Engineering Mathematics Workbook_Final

Calculus

2
2
(x + 1 + ) 1 = 2 . The integral 308. The double integral
) ( y −
evaluates to   1 2 x 2x f ( , x y )dy dx under the

17
=
(a) 17 2 (b) transformation x = ( 4 1 v− ) , y uv is
2
transformed into
2
(c) (d) 0 2/3 ( 2 1 v− )
17 (a)  1/ 2  1/ (1 v− ) ( f u uv− ,uv )du dx
2/3 2 ( / 1 v )
−
−
(b)  1/ 2  1/ − ) ( f u uv ,uv )du dx
(1 v
  − 2 2 ) )
−
306. The value of   e ( x + y dxdy = (c)  1  2 ( / 1 v ( f u uv ,uv )u du dx
−
(1 v
0 0 2/3 1/ − )
−
 (d)  2/3  2 ( / 1 v ) ( f u uv ,uv )u du dx
−
(a) (b)  1/ 2 1/ − )
(1 v
2
 309. By change the order of integration
(c)  (d)
4  2 2x f ( ,x y )dy dx may be represented
2 
307. The evaluate the double integral 1 x
as
)+
y
8  ( / 2 1  2x y  − 
     dx dy , we make the 2 2x

)
  2   (a)   f ( , x y dy dx
0  y / 2 
0 x 2
 2x −  y y y
substitution u =   and v = . 2
)
 2  2 (b)   f ( , x y dy dx
0 y
The integral will reduce to
4 y
)
4  2  (c)   f ( ,x y dy dx
 

(a)    2udu dv
0  0  0 y / 2
x
2 2
)
4  1  (d)   f ( ,x y dy dx [GATE]
 

(b)    2udu dv x 2 0
0  0 
310. Changing the order of the integration
4  1  in the double integral
 
(c)    udu dv

0  0 
4  2 
 

(d)    udu dv [GATE-14-EE-SET2]
0  0 

49

Calculus

8 2 1 1
)
I =   f ( , x y dydx leads to (a) 4 (b) 2
0 / 4
x
s q 1
)
I =   f ( , x y dxdy . What is q? (c) (d) 1
3
r p
316. The value of
2
(a) 4y (b) 16y (c) x (d)8 2 sin y
[GATE-2005] 0   x   0  dz dy dx =
y
1 x 2
311. The value of   e / y x dy dx = (a) -2 (b) 2
0 0
(c) -4 (d) 4
_________
[GATE]
[JNU]


 / 2 / 2 cos y
312. The value of   dydx = 317.   1 1 y 1 x dxdy = 3
+
0 x y 0 y
______ 2 2 1
−
(a) 2 2 (b)
[CSIR] 2

1 2 2 2 1 2 2 1
−
−
2
x
313. The value of   e dx dy = (c) (d)
0 2 y 8 9
__________ [IISC]
x
4
4
3
(a) e (b) e − 1 318. The integral 0  1 x  x 2 e − x y dy dx =
y
e − 1 e 4
4
(c) (d) ____
4 4
e − 2 e − 1
314. Let (a) (b)
e 2e
1 1 1 b
 xy sin xy  xy sin xydx dy e − 1 e − 1
( )dx dy =
0 y 0 a (c) 2 (d) 2e
then

(a) a = 0, b = x (b) a = 1, b = x 319. The volume enclosed by the surface
)
x
f ( ,x y = e over the triangle
x
(c) a = 0, b = 1 (d) a = − 1,b =
bounded by the lines
 
3
315.   x 4 .e − x y dx dy =__________ x = ; y x = 0; y = 1in the xy plane is
0 1/ 4
_________




50

Calculus

1
320. The integral   (x + y + 10 )dxdy
2 D

where D denotes the disc: x + 2 y  2 4

, evaluates to __________

[GATE-16-EC-SET 1]
[GATE-20-EC]
321. The region specified by 325. The area of an ellipse represented by
    2 2

 
z
 ( , ,Z ) :3   5,    , 3   4.5  x y
 8 4  an equation a 2 + b 2 = 1 is
in cylindrical coordinates has volume  ab

of __________ (a) ab (b)
4

[GATE-16-EC-SET 1] (c)  ab (d) 4 ab
2 3
322. A triangle in the xy-plane is bounded [GATE-20-CE-SET1]
by the straight lines 2x = 3 , y y = 0 326. The volume of
and x = 3. The volume above the  1 dx dydz where R is

triangle and under the plane ( x + 2 y + 2 z 2 ) 3/ 2
+ +
x y z = 6 is __________ the region bounded by
2
2
2
2
x + y + z = b
[GATE-16-EC-SET 3]
)

( /
ln
(a) ( / b a (b) 4 ln b a )
)
)


( /
( /
323. A three dimensional region R of (c) 2 ln b a (d) 8 ln b a
finite volume is described by 327. The volume of
 ( x + 2 y + 2 z 2 ) dx dydz taken over the
2
2
z
x + y  z 3 ;0   .
1
2
volume enclosed by x + y + z = 1 is _
2
2
Where , , y z are real. The volume of R 4 8
x
(upto two decimal places) is ______ (a) 5 (b) 5
2 
[GATE-17-EC; SESSION 1] (c) (d)
5 5
324. For the solid S shown below, the 328. The volume of the solid surrounded
value of  xdxdydz (rounded off to by the surface 2/3
2/3
2/3
s  x  +  y  +  z  = 1
 
 
 
two decimal places) is  a   b   c 

4 abc abc
(a) (b)
35 35

2 abc  abc
(c) (d)
35 35
[EE, ESE-2019]

51

3





















Vector






















Calculus

Vector Calculus
r r
)
$
$
$
2
2
z
1. Let r = ( x i + y j + zk and r = r . If x + y − = 0 and the plane
1 x z + 3 = at the point (1, 1, 2) passes
f r = g , r  0,
( ) ln r and ( ) r =
r through
satisfy 2 f  + h ( ) r  = , then h(r) is (a) (-1, -2, 4) (b) (-1, 4, 4)
0
g
(c) (3, 4, 4) (d) (-1, 4, 0)
1
(a) r (b)
r [JAM MA 2019]
(c) 2r (d) 3
R
5. Let : f R → be defined by
[JAM MA 2016] y
( , , y z =
2
f x ) sin x + 2e + z
2
2. The tangent plane to the surface
z = x + 2 3y at (1, 1, 2) is given by The maximum of change of f at
2
  

+
(a) x − 3y z = 0   4 ,0,1 , correct upto three decimal

 


(b) x + 3y − 2z = 0 places, is _______
0
(c) 2x + 4y − 3z = [JAM MA 2015]
0
(d) 3x − 7y + 2z = 6. Let
[JAM MA 2018]  x + y 3 , x − 2 y  2 0
3

 
)
2
3
3. In R , the cosine of the acute angle f ( , x y =  x − y 2

between the surfaces  0, x − 2 y = 2 0


x + y + z − =
2
2
2
9 0 and
−
z x − y + = Then the directional derivative of f at
2
2
3 0 at the point
4 $ 3 $
(0, 0) in the direction of i + j is
(2, 1, 2) is 5 5
8 10 ________.
(a) (b)
5 21 5 21 [JAM MA 2019]
8 10 r r
(c) (d) 7. For a constant vector a and r
3 21 3 21 r $ $ $
r = x i + y j + zk , consider the
[JAM MA 2018] following statements:
r r r r
(

curl a r =
4. The tangent line to the curve of (I) (  ) 2grad a r )
intersection of the surface


53

Vector Calculus
 ( r r r  ( r r ) and y + z = 2, then which of the


) 
(II) div a r r = a r
  ur r

following is (are) equal to C  F dr
then,
?
(a) both statements are true 2
(b) only statement I is true (a) 0  0  1 (1 2 sin+ r )r dr d

(c) only statement II is true  1 2 
(d) both statements are false (b) 0  2 0 1  2 + 3 sin      d


[JAM GP 2010] 2 1  

3
3
8. Let : f R → R be a scalar fields (c) 0  0  (1 2 sin + r )dr d
r
: v R → R be a vector field and let (d) 0  2 (1 sin + )d
3
3
r r
a  R be a constant vector. If r
3
represent the position vector [JAM MA 2015]
$
$
+
x i + $ y j zk , then which one of the 10. Consider the vector field
ur
)
$
$
−
following is FALSE? F = r  ( y i x j , WHERE   ,
R
r
r
r
( )
$
(a) curl f v = grad ( ) f  + f curl v r x i + $ y j and r = r
v
( )
r =
r . If the
(b) absolute value of the line integral
ur
r
 F  dr along the closed curve
  2  2  2 
div ( grad f   + + 2    f C
( )) = 
2
  x 2  y 2  z  : c x + 2 y = 2 a (oriented counter
(c) ( r r ) r r clockwise) is 2 , then  is

curl a r = 2 a r
(a) -2 (b) -1
   r  r r
0
(d) div   r 2 r   = 0 , for r  (c) 1 (d) 2
  r  
  [JAM MA 2016]
11. The values of the line integral
[JAM MA 2018]  
3
2
ur  (3x + 2xy ) dx + ( x + x 2 ) dy  
9. Let F be a vector field given by  
ur from M (0, 0) to N (1, 1) along the paths
)
$
$
$
3
F ( , ,x y z = − y i + 2xy j + z k , for 2
: =
C 1 : y = x and C y x are,
2
( , ,x y z ) R 3 . If C is the curve of respectively
intersection of the surfaces x + 2 y 2 1 = (a) 2 and -1 (b) 3 and 3

54

Vector Calculus

(c) -1 and 3 (d) 2 and 2 along the path : c x t , y t = 3 ,
=
2
2
2
12. Let C be the circle (x − ) 1 + y = 1, 0 t   1 is _____________
oriented counter clockwise. Then, the [JAM MA 2016]
value of the line integral

C  − 4 xy dx + 3 x dy is 16. Let C be the boundary of the region
4
2
3 enclosed by y x= , y = x + 2 and
x = 0. Then the value of the line
(a) 6 (b) 8
3
integral C  ( xy − y 2 ) dx − x dy ,
(c) 12 (d) 14 where C is traversed in the counter

[JAM MA 2019] clockwise direction, is ____________
ur [JAM MA 2016]
$
$
2
) 2y i +
F x
13. Let ( , , y z = $ x j + xyk
and let C be the curve of intersection of 17. Let T be the smallest positive real
the plane x + + = 1 and cylinder number such that the tangent to the helix
y
z
t
$
x + 2 y 2 1 = . Then the value of cost i + $ sint j + $ 2 k at t = T is
C  ur  r orthogonal to the tangent at t= =0. Then
F dr is
ur
$
the line integral of F = x i − $ y j along
3 the section of the helix from t = 0 to t =
(a)  (b) T is ________
2
[JAM MA 2017]
(c) 2 (d) 3 ur $ $
)
+
F
[JAM MA 2019] 18. Let ( , x y = − y i x j and let C be
14. Evaluate C  yzdx + zxdy + xydz the ellipse x 2 + y 2 = 1 oriented
16
9
=
where C is the are of curve x b cost , counter clockwise. Then the value of
at ur r

=
y b sint , z = from the point C  F dr (round off to 2 decimal
2
0
intersects z = to the point it intersects places) is ________
=
z a. [JAM MA 2019]
ur
$
$
15. Let F = x i + ( x + y 3 ) j be a vector 19. Let C be the boundary of region x 1 y 2 
)
R =
y 
: 1
−
−
1,0  
(  ,x y 
2
R
)
0
field for all ( ,x y with x and oriented in the counter-clockwise
r direction. Then the value of
$
r = x i + $ y j . Then the value of the line  ydx + 2xdy is
r
ur
C   C
integral F dr from (0, 0) to (1, 1)

