Search Freshers Jobs

# CBSE Set Qa1a Q23 Sample Test Papers For Class 12th for students online

Latest for students online. All these are just samples for prepration for exams only. These are not actual papers.

Q23) Show that the curve x = y2 and xy = K cut at right angle, if 8K2 = 1
Ans23) Let (a, b) be the points of intersection of the curves
x = y2    - (i)
xy = K   - (ii)
...  the two curves are orthogonal
... [(dy/dx)1 . (dy/dx)2]at(a, b) = -1   - (iii)
differentiating (i) w.r.t. x
2y dy/dx = 1
... (dy/dx)at (a, b) = 1/2b
differentiating (ii) w.r.t. x
y + x dy/dx = 0
... dy/dx = -(y/x)
(dy/dx)at (a, b) = -b/a
hence from (iii)
1/2b . (-b/a) = -1
= a = 1/2
... from (i)  1/2 = b2
= b = +1/2
... from (ii)
ab = K
= 1/2(+ 1/2) = K
squaring
1/8 = K2
or 8K2 = 1

Q24) Find the intervals in which the function f(x) is (a) increasing; (b) decreasing:
f(x) = 2x3 + 9x2 + 12x + 20
Ans24) f(x) = 2x3 + 9x2 + 12x + 20
... f '(x) = 6x2 + 18x + 12
= 6(x2 + 3x + 2)
= 6(x2 + 2x + x + 2)
= 6(x + 1)(x + 2)
putting f '(x) = 0
= x = -2 , -1
which divide the number line in following intervals
(-, -2) , (-2, -1) and (-1, )
Let us see the sign of f'(x) in these intervals

 Interval sign of  (x + 1) sign of (x + 2) sign of  f '(x) = 6(x + 1)(x + 2) Nature of f '(x) (-, -2) - - + (-2, -1) - + - (-1, ) + + +

... f(x) is increasing on the (-, -2) U (1, ) and decreasing on (-2, -1)

Q25) For observations of pairs (x, y) of the variables x and y, the following results are obtained.
x = 100, y = 150, x2 = 1500, xy = 1200 and n = 50. Find the equation of the line of regression of y on x. Estimate the value of y when x = 15
Ans25) We have; x = 100, y = 150, x2 = 1500, xy = 1200 and n = 50
= x/n = 100/50 = 2
y = y/n = 150/50 = 3
Also byx = (xy - (1/n)xy)/(x2 - (1/n)(x)2)
= (1200 - (1/50) x 10 x 150)/(1500 - (1/50) x 100 x 150) = (1200 - 300)/(1500 - 200)
... The eq of the line of regression of y on x is
y - y = bxy (x - )
= y - 3 = 9/13 (x - 2)
when x = 15, y = 3 + 9/13 (15 - 2) = 12

Q26) Using integration find the area of the region bounded between the line x = 2 and the parabola y2 = 8x
Ans26)

 The reqd area (area OABC) = 2 2   ydx  (... y2 = 8x is curve symmetrical w.r.t. x axis) 0 = 2 2   (8xdx)  (area OABC = 2 are OAB) 0 = 42 x3/2 3/2 2 0

= (42 . 2)/3 [23/2 - 0]
= 8/3 . (21/2 . 23/2) = (8 . 4)/3 = 32/3 sq units

Q27) Evaluate

 /2     (cos x dx)         sin x + (cos x) 0

State that property of definite which you have used
Ans27)
Let I =

 /2     (cos x dx)            (i)    (sin x) + (cos x) 0

we use the property

 a  f(x)dx = 0 a  f(a - x)dx 0

Using it (i) =

 /2                (cos (/2 - x))            dx    (sin (/2 - x)) + (cos (/2 - x)) 0 /2          (sin x)    dx    (ii)    (cos x) + (sin x) 0 Now (i) + (ii) /2    (cos x) + (sin x)dx    (cos x) + (sin x) 0 /2    1 . dx  = [x]o/2  = /2 0 = I = /4

Q28) Evaluate

 dx     x(x4 + 1)

Ans28)

 dx      =  x(x4 + 1) x3dx     x4(x4 + 1)
 = 1/4 4x3 dx     x4(x4 + 1)

Putting x4 = t
4x3dx = dt

 = 1/4 dt     t(t + 1) = 1/4 (t + 1) - t dt    t(t + 1) = 1/4 (1/t - 1/(t + 1)) dt

