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St. JOSEPHS COLLEGE OF ENGINEERING, CHENNAI- 119
St. JOSEPHS INSTITUTE OF TECHNOLOGY, CHENNAI- 119
I-YEAR B.E./ B.TECH. ( COMMON TO ALL BRANCHES )
MATHEMATICS I (MA6151)
CYCLE TEST III (QUESTIONS WITH ANSWERS)
UNIT -1 MATRICES
1. If
1
2
1
is an eigen vector of
021
612
322
, find the corresponding eigen value.
Ans: (A - I)X = 0
21
612
322
0
0
0
x
x
x
3
2
1
21
612
322
0
0
0
1
2
1
(2)(1)+2(2)+(-3)(-1) = 0 = 5.
2. Determine so that (x2 + y2 +z2) + 2xy 2xz + 2zy is positive definite.
Ans: The matrix of the given quadratic form is
11
11
11
A
D1 = , D2 = )1)(1(11
1 2
& D3 = |A| = (+1)2( -2)
The Quadratic form is positive definite if D1, D2 & D3> 0 > 2
3. State Cayley Hamilton theorem.
Ans: Every square matrix satisfies its own characteristics equation.
4. Find the sum and product of the eigen values of the matrix
312
301
221
Ans: Sum of the eigen values = Sum of the main diagonal elements = 1+0+3 = 4
Product of the eigen values = |A|= 13
5. One of the eigen values of
814
184
4-47
is -9, Find the other two eigen values.
Ans: If 1 , 2be the other two eigen values, then
-
1 + 2 9 = 7 8 8 = 9 (since sum of the eigen values = sum of the leading diagonal elements) 1 + 2 = 0 =>1 = 2 (1) 912 = |A| = 441 ( since product of the eigen values = | A | )
12 = 49 =>2
1
49
(2)
substitute in (1) we get, 72 (1) => 71 . Hence the other two eigen values are 7 and -7.
UNIT -2 SEQUENCES & SERIES
6. Test the convergence of 1
1
n log n
Ans: The series 1
1
n log n
is a positive term series decreases as n increases after n 2.
So we can apply integral test.
22 2
1 (1/ x)dx dx log (log x)
x log x log x
By integral test, the series diverges.
7. Define Monotonically increasing and Monotonically decreasing sequence with examples
Ans: A sequence { an } is said to be monotonically increasing if an an+1 , for every n
A sequence { an } is said to be monotonically decreasing if an+1 an , for every n
Eg. an = { n } is monotonically increasing
an = { n } is monotonically decreasing
8. Define Absolute convergence.
Ans: If the series of arbitrary terms 1 2 3 .. .nu u u u be such that the series
1 2 3 . nu u u u is convergent, then the series nu is said to be absolute convergent
9. Test the convergence of
n
n1
n! 2
n
Ans: Let
-
n (n 1)
n n 1n (n 1)
(n 1) nn 1
(n 1) nn nn
n
n
n
n
n! 2 (n 1)! 2u , u
n (n 1)
u (n 1)! 2 nlim lim
u (n 1) n! 2
nlim 2
n 1
1 2lim 2 1
1 e1n
By Ratio Test,
n
n
n! 2
n
converges.
10. Define Conditional convergence with example
Ans: If nu is convergent and | |nu is divergent, then nu is said to be conditionally convergent.
ex: 1 1 1
1 ...2 3 4
UNIT III APPLICATIONS OF DIFFERENTIAL CALCULUS
11. Find the centre of curvature of the curve y = x2 at the point (1,1)
Ans: 2121
1
2 2
1 31 4;
2
yyx x y y y
y y
Centre of curvature is 3
4,2
12. Find the envelope of the family of lines cos sin 1x y
a b , where being the parameter.
Ans: (x/a) cos + (y/b) sin =1 --------(1)
Diff. w.r.to partially, we get
(x/a) sin + (y/b) cos =0 --------(2)
(1)2 + (2)2 2
2
a
x (cos2 + sin2) + 2
2
b
y(sin2 + cos2) =1 2
2
2
2
b
y
a
x = 1
13. Find the curvature of the curve 2 22 2 5 2 1 0x y x y
-
Ans: Radius of the circle 2 2 25 1 1 21
16 4 2 4f g c radius of curvature
Curvature1 4
21
14.