55

Vector Calculus

4 − 2 − 5
(a) (b) (a) (b) 3
3 3 2


2 4 (c) 4 (d) 5
(c) (d)
3 3 [JAM MA 2018]
ur u r
[JAM MA 2014]
23. The value of the integral   F  dS ,
20. The value of the integral ur S
$
$
$
 (x + ) y dx x dy , where C is the where F = 3x i + 2y j + zk and S is
+
2
C the closed surface given by the planes
triangle with vertices (0, 0), (2, 0) and x = 0, x =
0
0
2
(2, 4) in the anticlockwise direction is 1, y = , y = , z = and
z =
(a) 5/3 (b) 10/3 3 is
(c) 20/3 (d) 40/3 (a) 6 (b) 18
[JAM GP 2010] (c) 24 (d) 36

$
$
ur x i − y j [JAM GP 2009]
21. For a > 0, b > 0, let F =
b x + a y 2 ur $ $ 2 $
2
2 2
−
+
be a planar vector field. 24. The flux of F = y i x j z k along
the outward normal, across the surface of
Let the solid
C = (  , x ) y  R x + y = a + b 2 
2
2
2
2
)
−
x
2
z
x
be the circle oriented anti-clockwise.  ( , , y z  R 3 0   1, 0  y  1, 0   2 x − y 2 
ur
r
C  F dr =  is equal to
Then
2 2 5
(a) (b) 2 (a) (b)
ab 3 3
(c) 2 ab (d) 0 (c) 8 (d) 4
3 3
22. If
ur
$
$
) (3x −
F ( , x y = 8y ) i + (4y − 6xy ) j [JAM MA 2017]
r
for ( ,x y ) R 2 , then  ur dr , 25. Let S be the closed surface forming the
F 
C boundary of the region V bounded by
where C is the boundary of the triangular 2 2
y =
6
0
0
region bounded by the lines x = , x + ur 3, z = , z = . A vector
y = 0 and x + y = 1 oriented in the
field F is defined over V with
anti-clockwise direction, is ur
+
 F = 2y z + 1. Then the value of

56

Vector Calculus

1   ur $ $ F ( , x y = ) y i − x j . Let
F 
 S n ds , where n is the unit x + 2 y 2 x + 2 y 2

outward drawn normal to the surface S,  , : 0,1 → 2

  R be defined by
is _____.
t =
 ( ) (8cos ,17sin2 t  ) and
t 
[JAM MA 2016]
 ( ) (26cos2 , 10sin2 t  ). If
t =
t  −
26. The line integral of the vector field
ur 3   F  dr − 4  F  dr = 2m ,
$
$
$
+
F = zx i xy j + yzk along the
then m =
boundary of the triangle with vertices (1,
0, 0), (0,1,0) and (0,0,1), oriented 29. Consider the unit sphere
)
anticlockwise, when viewed from the S = (  , , y z  R 3 : x + y + z =  1
2
2
2
x
point (2,2,2) is
)
and the unit normal vector n = ( , , y z
x
1 −
(a) (b) -2 at each point (x,y,z) on S. The value of
2 the surface integral
1   2x   y   2z   
2 

z


(c) (d) 2   S   + sin y 2  x +    e −    y +  + sin y z d =
 
 
 
2             

[JAM MA 2017]
30. Let
27. Let S be the surface of he cone
)
 = (  , , y z  R 3 : 1 x y z   1
− 
x
, ,
z = x + 2 y bounded by the planes
2
z = 0 and z = 3. Further, let C be the and :   → R be a function whose all
closed curve forming the boundary of second order partial derivatives exist and
ur are continuous. If  satisfied the
the surface S. A vector field F is such

ur Laplace equation  = 2 0 for all
$
$
that  F = − x i − y j . The absolute ( , ,x y z   , then which one of the
)
ur
r
value of the line integral  F  dr , following statements TRUE in  ?
C
where
(a)   is solenoidal but not irrotational
(a) 0 (b) 9
(b)   is irrotational but not solenoidal
(c) 15 (d)
(c)   is both solenoidal and
[JAM MA 2016] irrotational
28. Let F be a vector field defined on (d)  is neither solenoidal nor
R 2 / 0,0 irrotational
  by

57

Vector Calculus
) )
31. Let C be the boundary of the triangle  (  sin x i + 2y j − z (1 sin2x k  nds = ____
$
$
$
2 $
−
formed by the points (1,0,0), (0,1,0) and S
(0,0,1). Then the value of the line 
integral (a) 1 (b) 2
 − 2ydx + (3x − 4y 2 ) dy + ( z + 3y ) dz
2
C (c)  (d) 2
is _______

(a) 0 (b) 1 35. The value of  − ydx + xdy where C
2
2
C x + y
(c) 2 (d) 4
1
is z = is _____
32. Let,
−
B = (  , , y z ): , , y z  R & x + y + z   4 (a) 2 (b) 2
2
2
2
x
x
=
+
. Let, r xi y j + 3k be vector (c) 0 (d) none
valued function defined on B. If 36. The line integral of F = 3i x j + yk
+
r = x + y + z then the value of on the circle x + 2 y = 2 2
2
2
2
2
0
    r r dv is ______. a , z =
2
S described in clockwise is _______
37. For the vector field
(a) 16 (b) 32
F = − 2 y 2 i + 2 x 2 j + sin y cos zk
3
2
(c) 64 (d) 128 x + y x + y
, the value of  F  dr where C is
33. The maximum magnitude of the C
directional derivative for the surface closed contour in xy plane consisting of
2
x + xy yz = 9 at the point (1,2,3) is parabolas y =  ( x + ) 1 and straight
+
2
along the direction __________. lines x =  1.
+
+
2
(a) i + j k (b) 2i + 2 j k 38. Let D be the triangular region in R
bounded by the y-axis, the line y = 1
=
(c) i + 2 j + 3k (d) i − 2 j + 3k and the line y x . Let C be the
boundary curve of the region and C be
34. Let S be the surface bounding the region oriented counter clockwise. Then the
x + 2 y  2 1, x , y  and n is value of the line integral
0
$
0
2
2
1
outward drawn unit normal to S. z   ( x + 2sin x cos ) x dx + (4x + y 2 ) dy
C
Then is
(a) 1 (b) 2

(c) 4 (d) 8


58

Vector Calculus

39. Let ∫ (2 − ) − − where C is

C = (  , x ) y  R 2 :max  , x y = 1 the circle x + 2 y 2 1 = , z = , oriented
0
  .
The value of the line integral counter clockwise. [JAM 2005]
( ))
( ))
(  xy + 2y + sin e x dx + ( x y + cos e y dy
2
2
C 43. The vector field
) ( x −
4
is F = ( 2xy − y + 3 i + 2 4xy 3 ) j
(a) 0 (b) -2 is conservative then its potential and also
(c) -4 (d) -8 work done in moving particle from (1, 0)
)

+
+
=
x
40. Let r xi y j zk . If ( , , y z is to (2, 1) along some curve.
a solution of the Laplace equation then [JAM 2005]
the vector field (   r +
) is _________
44. Let C be the circle x + 2 y 2 1 = taken in
the anticlock wise sence. Then the value
(a) neither solenoidal nor irrotational
of the integral
(b) solenoidal but not irrotational
 (2xy + ) y dx + (3x y + 2x ) dy =
3
2
2
(c) both solenoidal and irrotational C
_________
(d) irrotational but not solenoidal

[JAM 2005] (a) 1 (b)
2

(c)  (d) 0

41. Let F = xi + 2y j + 3zk , S be the [JAM 2006]

surface of the sphere x + y + z = 1 45. Let r be the distance of a point P(x, y, z)
2
2
2

and be the inward unit normal vector from the origin O. Then r is a vector
⋀
to S then   F  $ ________.
n ds is equal to
S
_______ (a) orthogonal to OP

(a) 4 (b) 4− (b) Normal to the level surface of r at p.
(c) normal to the surface of revolution
(c) 8 (d) 8−
(d) normal to the surface of revolution
[JAM 2005] generated by OP about y axis.

2
2
2
42. Let S be the surface x + y + z = 1, [JAM 2006]
z = 0 use stoke’s theorem to evaluate





59

Vector Calculus

46. Let S be a closed surface for and let ( , ) = + or ( , x y 
) D
2
which∬ . = 1. Then the volume
̅
̂
$
. If n is the outward unit normal to C,
enclosed by the surface is __________
nds , evaluated counter-
then     $
1 C
(a) 1 (b) clockwise over C, is equal to _______
3
2 (a) 0 (b)  − 2
(c) (d) 3
3 (c)  (d)  + 2


[JAM 2006] [JAM 2007]

2
R
47. Let : f R → be thrice differentiable 50. The work done by the force
+
2
and vanish on the boundary of the region F = 4yi − 3xy j z k in moving
)
( 1,1  −
 = − ) ( 1,1 . Then particle over the circular path
0
)
  1 1 div (grad f )( , x y dx dy is x + 2 y 2 1 = , z = from (1, 0, 0) to (0,
− 1 − 1 1, 0) is _______
_______
(a)  + 1 (b)  − 1
(a) never 0 (b) 1
(c)  − 1 + (d)  − 1 −
(c) 0 (d) depends on f
[JAM 2008]
[IISC 2007]
3
,
48. Let 51. Let V = ( { , , y z  x ) R
+
u = (ae x sin y − 4x i + + x cos ) y j azk 1  x + y + z  1} and
) (2y e
2
2
2
, where a is a constant. If the line integer 4
 u dr over every closed curve C is F = xi + y j + zk for

C ( x + 2 y + 2 z 2 ) 2
zero, then a is equal to ___________
$
( , ,x y z ) V Let n denoted the
−
(a) 2− (b) 1
outward unit normal to the boundary of
(c) 0 (d) 1 V and S denoted the part
  1  
) R x +
[JAM 2007]  ( , , y z  3 : 2 y + z =  2
2
x
  4 
49. Let C denote boundary of semi-circular
disc of the boundary of V. Then
$
)
D = (  ,x y  R 2 ;x + y  1, y   0   S F  nds _______
2
2


60

Vector Calculus
−
−
(a) 8 (b) 4 (c)  r  dr = 0 for every piece wise
C
(c) 4 (d) 8 smooth closed curve C in D.

