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

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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

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