z - 3)/1 =
Then, x = (3 + 3); y = (- - 8) and z = ( + 3)
Thus a general point on this line is
P(3 + 3, - - 8, + 3)
Again, Let (x + 3)/-3 = (y + 7)/2 = (z - 6)/4 =
Then x = (-3 - 3); y = (2 - 7) and z = (4 + 6)
Thus a general point on this line is 
Q(-3 - 3, 2 - 7, 4 + 6)
Direction ratios of PQ are - (-6 - 3 - 3, 2 + - 15, 4 - + 3)
Let us choose and in such a way that PQ is perpendicular to each of the given lines, then
3(-6 - 3 - 3) - 1(2 + - 15) + 1(4 - + 3) = 0
and -3(-6 - 3 - 3) + 2(2 + - 15) + 4(4 - + 3) = 0
These equations on simplification gives
11 + 7 = 0
7 + 29 = 0
solving these equations, we get = 0, = 0 so the required points where the line of shortest distance meets the given lines are P(3, 8, 3) and Q(-3, -7, 6)
Shortest distance = PQ = (62 + 152 + (-3)2) = 330
The lines of S.D. passes through P and Q
.^.. its equation is
(x - 3)/(-3 - 3) = (y - 8)/(-15) = (z - 3)/(6 - 3)
= (x - 3)/-6 = (y - 8)/-15 = (z - 3)/3  = (x - 3)/2 = (y - 8)/15 = (z - 3)/-1
= = (3 + 8 + 3) + (2 + 5 - )

Q27) Find the karl Pearson's coefficient of correlation between X and Y for the following data:

Also state whether Y increases or decreases with increase in X
Ans27) Here n = 7
Taking a = 11, b = 8, we have
X = x - a = x - 11 and Y = y - b = y - 8

Now P = (XY - (1/3)XY)/(X² - (1/n)(X)²) . (Y² - (1/n)(Y)²)
= (13 - (1/7) . (-1) . 0)/((63 - 1/7) . 1 (38 - (1/7)0)) = 13/((440/7 . 38) = 0.27
.^.. P(x, y) is positive, .^.. as x increases y also increases

Q28) If 

A =

1  -2  0 2  1  3 0  -2  1

find A^-1. Using A^-1, solve the system of linear equations
x - 2y = 10
2x + y + 3z = 8
-2y + z = 7
Ans28) Given

A =

1  -2  0 2  1  3 0  -2  1

= |A| = 1(1 + 6) + 2(2 - 0) = 11 0
= A^-1 exists

and A-1 = 1/|A| adj A = 1/11

7  2  -6 -2  1  -3 -4  2  5

the given system of linear eqs.

1  -2  0 2  1  3 0  -2  1

x y z

=

10 8 7

= n

x y z

=

10 8 7

=

x y z

= A-1

10 8 7

= 1/11

7  2  -6 -2  1  -3 -4  2  5

10 8 7

= 1/11

44 -33 11

=

4 -3 1

= x = 4, y = -3, z = 1

Q29) Find the point on the curve y² = 4x which is nearest to the point (2, -8)
Ans29) Let P(x, y) be the point nearest to the point A(2, -8). Let S = PA then
S = ((x - 2)² + (y + 8)²)
for S to be minimum , S² has to be minimum
Now S² = (x - 2)² + (y + 8)²
or S² = (y²/4 - 2)² + (y + 8)²   (^..^. (x, y) lies on the parabola y² = 4x)
for S² to be minimum
ds²/dy = 0
Now ds²/dy = 2(y²/4 - 2)(2y/4) + 2(y + 8)
= y³/4 - 2y + 2y + 16
= y³/4 - 16 = (y³ + 64)/4
putting ds²/dy = 0 = y³ + 64 = 0
= y³ = -64
= y = -4
and thus 16 = 4x = x = 4
.^.. Pt is (4, -4)
also d²s²/dy² = 3y²/4 (+ve)
= y = -4 is the point of minima
hence (4, -4) is the point nearest to the point (2, -8)

Q30) Draw a rough sketch of the region {(x, y) : y² 3x, 3x² + 3y² 16} and find the area enclosed by the region using method of integration.
Ans30) The given curves are
y² = 3x   - (i)
x² + y² = 16/3   - (ii)
These intersect where
x² + 3x = 16/3 = 3x² + 9x - 16 = 0
= x = (-9 + 273)/6 = a (say)
.^.. the required area = area OABCO = 2(area OABO)

= 2

a   y1 dx + 0

4/3   y2 a

= 2

a   (3x) dx +  0

4/3   (16/3 - x2) dx a

= 2 [3 . 2/3 {x3/2}

a 0

+ {(x(16/3) - x2)/2 + (16/3)/2 sin-1 x/(4/3)}

4/3  a

]

= (4/3) a^2/3 + 16/3 . sin^-1 (1 - a)(16/3) - a² - 16/3 sin^-1 (3a/4)
where a = (-9 + 273)/6