[JAM 2008] (d) The differential equation
+
pdx qdy is exact in D.
) xy +
−
2 2
T x
52. Let ( , , y z = 2 2z x z be
[JAM 2009]
the temperature at the point (x, y, z). The
unit vector in the direction in which the 54. The value of C for which there exists a
temperature decreases most rapidly at twice differentiable vector field F with
)
(1,0, 1 − is ________
+
curl F = 2xi − 7y j czk is _______
1 $ 2 $
(a) − i + k (a) 0 (b) 2
5 5
(c) 5 (d) 7
1 $ 2 $
(b) i − k [JAM 2009]
5 5
x $
$
$
2
2
2
x
x
2 $ 3 $ 1 $ 55. Let F = 2xyze i + ze j + ye k be
(c) i + j + k the gradient of a scalar function. The
14 14 14

 2 $ 3 $ 1 $  value of  F dr along the oriented

 
(d) −   i + j + k C
 14 14 14  path L from (0, 0, 0) to (1, 0, 2) and then
to (1, 1, 2) is ______
[JAM 2009]
(a) 0 (b) 2e
$
)
53. Suppose V = P ( , x y )i + $ q ( , x y j is 2
(c) e (d) e
a continuously differentiable vector field
2
defined in a domain in R . Which one [JAM 2010]
of the following statements is not Let F = + −
equivalent to the remaining ones? 56. xyi y j yzk denote the
force field on a particle traversing the
(a) There exists a function ( ,x y ) such path L from (0, 0, 0) to C (1, 1, 1) along
the curve of intersection of the cylinder
 
)
=
2
that = p ( ,x y and y x and the plane z x. The work
=
 x
  = q ( ,x y for all ( , x y  done by F is __________
)
) R
 y
1
(a) 0 (b)
 p  q 4
(b) = holds at all points of D.
 y  x

61

Vector Calculus

1 (d) a + b + c = 0 [JAM 2012]
2
2
2
(c) (d) 1
2
3
60. If C is a smooth curve in R from (0, 0,
[JAM 2010] )
0) to (2,1, 1 − , then the value of
)
$
$
$
+
)
)
+
+
2
+
57. Let F = ayi + z j xk and C is the  (2xy z dx + ( z x dy + (x y dz is
C
positive oriented closed curve given by
__________
x + 2 y 2 1 = , z = . If  F  dr  −
0
=
C (a) 1 (b) 0
then the value of a is _________ (c) 1 (d) 2
−
(a) 1 (b) 0 [JAM 2012]
1 61. The value of n for which the divergence
(c) (d) 1
2 r
of the function F = n , where
[JAM 2011] r
+
+
r xi y j zk , r  0 vanishes is
=
58. Consider the vector field
$
$
$
F = (ax + + ) a i + − (x + ) y k , _________
j
y
−
where a is constant. (a) 1 (b) 1
If F  curlF = 0 then the value of a is
___________ (c) 3 (d) 3 −
[JAM 2013]
(a) 1− (b) 0
2
3 62. Let  be the triangle R with vertices
(c) 1 (d) (0, 0) (1, 0) and (0, 1) and let
2
)
4
F ( , x y = − xyi + y + 1 j . The line
[JAM 2011]

integral  F dr taking anticlock
$
$
$
+
59. For C > 0, if ai + b j ck is the unit 
( ) wise orientation of  
normal vector at 1,1, 2 to the cone
1 −
z = x + 2 y , then ___________ (a) 6 (b) 0
2
1
2
2
2
(a) a + b − c = 0 (c) 6 (d) 6
[IISC 2006]
2
2
(b) a− 2 + b + c = 0
2
2
2
(c) a − b + c = 0

62

Vector Calculus


63. The  yzdx + (xz + ) 1 dy + xydz , (c) x + 1 2 x + 2 2 2x x
1 3
C
where C is a simple closed curve, equals (d) (2 , x x + 3 , x 1 ) x [IISC 2002]
1
1
____________
67. A unit normal vector to the curve
(a) 0
( 
, , 
2
C : x x x R ) in the plane R at
2
(b) 3xyz + y
the point (0, 0) is given by _________
(c) length of C
)
−
(a) (0, 1 − ) (b) ( 1,0
(d) area enclosed by C [IISC 2005]
 1 1 
2
64. Let D be the square in R with vertices (c)   ,     (d) (1, 0)
(0, 0) (1, 0) (0, 1) (1, 1). The integral  2 2 

∫ where D is the boundary of
[IISC 2002]
the square, is equal to ________
3
F
(a) 0 (b) 0.5 68. Let : R − 3   0 → R be the vector
x
F
(c) 1 (d) 1.5 field defined by ( ) x = where
x
[IISC 2005]
x = ( , , x x  3   0 and
) R −
x
1
3
2
65. Let
2
2
2
) ( x y +
V = 2xyzi + ( x z y j + 2 3z 2 ) k . x = x + x + x . Then the
+
2
3
2
1
divergence of F (x) is ____
̅
Then the magnitude of curl at (1, 1, 1)
is ____ 1
(a) x (b)
(a) not defined x
(b) 1 2
(c) (d) 2  x
(c) 0 x
(d) strictly greater than 1 [IISC 2005]
[IISC 2003]
3
3
66. Let :V R → R be the vector field 69. Let G be the tetrahedron in R with
3
defined by vertices (0, 0, 0) (1, 0, 0) (0, 1, 0) and (0,
: x +
2
V ( , ,x x x 3 ) ( 1 2 x 2 2 , x x + x 2 3 , x x + x x ) 0, 1). The outward flux of the vector
2
1
1 2
1 3
2
the divergence of V is ________ field
(a) 4x + x V ( , ,x y z ) 2 cosz= ( ) ( z− 2 + ) 1 j yz+ 2 sin xy
xy i
( )k
1
3
across the boundary of G is _________
(b) 0

63

Vector Calculus

1 − (b) 2i + 4 j + k
(a) (b) 0
6
  1  

1 (c)     (2i + 4 j − ) k
(c) (d) 6  21 
6
(d) none
[IISC 2006]
73. If the temperature at any point in space
3 is given by T = xy + +
R
70. Let : f R → be defined yz zx , then the
( , , y z =
2
4
2
f x ) x + 2xy + 5y − z − 1. directional derivative of T in the
direction of the vector 3i − 4k at
The unit vector u which gives the
maximum value for the directional (1, 1, 1) is
derivative D f at the point (1, 0, 1) is
u 2 1
_______ (a) − (b)
5 5
(a) u = (1, 0, 0)
5 5
(b) u = (0, 0, 1) (c) (d) −
2 2
− 1
)
(c) u = (1,0,1 In what direction from (3, 1, −
2 74. ) 2 ,
the directional derivative of
1 2 2 4
x
(d) u = (1,1, 2− ) [IISC 2006]  ( , , y z =
) x y z is maximum?
6
(a) (
71. Find the gradient of the function 96 i + 3j − 3k )
 = x − 2xy + z at the point
2
2
96 i −
(2, 1,1− ) . (b) ( 3j + 3k )


96 i −
(a) 6i − 4 j + 2k (b) 4i − 6 j + 2k (c) ( 3j − 3k )


(c) 6i + 4 j − 2k (d) 6i − 4 j − 2k (d) (
48 i − 3 j + 3k )
72. A unit normal to the surface z = 2xy at

the point (2, 1, 4) is Divergence

75. The divergence of
+
x zi − 2 2y z j xy k at the point
2
3
2

(1,1, 1− ) is
−
(a) 2i + 4 j k

64

Vector Calculus

2
(a) 8 − (b) 3 (a)  (b) 

2
(c) 3 (d) 2 (c) 2 (d) 2
76. The value of p for which the vector field Vector Identities
V = (2x + y i + 2z ) j + (x + pz )k
) (3x −
j
y
is solenodial 80. If F = (x + + ) 1 i + − (x + ) y k
then  (
(a) 0 (b) 2  F ) =

(c) 2− (d) 1 (a) zero (b) F

(c) 2 (d) None

Curl
+
+
=
81. If r xi y j zk and r = r then
77. If the velocity vector in a two- grad (1/ r) is
dimensional flow fluid is given by
V = 2xy i + (2y − x 2 ) j then the curl (a) r (b) r
2
r 2 r 3
V will be
r r
(c) − (d) −
(a) 2y j (b) 6yk r 2 r 3


−
(c) zero (d) 4xk If r xi y j zk= + + and r =
82. r then
78. The value of a, b, c for which  r 

V = (x + + ) (bx + 3y − ) z j div   r 3    =
y az i +
)
+
+
+ (3x cy z k is an irrotational.  
(a) 0 (b) 1
(a) a = 1, b = 3, c = 1
−
(c) 1 (d) 2
3
(b) a = 1, b = 1, c =
+
=
+
83. If r xi y j zk and r = r then
(c) a = 3, b = 1, c = − 1
( )
n
curl r r =
(d) None
79. A rigid body is rotating with constant (a) 0 (b) 1−
angular velocity  about a fixed axis. If
V is the velocity of a point of the body (c) r (d) r
then curl V is equal to





65

Vector Calculus

84. If A & B are irrotational vectors then 88. Evaluate  f  dr where

the divergence of A B is ______ C
f = ( x − yz i + 2 xz ) j + ( z − xy )k
) ( y −
2
2
−
(a) 1 (b) 0 from (1, 1, 0) to (2, 0, 1).
−
(c) 2 (d) 3− (a) 7/3 (b) 6 −
)

x
85. Let r xi y j zk= + + . If ( , , y z is (c) 0 (d) 8
a solution of the laplace equation then Greens Theorem
r
the vector field V + is

89. The value of  A dr where
(a) neither solenoidal nor irrotational C
A = (2y − y i + ) y j and C is
) (2x +
2
(b) solenoidal but NOT irrotational
the closed curve of the region bounded
(c) both solenoidal and irrotational by y = x and x = y , is
2
2
(d) irrotational but NOT solenoidal
(a) 1/2 (b) 3/10
Line Integral (c) 1/6 (d) 4/5




86. The value of the line integral F dr 90. The value of the line integral
2
2
+
, where F = 3x i + (2xz − ) y j zk  c ( y dx + x dy ) where C is the
2
and C is the straight line from (0,1, 1 − ) boundary of the square bounded by
=
x = 0 , x a , y = , y a is
=
0
to (1, 2, 0) is
)
+
−
(a) 1 (b) 3 (a) 0 (b) ( 2 x y
(c) 0 (d) None (c) 4 (d) 4/3
87. The work done in moving a particle in  − )
the force field 91. Find the value of (x dy y dx
) (2xy +
f = ( x − y + x i − ) y j around the circle x + 2 y = 2 1
2
2
x
2
along the parabola y = from (0, 0) to
(a)  (b) 2
(1, 1) is
−
(c) 0 (d) 
(a) 24 (b) 18
(c) 3/2 (d) 2/ 3− Surface Integral

92. Evaluate  f  nds or   f  n ds
S S
over the surface of the cylinder


66

Vector Calculus

x + 2 y = 2 9 included in the first octant (c) (4/3  + + ) 2
) (a b c
5
0
between z = and z = where
(d) none
+
f = zi x j yzk .
−
96. Evaluate
(a) 40 (b) 60

(   x − yz ) i − 2x y j + 2k  n ds
2
3
(c) 80 (d) 100 S
where S denotes the surface of the
Volume Integral rectangular parallelepiped 0 x a  ,



0  y b, 0 z c is
93. Find     dV where  = xyz and
V 2
‘V’ is the volume of the region bounded (a) a bc (b) abc
by x = , y = , y = , z = x , z = . 3 3
0
2
0
6
4
3
ab c abc
(a) 162 (b) 172 (c) (d)
3 3
(c) 182 (d) 192


Stokes Theorem
Gauss – Divergence Theorem

−
97. Evaluate (e dx + x 2y dy dz ) where
94. Evaluate  f  n ds where S is the C
S ‘C’ is the curve x + 2 y = 2 4 and z = 2.
0
surface of tetrahedron bounded by x = ,
y = 0, z = and the place (a) 0 (b) 1
0
x + + = 1 and f = xyi z j + 2yzk (c) 2 (d) 3
+
2
z
y
3 1 98. Evaluate  f  dr where
(a) (b) C
8 8 f = (2x y i yz i yz j y zk− ) − 2 − 2 − 2 and

8 3 ‘C’ is the boundary of the upper half of
(c) (d) 2 2 2
2 5 surface of the sphere x + y + z = 1
above the xy-plane.
+
+
95. If f = ax i by j czk where a, b , c
(a) 0 (b) 1
are constants and ‘s’ is the surface of a
unit sphere then   f  n ds = (c) 2 (d) none
S
99. If  = 3x y y z then grade at (1, 2, 1− − )
−
3 2
2
(a) (4/3 ) (a b c + + )
is
(b) 0 a) 12i− − 9 j − 16k