= 1/4 log (t/(t + 1)) + c
= 1/4 log |x4/(x4 + 1)|

Q29) Find the value of

 2   (x + 4)dx  as limit of sums 0

Ans29) Here f(x) = x + 4, a = 0, b = 2
... nh = b - a = 2
By definition of limit of sums,

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

where nh = 2

 = 2   (x + 4) dx  0
 = h[4 + (h + 4) + (2h + 4) + ... + (n - 1)n + 4] = h[4n + h(1 + 2 + 3 + ... + (n - 1)] = h[4n + h . (n - 1)n/2] = 4nh + 1/2 nh(nh - h)] , where nh = 2

= 4 x 2 + 1/2 x 2 x (2 - 0) = 10

Q30) Solve the differential equation x(1 + y2)dx - y(1 + x2)dy = 0. Given that y = 0 when x = 1
Ans30) The given D.E is
x(1 + y2)dx - y(1 + x2)dy = 0   - (i)
= x/(1 + x2) dx - y/(1 + y2) dy = 0
= x/(1 + x2) dx = y/(1 + y2) dy   - (ii)
integrating

 1/2 2x/(1 + x2) dx = 1/2 2y/(1 + y2) dy

= 1/2 log (1 + x2) = 1/2 log (1 + y2) + log c
= 1/2 [log (1 + x2) - log (1 + y2)] = log c
= log ((1 +x2)/(1 + y2)) = log c
= (1 + x2) = c(1 + y2)
given at x = 1 , y = 0
= 2 = c
... the reqd sol. is
(1 + x2) = (2(1 = y2))
or (1 + x2) = 2(1 + y2

Q31) 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:
= (8 + 3) - (9 + 16) + (10 + 7)
and = 15 + 29 + 5 + (3 + 8 - 5)
Ans31) The lines of shortest distance is the line whose intercept between given two lines is the shortest distance between these skew lines.
Here the two lines are
= (8 + 3) - (9 + 16) + (10 + 7)
and = 15 + 29 + 5 + (3 + 8 - 5)
or (x - 8)/3 = (y + 9)/-16 = (z - 10)/7 =
and (x - 15)/3 = (y - 29)/8 = (z - 5)/-5 =

Let AB be the intercept of line of S.D. between given two lines. Co-ordinates of A are (3 + 15, -16 - 9, 7 + 10)
and co-ordinates of B are (3 + 15, 8 + 29, -5 + 5)
... Dr's of AB are
3 - 3 - 7 : -16 - 8 - 38 : 7 + 5 + 5
... AB is  L1
3(3 - 3 - 7) - 16(-16 - 8 - 38) + 7(7 + 5 + 5) = 0
= 9 - 9 - 21 + 256 + 128 + 608 + 49 + 35 + 35 = 0
= 314 + 154 + 622 = 0
= 157 + 77 + 311 = 0   - (i)
Also AB  L2
= 3(3 - 3 - 7) + 8(-16 - 8 - 38) - 5(7 + 5 + 5) = 0
= 9 - 9 - 21 - 128 - 64 - 304 - 35 - 25 - 25 = 0
or -154 - 98 - 350 = 0
or 11 + 7 + 25 = 0   - (ii)
Solving (i) and (ii)
157 + 77 + 311 = 0
11 + 7 + 25 = 0
or /(1925 - 2177) = -/(3925 - 3421) = 1/(847 - 1099)
= /-252 = -/504 = 1/-252
= = 1, = 2
... Co-ordinates of A are (11, -25, 17)
and co-ordinates of B are (21, 45, -5)
... shortest distance = AB
= ((10)2 + (70)2 + (-22)2)
= (100 + 4900 + 484)
= 5484
= (4 x 1371) = (4 x 3 x 457)
= 2 (1371)
and vector line of AB is
= (11 - 25 + 17) + l(10 + 70 - 22)

Q32) Solve the following system of equation using matrix method
3x - 4y + 2z = -1
2x + 3y + 5z = 7
x + z = 2
Ans32) Given eq are
3x - 4y + 2z = -1
2x + 3y + 5z = 7
x + z = 2
These imply Ax = B   - (i)

 where A 3  -4  2 2  3  5 1  0  1 ; x = x y z and B -1 7 2
 = |A| = 3  -4  2 2  3  5 1  0  1

= 1[-20 - 6] + 1[9 + 8] = -9 0
= A-1 exists and
= A-1 = 1/|A| adj A

 = -1/9 3   4  -26 3   1  -11 -3 -4   17

... (i) = x = A-1 B

 = -1/9 3   4  -26 3   1  -11 -3 -4   17 -1 7 2
 = -1/9 -3   +28   -52 -3    +7   -22  3   -28   +34 = -1/9 -27 -18 9 = 3 2 -1

= x = 3; y = 2; z = -1 is the reqd sol.