Find the evolute of the curve whose centre of curvature of the curve is
2 32 3 , 2x a at y at
Ans: 3 2
3262 27 4 23 2
x a yt a y x a
a a
Equation of the evolute is 3227 4 2ay x a
15. For the catenary y = c cosh (x/c), find the curvature.
Ans: 1 sinhx
yc
, 2
1cosh
xy
c c
3 / 2
2
2
2
1 sinh1
1cosh
x
c y c
x c y
c c
UNIT IV DIFFERENTIAL CALCULUS OF SEVERAL VARIABLES
16.
If , ,x y z u y z uv z uvw find ( , , )
, ,
x y z
u v w
Ans: 1 ; 1 ;y uv z uv w x u uv u v z uvw
J =
uvuwvw
uvwuwv
uv
w
z
v
z
u
zw
y
v
y
u
yw
x
v
x
u
x
wvu
zyx
)1()1(
01
),,(
),,( = u2v
17. If cos , sinx r y r find ( , )
( , )
r
x y
Ans:
cos sin( , )
sin cos( , )
rx yr
rr
( , ) 1 1
( , )( , )( , )
r
x yx y rr
18. If y xx y c , find
dx
dy
-
Ans: Let , y xf x y x y c
/
/
dy f x
dx f y
1
1
log
log
y x
y x
yx y y
x x xy
19. Find Taylors series expansion of e
xsin y near the point 1,4
up to first degree terms.
Ans: f (x , y) = ex sin y , f x(x , y) = ex sin y, fy(x ,y) = e
x cosy
f(1,4
) =2
1
e, fx(1,
4
) =2
1
e, fy(1,
4
)=2
1
e .
f(x ,y) = f(a ,b) + [(xa) f x(a ,b) + (yb) f y(a ,b)] =
4)1(1
2
1 yx
e
20. Find the maxima and minima of f(x, y) = 3x2 + y2 + 12x +36
Ans: fx= 6x +12 = 0 x = 2; fy= 2y = 0 y = 0 .
The stationary point is (2,0).
A = f xx = 6 ,B = f xy = 0 , C = f yy = 2, AC B2 = 12 > 0 and A > 0 .
f is minimum at (2,0) and the minimum value is f (2,0) = 24.
UNIT V MULTIPLE INTEGRALS
21. Evaluate .
22 x
1 0
x dy dx
Ans:
2
22 2
01 0 1
xx
x dy dx x y dx = 2
3
1
15
4x dx
22. Find the area bounded by the lines x = 0, y = 1, y = x using double integration.
Ans: Area = 2
1
2
1
0
21
0
0
1
0 0
ydyxdxdydydx
yy
R
23. Evaluate
a b
2 2
dxdy.
xy
Ans: I = 1 log22
bax
y dy =
y
dylog2][logb
a
2
= ydy
2
blog
a
2
=
2log
b
a2logy
=
2log
b]2log[log a =
2log
b
2log
a
24. Shade the region of integration in
2 2
2
a a x
0 ax x
dxdy.
-
Ans: y = 2xax x2 + y2 ax 0 which is a circle
with centre at ( a 2 , 0) and radius a 2
y = 22 xa x2 + y2 = a2 which is a circle
with centre at (0,0) and radius a
25. Change the order of integration
1
0
x2
x2
dxdyy),(xf .
Ans:
Given, I = 1
0
x2
x 2
dxdy)y,x(f
After changing order of integration
y 2 y1 2
0 0 1 0
I f(x,y) dx dy f(x,y) dx dy
26. Evaluate R
22 dx)dyy(x over the region R for which x, y 0, x+y 1.