67

Vector Calculus

b) 12i − 9 j − 16k of 2i − j − 2k is

c) 12i + 9 j + 16k 37 14
(a) (b)
−
d) 12i + 9 j + 16k 3 3
100. The unit normal vector to the surface (c) 28 (d) 2

 : x − y + x zat (1,1, 2 is 3
)
3
3
2
−
107. The directional derivative of
i − − 3 j k i − 3 j + k
+
(a) (b)  ( ,x y = ) x )
11 11 x + y 2 at the point (0,2 along a
2
−
i + 3 j + k i + 3 j k 0
(c) (d) line making angle 30 . With the positive
11 11
direction of x axis is
101. The greatest value of directional derivative
3 3
)
of xy + yz at (2, 1,1 is (a) 8 (b) 4
2
3
−
(a) 19 (b) 2 19 3 1
(c) (d) 8
(c) 11 (d) 2 11 2
2 2
 1  108. The directional derivative of x + y at
3
3
102. The value of  2       = _____ where r is the
 r 
8,8
( ) along the line y = directed away
x
position vector of any point.
from the origin is ____
(a) 0 (b) 1
3 3
(c) 2 (d)3 (a) (b)
2 2
103. If F = ( x + 3y ) i + ( y + 2z ) j + ( x + pz ) k is
2 2
solenoidal vector then the value of P =____ (c) (d)
3 3
(a) 0 (b) 1
109. Let ( , x y = − 3 − 3
f
) kxy x y xy where K is real
(c) 2 (d) -2
n
104. If F = r r is solvenoidal then n = ______ constant. If the directional derivative of f at
(a) -1 (b) -2 point (1,2) in the direction of a unit vector
(c) -3 (d) -4 1 1 15
− i − j is then the value of K is

105. If F = r and .F = 0then n = ___ 2 2 2
n
____
(a) -1 (b) -2
(a) 2 (b) 4
(c) -3 (d) -4
(c) 1 (d) -2
106. The directional derivative of surface

2
 : x yz + 2 4xz at (1, 2, 1− − )in the direction

68

Vector Calculus


110. The value of  ˆ . F n ds where S is the (c) 4 (d) 8
S 3 3
surface of the sphere x + y + z = 4where
2
2
2
2
ˆ
114. The value of  ( 4xi − 2y j + 2 z k ).n ds
=
+
ˆ n is the unit normal and F xi + yj zk is s
where S is bounted by x + y = 4,z = and
2
2
0
_____
z = 3is
(a) 32 (b) 16
(a) 16 (b) 48
(c) 8 (d) 64
(c) 32 (d) 84
111. The flux of the vector filed
115. The value of
+
F xi + yj zk flowing out through the
=
s  ( ( x − yz ) i − 2x yj + zk ) ˆ .nds where S is
2
3
x 2 y 2 z 2
surface of the ellipsoid + + = 1
a 2 b 2 c 2 the surface of the cube bounded by
= =
a  b   0is x y z = 2 and co-ordinate planes is ___
c
(a) abc (b) 3 abc


(c) 2 abc (d) 4 abc
)
112. Let w = (  , ,x y z  R 3 :1 x + y + z   4 32 56
2
2
2
(a) (b)
and F w → defined by 3 3
R
3
:
x
( , , y z ) (c) 16 (d) 64
F = 3 for ( , , y z  3 3
x
) w . If 
( x + 2 y + 2 z 2 ) 2 y
116. Let I =  e dx + (e / y n x + ) x dy where C is
,
denotes the boundary of oriented by the c x
the positive oriented boundary of the region
out ward normal ˆ nto , then  ˆ . F n ds is
 1
enclosed by Y = + 2 , 2
1 x y = and x =
(a) 0 (b) 4 2
(c) 8 (d) 12 then the value of I = _____
113. Let S be the sphere x + y + z = 1. The 1 5
2
2
2
(a) 8 (b) 24
value of surface integral
7 3
  ( sin ,cosx y 2 x ,2z z sin y ) ( . , ,x y z is (c) (d)
)
−
s 24 8
____ C = (  , x ) y  R 2 ,max 1
117. Let  , x y = the
 2
(a) (b) value of line integral
3 3

69

Vector Calculus
)
( ))
( ))
(  xy + 2y + sin e x dx + ( x y + cos e y dy the straight line joining (0,0,0 to (1,1,1 )
2
2
C
is ____ is ________
(a) 0 (b) -2 (a) 0 (b) 1
(c) -4 (d) -8 (c) 2 (d) 3
118. Let D be the triangular region in R bounded 122. The value of ( 2x − ) y i − yz j − y zk ) .
(

2
2
by the y axis, the line Y = 1 and the line b = C
dr . Where C is the upper half of the
x. Let C be the boundary of the curve of the
2
2
2
region and C is oriented in counter surface of the sphere x + y + z = 1above
clockwise. Then the value of xy plane is ______
 ( x + 2sin x cos ) x dx + (4x + y 2 )dy is
2
2
C
(a)  (b) −
_____
(c) 2 (d) 0
(a) 1 (b) 2
123. The value of  . F dr where
(c) 4 (d) 8 C

+
F = yi + zj xk and C is the curve
+
119. The value of ( 2x +  2 y 2 ) dx dy , where C bounded by x = cos y = sin z = is ___


0
C
is the boundary of the region in first (a)  (b) −
−
quadrant bounded by y = 0, x = 0 and (c) 2 (d) 2
+
x + 2 y = 2 1 ______ 124. If r = xi + yj zk then  . r dr is___
C
2
(a) -1 (b) − (a) 0 (b) 1
3
(c) 2 (d) 3
2
(c) (d) 1 125. The value of  e dx + x 2ydy dz where C is
−
3 C
2
120. The value of  yzdx xzdy xydx where C the curve x + 2 y = 2 0 and z = is ___
+
+
C
)
)(
is the line segment (1,1,0 2,3,2 is _ (a) 0 (b) 1
(c)  (d) 2
(a) 0 (b) 7
126. The value of ( x +  2 y 2 ) dx + 3xy dy where
2
(c) 9 (d) 12 C
121. The work done of a moving particle in the C is the circle x + 2 y = 2 4 in xy plane is ___
force field F = 5x i + (xz y j− ) + 3zk along (a) 4 (b) 8
2

(c) 12 (d) 16


70

Vector Calculus


127. The value of  Fdr where (d) R R = d R

C
dt
=
+
F xyi + yzj zxk and C is the curve
[GATE-2005 ; (IN)]
+
=
2
3
+
r ti t j t k from t = to t = 1is ____ 131. The area of a triangle formed by the tips
0
28 27 of vectors a , b and c is
(a) (b)
27 28
1 ( ) ( )
−
−
29 (a) a b  a c
(c) (d)1 2
28
) (
)
1
−
−
128. Let p = 3i + 2 j k and q i = + 2j + 3k be (b) ( a b  a c
+
2
+
+
vectors in R .Suppose V = ai bj ck is a
3
1


=
unit vector such that v .p = 0 v .q The (c) 2 a b c
value of a b c = _____ 1
+ +

(d) ( a b  ) c
(a) 6 (b) 3 2
(c) 1 (d) 0 [GATE-2007-ME]
129. If P, Q and R are three points having
coordinates (3, -2, -1), (1, 3, 4), (2, 1, -2) 132. If a and b are two arbitrary vectors
in XYZ space (o being the origin of the with magnitudes a and b, respectively,
2
coordinate system) then distance of point a b will be equal to

P from plane OQR is
(a) 3 (b) 7 2 2 ( ) 2

(a) a b − a b
(c) 5 (d) 9

−
(b) ab a b
[GATE-2003]

R
130. If a vector ( ) t has a constant (c) a b + 2 2 ( a b ) 2
magnitude than
+
(d) ab a b [GATE-2011-CE]
d R
(a) R  = 0
dt 133. If A (0, 4, 3), B (0, 0, 0) and C (3, 0, 4)
are three points defined in x, y, z
d R
(b) R  = 0 coordinate system, then which one of the
dt following vectors is perpendicular to

both the vectors AB and BC
d R
(c) R R = 0
dt


71

Vector Calculus


(a) 16i − 9 j − 12k 137. The derivative of f(x, y) at point (1, 2) in
the direction of vector i + j is 2 2

−
(b) 16i − 9 j + 12k and in the direction of the vector 2 j is
-3. Then the derivative of f(x, y) in
(c) 16i − 9 j − 12k
direction i − − 2 j is
−
(d) 16i − 9 j + 12k
−
(a) 2 2 + 3 / 2 (b) 7 / 5
[GATE-2011 (PI)]
−
−
(c) 2 2 3 / 2 (d) 1/ 5
134. A particle, starting from origin at t = 0s,
is traveling along x-axis with velocity [GATE-95]

    138. The magnitude of the directional

V = cos   t m / s derivative of the function
 

2  2  f ( , x y = ) x + 2 3y in a direction
2
At t = 3s, the difference between the normal to the circle x + 2 y = 2 2 , at the
distance covered by the particle and the point (1, 1) is
magnitude of displacement from the
origin is _______ (a) 4 2 (b) 5 2

[GATE-2014]
(c) 7 2 (d) 9 2

[GATE-2000 (CE)]
135. A particle move along a curve whose
=
parametric equations are: x t + 3 2t , 139. The maximum value of the directional
derivative of the function
y = − 3e − 2t and z = 2 sin (5t), where x,  = 2x + 3y + 5z at a point
2
2
2
y and z show variations of the distance
covered by the particle (in cm) with time (1, 1, -1) is
t (in s). The magnitude of the
acceleration of the particle ( in cm s 2 ) (a) 10 (b) -4
/

at t = 0 is _________ (c) 152 (d) 152


[GATE-2014 (PI-SET 1)] [GATE-2002]
136. The smaller angle (in degrees) between 140. A scalar field is given by
+
the planes x + y z = 1 and f = x 2/3 + y 2/3 , where x and y are the
2x − + 2z = is _____
0
y
Cartesian coordinates. The derivate of ‘f’
[GATE-2017-EC SESSION-II] along the line y = x directed way from
the origin at the point (8, 8) is






72

Vector Calculus

2 3 (c) -3 (d) 3
(a) (b)
3 2 [GATE-99]

2 3 If F = n
(c) (d) 144. r r is solenoidal then n =
3 2 ______


[GATE-2005 (IN)] (a) -1 (b) -2

141. A sphere of unit radius is centred at the (c) -3 (d) -4
origin. The unit normal at a point (x, y, [GATE]
z) on the surface of the sphere is the
vector.  1 
145. The value of  2       = ______ where
(a) (x, y, z)  r 

 1 1 1  r is the position vector of any point. If
(b)   , ,     F = r and  F = 0 then n =
n
 3 3 3 
_________ [GATE]
 x y z 
(c)   , ,     If r = xa x + ya y + za r
$
$
$
 3 3 3  146. z and r = ,
 x y z  then div r  2 (ln r = ) _____
(d)   , ,    
 2 2 2  [GATE-2014 (EC-SET 2)]


[GATE-2009 (IN)]

$
$
$
142. The directional derivative of 147. For a position vector r = x i + y j + zk
f ( , x y = ) xy (x + ) y at (1, 1) in the the norm of the vector can be defined as
2 r − x + y + 2
2
2
direction of the unit vector at an angle of z . Given a function
  = 
with y-axis, is given by ….. ln r , its gradient  is
4

[2014-EC-SEC 4] (a) r (b) r
r
3
143. For the function  = ax y − 2 y to
represent the velocity potential of an r r
ideal fluid,  2 should be equal to zero. (c) r r (d) r 3

In that case, the value of ‘a’ has to be

(a) -1 (b) 1 [GATE-2018 (ME-AFTERNOON
SESSION)]


73

Vector Calculus

148. Divergence of the vector field 21 151. The figures show diagrammatic

)
v ( , ,x y z = − ( cosx xy + y ) ( cosi + y xy ) j representations of vector fields, X , Y
 2 and Z respectively. Which one of the
+  ( sin z 2 ) x+ 2 + y  k
  following choices is true?