Q33) Solve the differential equation (x + y)dy + (x - y)dx = 0, given that y = 1 when x = 1
Ans33) The given D.E. is
(x + y)dy + (x - y)dx = 0   - (i)
= dy/dx = (y - x)/(y + x)   - (ii)
This is homogeneous differential equation
Putting y = vx
= dy/dx = y + x dv/dx   - (iii)
Now (ii) and (iii) =
v + x dv/dx = (vx - x)/(vx + x) = (v - 1)/(v + 1)
= x dv/dx = (v - 1)/(v + 1) - v = -(v2 + 1)/(v + 1)
or dx/x = -(v + 1)/(v2 + 1) dv
Integrating both sides

 dx/x  = - (v + 1)/(v2 + 1) dv
 or dx/x = - 1/2 2v/(v2 + 1) dv - 1/(v2 + 1) dv + c

or log x = -1/2 log |v2 + 1| - tan-1 v + c
or log x = -1/2 log |y2/x2 + 1| - tan-1 (y/x) + c
or log x = -1/2 log |(y2 + x2)/x2| - tan-1 (y/x) + c
putting x = 1 and y = 1
0 = -1/2 log |2/1| - /4 + c
or c = /4 + 1/2 log 2
... reqd sol is
log x = -1/2 log |(y2 + x2)/x2| - tan-1 (y/x) + /4 + 1/2 log 2
or log (x2 + y2) + 2tan-1 (y/x) = /4 + log 2

Q34) A wire of length 25 m is to be cut in to two pieces one of the pieces is to be made into a square and other into a circle. What should be the lengths of the two pieces so that the combined area of square and the circle is minimum?
Ans34) Let the two pieces be x and 25 - x meters
Let x length be made into square and 25 - x be made into a circle.
... x = 4a, where a is the side of a square and 25 - x = 2r where r is the radius of the circle
A = Area of square + Area of circle
= a2 + r2
= (x/4)2 + ((25 - x)/2)2 = x2/16 + (25 - x)2/4
= dA/dx = x/8 - (25 - x)/2   - (i)
= d2A/dx2 = 1/8 + 1/2   - (ii)
for max. or min. dA/dx = 0
... (i) = x/8 - (25 - x)/2 = 0
= x - 4(25 - x) = 0
= ( + 4)x = 100
= x = 100/( + 4)
for this value of x, d2A/dx2 0
... Area is minimum , when
x = 100/( + 4) and 25 - x = 25/( + 4)
Hence the two pieces are of length 100/( + 4) and 25/( + 4)

Q35) Calculate the co-efficient of correlation between x and  y

 x : 10 10 11 12 13 13 13 12 12 12 y : 11 13 14 16 16 16 15 14 13 13

Ans35) Here n = 10
Taking a = 12; b = 14; we have
u = x - a = x - 12 and  v = y - b = y - 14

 x y u = x - 12 v = y - 14 uv u2 v2 10 11 -2 -3 6 4 9 10 13 -2 -1 2 4 1 11 14 -1 0 0 1 0 12 16 0 2 0 0 4 13 16 1 2 2 1 4 13 16 1 2 2 1 4 13 15 1 1 1 1 1 12 14 0 0 0 0 0 12 13 0 -1 0 0 1 12 13 0 -1 0 0 1 Total u = -2 v = 1 uv = 13 u2 = 12 v2 = 25

P(x, y) = p(u, v) = (uv - (1/n)uv) / (u2 - (1/n)u2)(v2 - (1/n)v2)
= (13 - (1/10)(-2)(1)) / (12 - (1/10)(4)) (25 - (1/10) (1))
= (13 + 0.2) / (12 - 0.4) (25 - 0.1)
= 13.2 / (11.6) (24.9)
= 13.2/16.99 = 0.78