Ans: The region of integration is the triangle bounded by the
lines x 0, y 0 and x y 1
Limits of y : 0 to 1x ; Limits of x : 0 to 1
R
22 dxdy)yx( = 1 1
2 2
0 0
x
x y dydx
= dx3y
yx1
0
x1
0
32
1 32
0
(1 )(1 )
3
xx x dx
13 4 4
0
(1 )
3 4 12
x x x
=
6
1
12
1
4
1
3
1
27. Change the order of integration in
2 2a a x2 2
a 0
(x y )dxdy.
Ans: I = xddyyxax
ax
xay
y
22
0
22 )( ( Correct Form )
order)thechanging(afterdydx)y(x
ay
0y
yax
yax
22
22
22
28. Change the order of integration in
2 y1
0 y
xy dx dy
.
-
Ans: Given, I = 1
0
2 y
y
dydxxy
After changing order of integration
2
1
2
0
1
0 0
xxdxdyxydxdyxyI
29.
Compute the area enclosed by y2 = 4x, x + y = 3 and y = 0.
Ans: Area A =
R
dydx =
2
0
3
42y
y
yx
dydx = 22
3
/ 40
y
yy
x dy
2
0
322
0
2
1223
43
yyydy
yy
y
= 6 2 12
8 = 4
3
2 =
3
10
30. Evaluate
a sin
0 0
rdr d.
Ans: a
0
sin
0
drdr =
dra
sin
00
2
2
=
ad
0
2
2
sin
=
2
sin2aa
4
1
31. Evaluate
52
0 0
r sin dr d.
Ans: I =
ddrr
0
5
0
2sin =
dr
0
5
0
22
2sin =
d
0
2sin2
25
=
d 0
2cos14
25 =
02
2sin
4
25
=
0
2
2sin
4
25 =
4
25
32. Evaluate
2
2 2 20 0
r dr d.
(r a )
Ans: I =
2
0 0222 )(
ar
ddrr =
22 12 2 22 ( )0 0
d rd
r a =
2
0 022
1
2
1
dar
-
=
2
02
10
2
1
da
= 2
2 0
1 1
2 a
= 24a
33. Evaluate / 2 sin
0 0
.
r dr d
Ans: / 2 sin
0 0
I r dr d
sin/ 2 2
0 02
rd
/ 2 2
0
sin0
2d
/ 2
2
0
1sin
2d
1 1
.2 2 2
8
34.
Evaluate
cos
0 0
r drd .
Ans: I = cos
2
20 0
r
d = 1
2
0
2cos d = 2
0
221 cos2
d = 1
2 2 4
35. Transform the integration
0
y
0
dydx into polar coordinates.
Ans: Let x = r cos and y = r sin , dxdy = r drd
2
4
00 0
r
yddrrdydx
36. Compute the entire area bounded by r2 = a2 cos2. Ans:
Area A = R
ddrr
4/
0
2cos
0
4
a
r
drdr
cos2/ 4 2
0 0
42
a
rd
= 4
4
0
2
2
2cos
d
a4
2 2
0
sin 22
2a a
37. Transform the integration from Cartesian to polar co-ordinates
22ax x2a 2 2(x y )dxdyx 0 y 0
.
Ans: Let x = r cos and y = r sin,dxdy = r drd
-
a
x
xax
y
dydxyx2
0
2
0
222
)( =2 cos/ 2
3
0 0
a
r dr d
38. Express the region bounded by 2 2 20, 0, 0, 1x y z x y z as a triple integral.
Ans: Here z varies from 0 to 2 21 x y , y varies from 0 to 21 x , x varies from 0 to 1
2 22 111
0 0 0
x yxI dz dy dx
39. Evaluate
1 1 1x y ze dxdydz.
0 0 0
Ans: I =
1
0
1
0 0
zyx dxdydze1
= dydzee zyzy 1
0
1
0
1
= 1
2 1
0
2z z ze e e dz 323 1133 eeee
40.
Evaluate
x y4 x
0 0 0
z dxdy dz.
Ans: I =
4
0
x
0
yx
0
dxdydzz = dydxzx
yx
4
00 0
2
2 = dydxyx
x 4
002
1
= 24
0 0
1
2 2
x
yxy dx
=
242
0
1
2 2
xx dx
= 1634
3
4
34
0
34
0
2
xdxx
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