2
(a) 2 cosz z


2
z
(b) sin xy + 2 cos z
x
(c) sin xy − cosz

(d) none of these [GATE-2007 (PI)]
(a)  X = 0, Y  0, Z  0
149. For a vector E, which one of the
following statement is NOT TRUE? X  0, Y = 0, Z 
(b)  0
(a) If  E = 0,E is called solenoidal
(c)  X  0, Y  0, Z  0
(b) If  E = 0,E is called
conservative (d)  X = 0, Y = 0, Z = 0

(c) If  E = 0,E is called irrotational [GATE-2017-EE SESSION-II]


(d) If  E = 0,E is called irrotational 152. Consider the two-dimensional velocity
field given by
$
[GATE-2013 (IN)] V = (5 a x b y i+ 1 + 1 ) + (4 a x b y j+ 2 + 2 ) $

where a b a and b are constants.
150. A vector P is given by 1 1 2 2
2
P = x ya x − x y 2 y a − x yza z . Which one of the following conditions
3
2
needs to be satisfied for the flow to be
Which one of the following statement is incompressible?
TRUE?
(a) a + 1 b = 1 0 (b) a + 1 b = 2 0
(a) P is solenoidal, but not irrotational
(c) a + 2 b = 2 0 (d) a + 2 b = 2 0
(b) P is irrotational, but not solenoidal
[GATE-2017-ME SESSION-1]
(c) P is neither solenoidal nor
irrotational


(d) P is both solenoidal and irrotational

[2015-EC-SET 1]





74

Vector Calculus
 3 3 (c) (x − x )( y − y
 x + y x − 2 y  2 0 2 1 2 1 )
 
2
153. f =  x − y 2 . The
 2 2
 0 x − 2 y = 2 0 (d) ( y − y 1 ) + (x − x 1 )

2
2

directional derivative off at (0, 0) in the [GATE-11 (PI)]
4 3
direction of i + j is _______ 2 $ 2 $
5 5 156. ( ,F x y ) ( x= + xy ) a + ( y + xy ) a . It’s
x
y
line integral over the straight line from
154. Consider points P and Q in xy – plane
with P = (1, 0) and Q = (0, 1). The line (x, y) = (0, 2) to (2, 0) evaluates to
integral 2  Q (xdx + ydy ) along the (a) -8 (b) 4
P (c) 8 (d) 0
semicircle with the line segment PQ as
its diameter [GATE-2009-EE]

(a) is -1 157. A path AB in the form of one quarter of
a circle of unit radius is shown in the
(b) is 0 2
figure. Integration of (x + ) y on path

(c) 1 AB traversed in a counter clockwise

(d) depends on the direction (clockwise sense is
(or) anti-clockwise) of the semicircle

[GATE-08 (EC)]

P

2
155. The line integral ( ydx x dy+ ) from
P 1
)
P 1 ( , x y to ( , y 2 ) along the
P x
2
1
2
1
semi-circle PP shown in the figure is
1 2
 
(a) − 1 (b) + 1
2 2

(c) (d) 1
2


[GATE-2009]

158. The value of the line integral
+
(a) x y − x y (  2xy dx + 2 2x ydy dz ) along a
2
1 1
2 2
(b) ( 2 2 2 2 path joining the origin (0, 0, 0) and the
x −
+
y − y 1 ) ( 2 x 1 ) point (1, 1, 1)
2

75

Vector Calculus

(a) 0 (b) 2 161. If f = 2x + 3y + 4z , the value of
2
3

)
(c) 4 (d) 6 line integral (grad f . Dr evaluated
C
[GATE-2016-EE-SET 2] over contour C formed by the segments

159. A scalar potential  has one following (-3, -3, 2) → (2, -3, 2) → (2, 6, 2) →
$
$
$
gradient;  = yz i + xz j + xyk . (2, 6, -1) is _______. [GATE-19-EE]

c    
Consider the integral dr on 162. Let a and a be unit vectors along x
y
x
$
$
$
the curve r = x i + y j + zk . The curve and y directions, respectively. A vector
−
C is parameterized as follows: function is given by F = a y a x .
y
x
 x t The line integral of the above function
=
  F  dl along the curve C, which
 

=
3
 y t 2 and 1 t 
 C
 z = 3t 2 follows the parabola y = x as shown

2

below is __________ (rounded off to 2
The value of the integral is _________
decimal places).
[GATE-2016]
160. As shown in the figure, C is the arc from
the point (3, 0) to the point (0, 3) on the
circle x + 2 y = 2 9. The value of the
integral
)
)
(  y + 2yx dx + ( 2xy x dy is
2
+
2
c
[GATE-20-EE]
_____________
163. Consider the line integral
(up to 2 decimal places).
 (xdy − y dx ) the integral being
C
taken in a counterclockwise direction
over the closed curve C that forms the
boundary of the region R shown in the
figure below. The region R is the area
enclosed by the union of a 2 3
rectangle and a semi-circle of radius 1.

The line integral evaluates to
[GATE-18-EE]









76

Vector Calculus



















2
(a) 0 (b)
+
+
(a) 6  / 2 (b) 8  3
+
+
(c) 12  (d) 16 2 (c) 1 (d) 2 3
[GATE-19-EC] [GATE-10 (EC)]

164. The value of the line integral 166. Let
 F  r ds , where C is a circle of C = (  , x ) y  R 2 ,max  , x y = 1 

C
4 the value of line integral
radius units is _______. Here,
( ))
( ))
 (  xy + 2y + sin e x dx + ( x y + cos e y dy
2
2
$
F ( , x y = ) y i + $ 2x j and r is the C
is ________ [CSIR]
UNIT tangent vector on the curve C at
an arc length s from a reference point on  e y ( y )
+
$ $ 167. Let I = dx + e ln x x dy ,
the curve. i and j are the basis vectors C x
in the x-y Cartesian reference. In where C is the positive oriented
evaluating the line integral, the curve has boundary of the region enclosed by
to be traversed in the counter-clockwise y = 1 x , y = 2 and x = 1
2
+
direction. [GATE-16-ME-SET 3] 2 then the
value of I = _______



(a) 1 (b) 5
8 24

u r ur 7 3
$

165. If A = xya x + x 2 $ y a then  A dr (c) 24 (d) 8
over the path shown in the figure is
[GATE]







77

Vector Calculus

168. Value of the integral [GATE-2014-EE-SET 1]
C  ( xy dy − y dx ) , where C is the $
2

178. The value of   curlv n ds where
square cut from the first quadrant by the S
2
lines x = 1 and y = 1 will be (use v = 2yi + 3x j − z k and S is the
Green’s theorem to change the line upper half surface of the sphere
integral into double integral)
$
2
x + y + z = 9 , n is the positive
2
2
1
(a) (b) 1 unit normal vector to s and c is its
2 boundary ____
3 5 (a) 3 (b) 9
(c) (d)
2 3
(c) 18 (d) 32

[GATE-2005] $
)
179. The value of  (   F  n ds
− yi + x j S
169. For a > 0, b > 0 let F = zi + x j +
b x + a y 2 where F = yk and S is a
2 2
2
be a planar vector field. Let hemisphere z = 1 x − y of unit
2
2
−
C = (  , x ) y  R 2 / x + y = a − b 2  radius above xy plane.
2
2
2
be the circle oriented anticlockwise. (a)  (b) 2
Then  F  dr = _____
C 
(c) (d) 4
2 2
(a) (b) 2
ab 180. The value of
 sin zdx − cos xdy + sin ydz

(c) 2 ab (d) 0 C

 xdy − ydx where c is the boundary of the rectangle


0 x  , 0 
y 
4
170. The value of 2 2 taken in 2 , z = .
C x + y
the positive direction over any closed (a) 1 (b) 2
continuous curve C with origin inside it.
(c) 3 (d) 4
171. The line integral of function F = yzi, in
the counter clockwise direction, along 181. The value of  F  dr where
the circle x + 2 y = 2 1 at z = 1 is C
2
F = y i + x j − (x + 2z )k where C
2
−
(a) 2 (b) − is the boundary of the triangle with
vertices at (0, 0, 0), (a, 0, 0), (a, a, 0).
(c)  (d) 2





78

Vector Calculus
 ydx zdy +
+
182. The value of xdz y $ j + z k
$
C ( x + 2 y + 2 z 2 ) 3/ 2 ( x + 2 y + 2 z 2 ) 3/ 2
where C is the intersection of
2
2
2
x + y + z = a and x z a . $ $ $
2
+
=
where i , j , k are unit vectors along
183. The value of the axes of a right-handed rectangular /
)
( (  2x − ) y i − yz j − y zk dr . Cartesian coordinate system. The surface
2
2
u r
u r
u r
C integral   f  dS (where dS is an
Where C is the boundary of upper half of elemental surface area vector) evaluated
the surface of the sphere over the inner and outer surfaces of a
2
2
2
x + y + z = 1 above xy plane is spherical shell formed by two concentric
spheres with origin as the center, and
−
(a)  (b)  internal and external radii of 1 and 2,
respectively, is
(c) 2 (d) 0
(a) 4 (b) 0
184. Given vector
)
1 (c) 2 (d) 8
$
$
3 $
$
3
3
u = ( − y i + x j + z k and n as
3 [GATE-20-ME-SET 1]
the unit normal vector to the surface of
the hemisphere 186. The value of
)
( x + y + z = 1;z  0 ) , the value of (  4xi − 2y j + z k  nds where S
$
2
2
2
2
2
S
)
integral (   u  n dS evaluated is bounded by x + 2 y = 2 4 , z = and
0
on the curved surface of the hemisphere z = 2 is
S is
(a) 16 (b) 48
 
(a) − (b) (c) 32 (d) 84
2 3
$
 187. The value of  F  nds where S is the
(c) (d)  S
2 surface of the sphere x + y + z = 4
2
2
2
$
[GATE-19-ME-SET 2] where n is the unit normal and
185. A vector field is defined as F = xi + y j + zk is ________
ur x
f ( , ,x y z = ) 3/ 2 $ i + (a) 32 (b) 16
( x + 2 y + 2 z 2 )
(c) 8 (d) 64





79

Vector Calculus

188. The flux of the vector filled (a) 2 (b) 4
F = xi + y j + zk flowing out through
(c) 8 (d) 12
the surface of the ellipsoid
x 2 + y 2 + z 2 =   0 [CSIR]
a 2 b 2 c 2 1. a b c  is




(a) abc (b) 3 abc 192. The surface integral   F  n ds
S
(c) 2 abc (d) 4 abc over the surface S of the sphere
2
2
2
189. The value of x + y + z = 9 , where
)
)
) (x z j +
S  ( ( x − yz i − 2x y j + zk  n ds F = (x + y i + + ) ( y + ) z k
$
3
2
where S is the surface of the cube and n is the unit outward surface normal,
2
z
y
bounded by x = = = and co- yields _____.
ordinate planes is ________ [GATE-17-ME]
32 56 193. Let a > 0 and let
(a) (b)
3
3 3 S = (  x , , y ) z  R x + y + z = a 2 
2
2
2
)
4
16 64 . Evaluate  S  ( x + 4 y + 4 z ds .
(c) (d)
3 3


