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 |
||
|
= 4 |
 |
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
|
|
|
State that property of definite
which you have used
Ans27) Let I =
|
|
|
we use the property
|
|
a |
|
a f(a - x)dx 0 |
Using it (i) =
|
|
|
|
|
| Now (i) + (ii) | |
|
 /2(cos
x) + (sin x)dx(cos
x) + (sin x)0 |
|
/2 1 . dx = [x]o /2
= /20 |
| = | I = /4 |
| |
dx |
Ans28)
| |
dx = |
|
x3dx x4(x4 + 1) |
| = 1/4 | |
4x3
dx |
Putting x4 = t
4x3dx = dt
| = 1/4 | |
dt |
| = 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)|
|
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 |
|
; x = | |
x y z |
|
and B | |
-1 7 2 |
|
| = |A| = | |
3 -4 2 |
|
= 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 |
|
-1 |
|
| = -1/9 | |
-3 +28
-52 |
|
= -1/9 | |
-27 |
|
= | |
3 |
|
= 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