2 a 6 12 a 4
2
2
2
190. Let S be the sphere x + y + z = 1. (a) (b)
5 5
The value of surface integral


12 a 8 12 a 6
  ( sin ,cos x ,2z z sin y  ) ( , , y z ds is (c) (d)
)
2
−
y
x
x
S 5 5
_______ 1 $
 2 194. Consider the function F = r 2 r , where
(a) (b)
3 3 r is distance from the origin and r is the
$
4 8 unit vector in the radial direction. The
(c) (d) divergence of this function over a sphere
3 3 of radius1, which includes the origin is
_____.
191. Let S be the unit sphere
2
x + y + z = 1. Then the value of (a) 0 (b) 2
2
2
surface integral (c) 4 (d) R
)
 S    (  2x + 3x − y + 5z  2  ds is
2
2

80

Vector Calculus

195. Let,
2
B = (  x , , y ) z  R 3 & x + y + z  4 
2
2
. Let, r = xi + y j + zk be vector
valued function defined on B. If
r = x + y + z then the value of
2
2
2
2
    r r dv is ______.
2
S
(a) 16 (b) 32

(c) 64 (d) 128







































81

4










Differential









Equations










&










Partial









Differential







Equations

Differential Equations & Partial Differential Equations

Differential Equations With real constant coefficients, then
the least possible value of n is
+
)) b
1. If y = ln (sin (x a + , where a
(a) 1 (b) 2
and b are constants, is the primitive,
then the corresponding lowest order (c) 3 (d) 4
differential equation is
[JAM CA 2011]
( ( ) 2 ) 4. The differential equation representing
11
1
(a) y = − 1+ y
the family of circles touching y-axis
at the origin is
2
1
y
(b) y = 11 y − 2 ( )
(a) Linear and of first order
2
1
y
(c) y = 11 1+ ( ) (b) Linear and of second order
2
(d) y = 11 y + 1 y [JAM CA 2005] (c) Non-linear and of first order
(d) Non-linear and of second order
2. Which one of the following
differential equations represent all [JAM MA 2006]
circles with radius a?
5. Solution of the differential equation
2
 dy  2 d y xy + 1 sin2y = x 3 sin y is
2
2
(a) 1+       + a − x 2 = 0
 dx  dx 2 (a) cot y = − x + cx
3
2
 dy  2 d y (b) 2cot y = x + 3 2cx
2
2
2
(b) 1+       + a − y 2 = 0
 dx  dx 2
2
3
(c) tan y = − x + cx
   dy  2  3  d y 2 3 2
2 

(c) 1+        + a 2   2    = 0 (d) 2tan y = x + 2cx
 dx    dx
       [JAM CA 2005]
   dy  2  3  d y 2 6. General solution of the differential
2 
(d) 1+         = a 2   2    − / y x
 dx    dx equation xdy = ( y + xe ) dx is
      
given by
[JAM CA 2008]
+
(a) e − / y x = ln x c
3. If y = x cosx is a solution of an nth
+
order linear differential equation (b) e / y x = ln x c
n
d y + a d n− 1 y + ...... a dy + a y = 0 (c) e − / x y = x c
+
+
dx n 1 dx n− 1 n− 1 dx n


83

Differential Equations & Partial Differential Equations

+
(d) e / x y = x c (b) x 2 ( 2x e − y 2 ) = c
+
7. The differential equation 2
+
2ydx − (3y − 2x )dy = (c) x ( 2x e − y ) = c
0

(a) exact and homogeneous but not (d) x ( 2x e − y 2 ) = c
−
linear
[JAM CA 2006]
(b) homogeneous and linear but not

exact 10. The solution of the initial value
problem xy − 1 y = 0 with
(c) exact and linear but not 1
y
homogeneous xy − y = 0 ( ) 1 = 1 is

x
y
(d) exact, homogeneous and linear (a) ( ) x =
[JAM CA 2006] 1
y
(b) ( ) x =
8. The general solution of the x
y x =
differential equation (c) ( ) 2x − 1
(x + − ) 3 dx − (2x + 2y + ) 1 dy =
0
y
is (d) ( ) x = 1 [JAM CA 2007]
y
2x − 1
k
(a) ln 3x + 3y − 2 + 3x + 6y =
11. The solution of the differential
equation
k
(b) ln 3x + 3y − 2 − 3x − 6y =   x sin y − y cos y   dx x cos dy = 0
y
+
k
(c) 7ln 3x + 3y − 2 + 3x + 6y =   x x     x
y
0
with initial condition ( ) 0 = is
k
(d) 7ln 3x + 3y − 2 − 3x + 6y = y

x
(a) sin = 1 (b) y =  n x
[JAM CA 2006] x
9. The general solution of the y
differential equation (c) y = x sin x (d) x = y
( 6x − e − y 2 ) dx + 2xye − y 2 dy = 0 is
2
[JAM CA 2008]



12. The differential equation
2
−
(a) x 2 ( 2x e − y 2 ) = c (2x + by 2 ) dx cxydy+ = 0 is made
exact by multiplying the integrating





84

Differential Equations & Partial Differential Equations

1 (c) 2 + x 2 = y c
factor . Then the relation between
x 2 x y
b and c is (d) 2 − 2 = c [JAM CA 2009]

=
=
(a) 2c b (b) b c 16. Consider the differential equation
dy 2 ( )
(c) 2b c+ = 0 (d) b + 2c = 0 dx − y = − y . Then lim y x is
x→ 
[JAM CA 2008] equal to

+
x
y
c
13. If e + xy + x sin y e = is the (a) -1 (b) 0
general solution of an exact (c) 1 (d) 
differential equation, then the
differential equation is [JAM CA 2010]

dy e − x y − sin y 17. An integrating factor of the
(a) = differential equation
−
dx e − y x x cos y
2
2xy dx + ( y − x 2 ) dy = is
0
dy e + x y + sin y
(b) = 1
+
dx e + y x x cos y (a) y (b)
y
x
y
dy − (e + + sin ) y 1 1
(c) = (c) (d)
+
dx e + y x x cos y y 2 y 3
x
y
dy − (e − − sin ) y [JAM CA 2010]
(d) =
−
dx e − y x x cos y 18. The general solution of the
differential equation
[JAM CA 2009] dy 2 )( − x 2 − 1 )
+
x
14. The general solution of dx = (1 y e − 2 tan y is
+
y 1 ( x + y 2 ) = y is (a) e x 2 tan y = 1 x c
−
+
(b) e − x 2 tan y = x c
=
=
2
(a) x cy + y (b) x cy − y 2
(c) e x tan y = x + 2 c
2
2
(c) y = cx + x (d) y = cx − x (d) e − x tan y = − 1 x + 3 c
[JAM CA 2011]
[JAM CA 2009]
)
g
0
19. If ( , x y dx + (x + ) y dy = is an
−
1
15. The general solution of y = 2 x y is exact differential equation and if
2
g ( ,0x ) = x , then the general
(a) 2 + − x 2 = − y c
solution of the differential equation is
(b) 2 − − x 2 = − y c
3
2
c
(a) 2x + 2xy + y =

85

Differential Equations & Partial Differential Equations

(b) 2x + 3 6xy + 3y = 2 c 23. One of the integrating factors of the
differential equation
2
2
2
c
(c) 2x + 2xy + y = ( y − 3xy ) dx + ( x − xy ) dy = 0 is
2
2
c
(d) x + 2xy + y = 1 1
(a) 2 2 ) (b) 2
x
[JAM CA 2011] ( x y ( ) y
20. The general solution of the 1 1
2
d y   dy   2 (c) ( ) 2 (d) ( )
xy
xy
differential equation =    
dx 2  dx 
is [JAM MA 2007]
=
y
(a) x c e − y + c e 24. Consider the differential equation
2
1
( ) dy =
2cos y 2 − sin y 2 0
( ) dx xy
=
(b) x c e + 1 y c (a) e is an integrating factor
x
2
=
−
(c) x c e − y + c (b) e is an integrating factor
x
2
1
=
(d) x c e + 1 y c y [JAM CA 2011] (c) 3x is an integrating factor
2
3
21. An integrating factor of (d) x is an integrating factor
dy
x + (3x + ) 1 y = xe − 2x is [JAM MA 2009]
dx
25. Consider the differential equation
x
3x
(a) xe (b) 3xe dy = ay by , where , a b  and
−
2
0
dx
3 x
x
(c) xe (d) x e y ( ) 0 = y . As x → +, the solution
0
[JAM MA 2005] y ( ) x tends to
22. If k is a constant such that (a) 0 (b) a/b
(x− ) 1 2
+
xy k = e 2 satisfies the (c) b/a (d) y
0
differential equation
dy [JAM MA 2010]
2
x
x = ( x − − ) 1 y + (x − ) 1 , then
dx 26. Consider the differential equation
k is equal to (x + + ) 1 dx + (2x + 2y + ) 1 dy = .
0
y
(a) 1 (b) 0 Which of the following statements is
true?
(c) -1 (d) -2
(a) The differential equation is linear
[JAM MA 2007]
(b) The differential equation is exact



86

Differential Equations & Partial Differential Equations

+
(c) e x y is an integrating factor of the (a) y 2 cos x + x sin y = 0
differential equation 
+
(b) y 2 sin x x cos y =
(d) A suitable substitution transforms 2
the differentiable equation to the (c) y 2 sin x + x sin y = 0
variables separable form 
(d) y 2 cos x + x cos y =
[JAM MA 2010] 2
30. Consider the differential equation
11
27. If y is an integrating factor of the dy
=

differential equation dx − 2x  ( ) x , x R , satisfying
2xy dx − (3x − y 2 ) dy = , then the  0, x  0
2
0


0
value of a is y ( ) 0 = , where ( ) x =  1, x  0


(a) -4 (b) 4 . This initial value problem
(c) -1 (d) 1 (a) has a continuous solution which is
0
[JAM MA 2011] not differentiable at x =
28. The solution of the differential (b) has a continuous solution which is
equation differentiable at x =
0
dy = − ( x x + 2 y − 2 10 ) , y ( ) 0 = 1 is (c) has a continuous solution which is
dx ( y x + 2 y + 2 ) 5 differentiable on R


(a) (d) does not have a continuous
4
=
2
2
2
x − 2x y − y − 20x − 10y + 11 0 solution on R.
2
4
(b)
2
4
2
=
x + 2x y + y + 20x + 10y − 11 0 31. If ( ) x satisfies dy + 2y = 2 e − x 2
4
2
2
+
y
(c) dx
( )
y
0
x + 2x y − y + 20x − 10y + 11 0 with ( ) 0 = , then lim y x equals
=
2
4
4
2
2
2
x→ 
(d)
=
x + 2x y + y − 20x + 10y − 11 0 (a) 0 (b) 1
2
4
4
2
2
2

(c) 2 (d) -1
[JAM MS 2008]
[JAM GP 2008]
29. The solution of the differential
equation 32. The solution of the initial value
dy = y 2 cos x + cos y ; y      = 0 problem dy = sin x , y ( ) 0 = 0 is
y
dx x sin y − 2 sin x   2     dx y + 2
is (a) ( y y + ) 2 = ( 4 1 cos x− )

87

Differential Equations & Partial Differential Equations
=
)
−
(b) y = 2 ( 2 1 cos x (b) r a cos
)
+
0
(c) ( y y + ) 4 + cos x = (c) r = a (1 cos
)
+
)
−
(d) ( y y + ) 4 = ( 2 1 cos x (d) r = a (1 sin
36. The solution of the differential
[JAM GP 2010]
equation y + 11 4y = 0 subject to
33. The orthogonal trajectories of the y 1 y 1 2
curves y = 3x + + are ( ) 0 = , ( ) 0 = is
2
3
x c
(a) 2tan 3x + − 1 3ln y = k (a) sin2x + 1
(b) cos2x + 2x
(b) 3tan 3x + − 1 2ln y = k
(c) sin2x + cos2x
(c) 3tan 3x − − 1 2ln y = k
(d) sin2x − cos2x [JAM CA 2005]

(d) 3ln x − 2tan 3y = − 1 k 37. A particular solution of the

differential equation
[JAM CA 2006]
2
4
( D + 2D − ) 3 y e is
=
x
34. Orthogonal trajectories of the family
x
x
2
2
of curves (x − ) 1 + y + 2ax = are (a) (x + ) 1 e (b) xe
0
the solution of the differential xe x xe x
equation (c) 4 (d) 8
dy
2
+
2
(a) x − y − 1 2xy = 0 [JAM CA 2005]
dx 38. A particular solution of the
dy differential equation
2
2
1 2xy
(b) x + y − + = 0
1
11
y
dx y 111 − 3y + 3y − = e x cos2x is
dx 1 1
−
2
2
(c) x − y − 1 2xy = 0 (a) − e x sin2x (b) e x sin2x
dy 8 8
dx 1 x x
−
2
2
(d) x + y + 1 2xy = 0 (c) e cos2x (d) e sin2x
8
dy
[JAM CA 2005]
[JAM CA 2008]
35. The orthogonal trajectory of the 39. If ( ) 3y x = 1 1 y 1 ( ) 4x + y 2 ( ) x and
( )
)
cardioid r = a (1 cos , a being the y x = 1 2 ( ) 4y x + 1 ( ) 3y x , then
+
2
parameter, is y 1 ( ) x is
(a) r = a (1 cos− ) x 7x
(a) c e + c e
1 2

88

Differential Equations & Partial Differential Equations

x
(b) c e + c e − 7x 42. Two linearly independent solutions of
2
1
the differential equation
11
x
1
−
x
7x
(c) c e + c e y − 2y + y = 0 are y = e and
1
1 2
y = xe . Then a particular solution
x
2
−
x
(d) c e + c e − 7x of y − 2y + = e x sin x is
11
1
y
2
1
40. The general solution of the (a) y cos x + y (sin x x cos ) x
−
differential equation 1 2
( cos x −
( ) 8y x =
x −
1
x
y 11 ( ) 4y x + ( ) 10 cos x (b) y + sin x + y x sin ) x
e
2
1
+
y x
(a) e 2x ( cos2x k 2 sin2x ) + (c) ( cos x − sin ) x − y 2 cos x
k
1
1
y x
1
e x (2cos x + sin ) x (d) ( sin x − cos ) x + y 2 cos x
[JAM CA 2008]
+
k
(b) e 2x ( cos2x k 2 sin2x ) +
( ))
1
( ),
W
43. Let x ( y x y x is the
1
2
e x (2cos x − sin ) x Wronskian formed for the solutions
y
y 1 ( ) x and ( ) x of the differential
2
+
k
1
11
(c) e − 2x ( cos2x k 2 sin2x ) − equation y + a y + a y = 0 . If
1
2
1
W  0 for some x = x in [a,b] then
0
e x (2cos x − sin ) x

(a) it vanishes for any x   , a b
+
k
(d) e − 2x ( cos2x k 2 sin2x ) + (b) it does not vanish only at x a
=
1
(c) it does not vanish for any

e x (2cos x + sin ) x x   , a b
=
[JAM CA 2006] (d) it does not vanish only at x b
[JAM CA 2009]
41. The general solution of the
differential equation 44. The general solution of
1
11
y
y 111 + y − y − = 0 is y − 11 m y = 2 0 is
+
c
(a) sinhmx c coshmx
1 2
+
c +
2
x
(a) ( 1 xc + 2 x c 3 )e (b) cosc 1 mx c 2 sinmx
+
(c) cosc 1 mx c 2 sinhmx
−
c +
2
x
(b) ( 1 xc + 2 x c 3 )e
+
(d) sinc 1 mx c 2 coshmx
−
x
(c) c e + x (c + xc )e
2
3
1
[JAM CA 2009]
x
(d) (c + 1 xc 2 )e + x c e
3

89

Differential Equations & Partial Differential Equations

1
45. The solution of the differential (a) ( xe + 4x − 4x −
2
d y 8 xe ) 1
x
equation − y = e satisfying
dx 2
1
dy 3 (b) ( xe − 4x − 4x +
0
y ( ) 0 = and ( ) 0 = is xe ) 1
dx 2 8
x 1   4x − 4x 1  
x
y x =
(a) ( ) sinh x + e (c)  e − xe +   
2 4  2 
x  
x
(b) ( ) x = y x cosh x + e (d) 1  xe − 4x e − 4x + 1 
2 4   2    
x
y x =
x
(c) ( ) sinh x − e [JAM CA 2011]
2
48. A general solution of the differential
x
y x =
x
(d) ( ) 2 cosh x − x e d y d y
3
2
2 equation − 3 + 4y = is
0
dx 3 dx 2
[JAM CA 2010]
2x
x
2x
(a) y = c e + c e + c xe
46. The solution of the differential 1 2 3
3
d y dy −
2x
x
2x
equation − 9 = cos x is (b) y = c e + c e + c xe
dx 3 dx 1 2 3
−
x
2x
−
x
1 (c) y = c e + c xe + c e
1
2
3
(a) ( ) x = C e + C e − 3x + C + sin x
3x
y
3
2
1
10
−
4x
x
x
(d) y = c e + c e + c e
1 1 2 3
3x
y
(b) ( ) x = C e + C e − 3x + C − sin x
3
1
2
10 [JAM CA 2011]

y
y
49. Let ( ) x and ( ) x be twice
1 1 2
3x
y
(c) ( ) x = C e + C e − 3x + C + cos x
3
1
2
10 differentiable functions on a interval I
satisfying the differential equations
1
y
(d) ( ) x = C e + C e − 3x + C − cos x dy
3x
x
1
3
2
10 1 − y − y = e and
dx 1 2
[JAM CA 2010] dy dy
y
0
2 1 + 2 − 6y = . Then ( ) x
47. A particular integral of the dx dx 1 1
differential equation is
2
d y − 16y = 4sinh 2x is
2
dx 2


90

Differential Equations & Partial Differential Equations

1 1
x
(a) c e − 2x + c e − 3x e (c) e x /2 + 2cos4x + 5sin4x
1
2
4 5
1 (d) 2cos4x + 5sin4x
x
(b) c e + 2x c e + 3x e
1
2
4
[JAM GP 2005]
1
x
(c) c e + 2x c e − 3x − e 52. Consider the differential equation
1
2
4
y + 6y + 25y = with initial
1
11
0
1
x
(d) c e − 2x + c e + 3x e condition ( ) 0 = . Then, the
y
0
1 2
4
general solution of the initial value
[JAM MA 2008] problem is
y
50. The solution ( ) x of the differential − 3x
+
(a) e ( cos4A x B sin4x )
2
d y dy
0
equation + 4 + 4y = − 3x
dx 2 dx (b) Be sin4x
4
y
satisfying the conditions ( ) 0 = , − 4x
(c) Ae sin3x
dy ( ) 0 = is
8
+
A
dx (d) e − 4x ( cos3x B sin3x )
2x
(a) 4e [JAM GP 2006]
(b) (16x + ) 4 e − 2x 53. The differential equation
2
d y + y = 0 satisfying ( ) 0y = 1,
(c) 4e − 2x + 16x dx 2
y  =
( ) 0 has
2x
(d) 4e − 2x + 16xe
(a) a unique solution
[JAM MA 2011]
(b) a singly infinite family of
51. The particular integral of the
following differential equation solutions
(c) no solution
y + 11 2y + 1 5y
(d) a doubly infinite family of
5
= e x / 2 x + 18cos4x − 71sin 4x is solutions
4 [JAM GP 2008]
54. The particular integral of the
5
(a) e x /2 + 5cos4x differential equation
4
y + y + 3y = 5cos (2x + ) 3 is
11
1
(b) 5cos4x + 2sin4x





91

Differential Equations & Partial Differential Equations
(
(a) 2cos 2x + ) 3 − sin (2x + ) 3 57. Consider the following differential
equation:
(
(b) 2sin 2x + ) 3 + cos (2x + ) 3 dy = − 5 ; initial condition: y = at
y
2
dt
(
(c) sin 2x + ) 3 − 2cos (2x + ) 3 t = 0
3
(
(d) 2sin 2x + ) 3 − cos (2x + ) 3 The value of y at t = is
−
(a) 5e − 10 (b) 2e − 10
[JAM GP 2009] −
−
2
(c) 2e 15 (d) 15e
2x
2x
55. If e and xe are particular [GATE-2015-ME-SET-2]
solutions of a second order 58. Consider the differential equation
homogeneous differential equation dx
−
with constant coefficients, then the dt = 10 0.2x with initial condition
equation is x ( ) 0 = 1. The response ( ) t for
x
t  0 is
2
d y dy
0
(a) − 4 + 4y = − 0.2t 0.2t
−
−
dx 2 dx (a) 2 e (b) 2 e
−
−
(c) 50 49e − 0.2t (d) 50 49e 0.2t
2
d y dy
(b) − 5 + 6y = 0 [GATE-2015-EC-SET-2]
dx 2 dx
59. Consider the following differential
equation:
2
d y
(c) − 4y = 0 y
dx 2 ( x ydx xdy )cos =
+
x
2
d y dy y
(d) − 3 + 2y = 0 ( y xdy − ydx )sin
dx 2 dx x
Which of the following is the solution
[JAM GP 2010] of the above equation (c is an
56. The general solution of the arbitrary constant)?
+
dy 1 cos2y x y
differential equation = (a) cos = c
−
dx 1 cos2x y x
is x y
(b) sin = c
=
(a) tan y − cot x c (c is a constant) y x y
=
(b) tan x − cot y c (c is a constant) (c) xy cos = c
x
=
(c) tan y + cot x c (c is a constant) (d) xy sin y = c
=
(d) tan x + cot y c (c is a constant) x
[GATE-2015-CE-SET-1]
[GATE-2015-EC-SET-2]



92

Differential Equations & Partial Differential Equations

60. The solution to 6yy − 1 25x = 0 (c) Both II and IV are true

represents a (d) Both III and IV are true

(a) family of circles [GATE-2017-EC-SECTION-1]
(b) family of ellipses
63. Which one of the following is the
(c) family of parabolas general solution of the first order
(d) family of hyperbolas differential equation

[GATE-2015 (PI)] dy
2
y
= (x + − ) 1 , where x, y are
di dx
61. A differential equation − 0.2i = 0 real?
dt
is applicable over 10 t  10. (a) y = + + tan − 1 (x c , where c

−
)
+
1 x
i −
i ( ) 4 = 10 , then ( ) 5 is _______.
is a constant
[GATE-2015-EE-SET-2]
)
+
1 x
(b) y = + + tan (x c , where c is
62. Consider the following statements a constant
about the linear dependence of the
)
+
1 x
real valued functions y = 1, y = (c) y = − + tan − 1 (x c , where c
x
1
2
and y = x , over the field of real is a constant
2
3
)
numbers. (d) y = − + tan (x c , where c is
+
1 x
I. y , y and y are linearly a constant
2
3
1
independent on 1 x−   0 [GATE-2017-EC-SECTION-1]
II. y , y and y are linearly 64. Consider the differential equation
1
3
2
dependent on 0 x  1 dy

(t − 81 ) + 5ty = sin ( ) t with
2
III. y , y and y are linearly y ( ) 1 = dt
2
1
3
dependent on 0 x  1 2 . There exists a unique
solution for this differential equation
IV. y , y and y are linearly when t belongs to the interval
1 2 3
independent on 1 x−   1
)
−
(a) ( 2,2 (b) ( 10,10− )
Which on among the following is
)
correct? (c) ( 10,2− ) (d) (0,10
(a) Both I and IV are true [GATE-2017-EE-SECTION-1]
(b) Both I and III are true


93

Differential Equations & Partial Differential Equations

65. The solution of the equation 68. The solution of the equation
dQ + Q = 1 with Q = at t = 0 is x dy + y = 0 passing through the
0
dt dx
point (1, 1) is
Q
(a) ( ) t = e − t − 1
2
(a) x (b) x
t −
Q t =
+
(b) ( ) 1 e
−
1
2
−
(c) x (d) x
Q t =
−
t
(c) ( ) 1 e [GATE-2018 (CE – Afternoon Session]
t −
Q t =
−
(d) ( ) 1 e 69. If y is the solution of the differential
dy
equation y 3 + x = 3 0, ( ) 0 = ,
1
y
[GATE-2017-CE-SECTION-1] dx
y −
66. For the initial value problem the value of ( ) 1 is
−
−
dx (a) 2 (b) 1
( ) ( ) 0 =
,
= sin t x 0
dt
(c) 0 (d) 1
the value of x at t =  , is [GATE-2018 (ME-Afternoon Session]
3
__________. 70. A curve passes through the point
( x = 1, y = ) 0 and satisfies the
[GATE-2017-(CH)]
differential equation
67. The solution of the differential dy x + y 2 y
2
equation = + .
dx 2y x
−
2
−
2
y 1 x dy + x 1 y dx = is
0
The equation that describes the curve
is
−
(a) 1 x = 2 c  y 2 


(a) ln 1+ 2    = x − 1

−
(b) 1 y = 2 c  x 
1   y 2  

−
2
c

(c) 1 x− 2 + 1 y = (b) ln 1+ 2   = x − 1
2  x 
2
+
c
(d) 1 x+ 2 + 1 y =   y  
(c) ln 1+    = x − 1

 x 
[ESE-2017 (EE)] 1  y 

(d) ln 1+     = x − 1

2  x 
[GATE-2018-(EC)]

94

Differential Equations & Partial Differential Equations

71. Consider the following second order Given ( ) 0 = 20 and ( ) 1 = 10 / e,
x
x
linear differential equation
x
2
d y = − 12x + 24x − 20. The where e = 2.718, the value of ( ) 2 is
2
dx 2 ___________.
boundary condition are at x = 0, [GATE-2015-EC-SET-3]
y = 5 and at x = 2, y = 21

The value of y at x = 1 is _________. 75. A solution of the ordinary differential
2
d y dy
[GATE-2015-CE-SET-2] equation + 5 + 6y = is
0
dt 2 dt
2
d y
72. Find the solution of = y which such that ( ) 0y = 2 and
dx 2
−
passes through origin and point y ( ) 1 = − 1 3e . The value of dy ( ) 0
 3  e 3 dt
 ln2, 
    is _________.
 4 
1 − [GATE-2015-EE-SET-1]
x
(a) y = e − x e
2
2
1
11
y
76. The solution to x y + xy − = 0 is
1
(b) y = (e + x e − x )
=
2 (a) y C x + 2 C x
−
3
1 2
1
(c) y = (e − x e − x ) (b) y C + C x
=
−
2
2 1 2
1
−
x
(d) y = e + x e (c) y C x + C 2
=
2 1 x
[GATE-2015-ME-SET-1]
+
4
=
(d) y C x C x
73. The solution of the differential 1 2
2
d y dy
equation + 2 + y = 0 with [GATE-2015-EC (PI)]
dt 2 dt
1
y
y
y ( ) 0 = y 1 ( ) 0 = 1 is 77. A function ( ) t , such that ( ) 0 =
−
1
y
)
)
−
t
t −
+
(a) (2 t e (b) (1 2t e and ( ) 1 = 3e , is a solution of the
differential equation
t
2
(c) (2 t e+ ) t − (d) (1 2t e− ) d y + 2 dy + y = 0. Then ( ) 2y is
dt 2 dt
[GATE-2015-EC-SET-1]
−
1
(a) 5e (b) 5e − 2
74. Consider the differential equation
−
−
2
1
2
d x ( ) t dx ( ) t (c) 7e (d) 7e
+ 3 + 2x ( ) 0t =
dt 2 dt [GATE-2016-EE-SET-1]

95

Differential Equations & Partial Differential Equations

78. If y = f ( ) t satisfies the boundary (a)
 
+
+

value problem y + 11 9y = 0,   c + c x c 3 sin 3x c 4 cos 3x

2
1

)
y  ) 2 , then ( / 4 is and 3x −    4 12x + 2 c
( / 2 =
y 


_______________.

[GATE-2016] (b) c X +    2 c 3 sin 3x + c 4 cos 3x



79. The particular solution of the initial and 5x −    4 12x + 2 c


value problem given below is

+

2
d y dy (c) c +    1 c 3 sin 3x c 4 cos 3x

0
+ 12 + 36y = with

dx 2 dx and 3x −   4 12x + 2 c

dy  
y ( ) 0 = 3 and = − 36
dx x= 0 (d)
 c + c x c sin 3x c 
+
+

)

−
(a) (3 18x e − 6x   1 2 3 4 cos 3x

and 5x −    4 12x + 2 c


)
+
(b) (3 25x e − 6x
)
+
(c) (3 20x e − 6x [GATE-2016-CE-SET-1]
82. What is the solution for the second
)
−
(d) (3 12x e − 6x [GATE-2016] order differential equation
2
d y
80. Let y(x) be the solution of the dx 2 + y = 0, with the initial
differential equation dy
2
d y − 4 dy + = conditions y x= 0 = 5 and dx = 10 ?
4 0 with initial
dx 2 dx x= 0
+
dy (a) y = 5 0sin x
0
y
conditions ( ) 0 = and = 1.
dx x= 0
−
Then the value of y(1) is _______. (b) y = 5cos 5sin x
[GATE-2016-EE-SET-2] (c) y = 5cosx + 10x
81. The respective expressions for (d) y = 5cosx + 10sin x
complimentary function and
particular integral part of the solution [GATE-2016 (CH)]
of the differential equation
2
4
d y + 3 d y = 108x are 83. The general solution of the
2
dx 4 dx 2 differential equation
2
d y + 2 dy − 5y = in terms of
0
dx 2 dx
arbitrary constant K and K is
1
2

96

Differential Equations & Partial Differential Equations

−
−
(a) K e ( 1+ ) 6 x + K e ( 1− ) 6 x 86. Consider the differential equation
( )
x +
0 with initial
( ) 27y x =
2
11
1
3y
1
y
y
0
−
−
(b) K e ( 1+ ) 8 x + K e ( 1− ) 8 x conditions ( ) 0 = and ( ) 0 = 2000 .
2
1
The value of y at x = 1 is
−
−
(c) K e ( 2+ ) 6 x + K e ( 2− ) 6 x ___________.
1 2
[GATE-2017-ME-SECTION-2]
−
−
(d) K e ( 2+ ) 8 x + K e ( 2− ) 8 x
1 2 87. The general solution of the
differential equation
[GATE-2017-EC-SECTION-2]
3
4
2
d y d y d y dy
84. Consider the following second-order 4 − 2 3 + 2 2 − 2 + y = 0
differential equation: dx dx dx dx
+
(a) y = (c − c 2 ) x e + c 3 cos x c 4 sin x
x
1
1
2
11
y − 4y + 3y = 2t − 3t
+
x
(b) y = (c + c 2 ) x e − c 3 cos x c 4 sin x
1
The particular solution of the
+
differential equation is (c) y = (c + c 2 ) x e + c 3 cos x c 4 sin x
x
1
−
− −
−
x
(a) 2 2t t − 2 (b) 2t t − 2 (d) y = (c + c 2 ) x e + c 3 cos x c 4 sin x
1
− −
2
2
(c) 2t − 3t (d) 2 2t − 3t [ESE-2017 (EE)]
[GATE-2017-CE-SECTION-2] 88. The solution of the differential
equation
85. The differential equation
2
d y + 16y = 0 for ( ) x with the d y − dy − 2y = 3e ,
2
2x
y
dx 2 dx 2 dx
dy
1
0
y
y
two boundary conditions = 1 where, ( ) 0 = and ( ) 0 = − 2 is
dx x= 0
dy (a) y e= − x − e + xe
2x
2x
and = − 1 has
dx x= 
2 (b) y e= x − e − 2x − xe
2x
(a) no solution
(c) y e= − x + e + xe
2x
2x
(b) exactly two solutions
(d) y e= x − e − 2x + xe
2x
(c) exactly one solution
(d) infinitely many solutions [ESE-2018 (EE)]
[GATE-2017-ME-SECTION-1]



97

Differential Equations & Partial Differential Equations

2
d y 93. An integrating factor of
89. If + y = 0 under the conditions
dt 2 dy + 2x
dy x (3x + ) 1 y = xe is ________
y = 1, = 0, when t = then y is dx
0
dt
equal to (a) xe (b) 3xe
x
3x
(a) sin t (b) cos t
x
3 x
(c) xe (d) x e
(c) tan t (d) cot t
[ESE 2018 (EE)] [JAM 2005]

90. The solution (up to three decimal 94. The general solution of −
2 11
place) at x = 1 of the differential 5 + 9 = 0 is __________
1
2
d y dy
equation + 2 + y + 0 subject to 3x
dx 2 dx (a) (c + c 2 ) x e
1
boundary conditions ( ) 0 = 1 and
y
3
(b) (c + c ln ) x x
dy ( ) 0 = − 1 is ___________. 1 2
dx
3
(c) (c + c 2 ) x x
1
[GATE-2018 (CE-Morning Session)]
3

91. Given the ordinary differential (d) ( + ) [JAM 2005]
2
1
equation
2
d y x )
+
y
1
2
d y dy 95. 2 − y = x (sin x e , ( ) 0 = ,
+ − 6y = dx
0
dx 2 dx
y 1 ( ) 0 = 1 [JAM 2005]
dy
with ( ) 0 = and ( ) 0 = , the
1
0
y
dx 96. Solve
value of y(1) is _____________ (2 sin x + y 3y 4 sin cos ) x dx −
x
(Correct to two decimal places).
(4y 3 cos x + 2 cos ) x dy = 0 .
[GATE-2018 (ME-Afternoon Session)]
92. The position of a particle y(t) is [JAM 2005]
described by the differential equation: (c + c ln ) x
97. If 1 2 is the general
2
d y dy 5y x
= − − .
dt 2 dt 4 solution of the differential equation
2
d y dy
2
0
The initial conditions are ( ) 0 = 1 x + kx + y = 0, x  then
y
dx 2 dx
dy k equals ______
and = 0 . The position (accurate
dt t= 0
to two decimal places) of the particle [JAM 2006]
at t = is _________.
[GATE-2018 (EC)]

98