Page 38 - Engineering Mathematics Workbook_Final

Calculus

f  f   1 ln x 
+
(b) and exist but not equal (c) lncot       + c
 x  y  2 
 ln x 
f  f  (d) lnsin 1+       + c
(c) exists but does not  2 
 x  y
f  f  [JAM CA 2005]
(d) exists but does not
 y  x 1
205. The value of 1  e e (1 ln x+ ) dx is
[JAM CA 2006] x 2

 x   y  (a) 1 (b) 1/e

202. Suppose z = x sin       + y sin       , (c) e (d) 0
 y   x 
z  z  206. If a real valued function f is given by
xy  0 . Then x + y is equal to f ( ) t
 x  y  x dt = 2 x b x  where a
+
,
0
a t 2
(a) -z (b) 0 > 0 and b are areal constants, then f(4) is
(c) z (d) 2z equal to

=
=
203. If z = e xy 3 , x t cost , y t sint then (a) 4 (b) 6
(c) 8 (d) 10
dz at t =  is
dt 2 [JAM CA 2010]

2
t −
cos x
f
(a)  3 / 8 (b)  3 / 4 207. Let ( ) x =  sin x e dt , then
−
(c)  3 / 2 (d)  3 / 8 ( ' f  / ) 4 equals
[JAM CA 2009]
(a) 1/ e (b) − 2 / e
 dx (c) 2 / e (d) − 1/ e
204. The value of
+
−
x 1 cos 2 (1 ln x )
[JAM MA 2006]
is
208. Let :f R R→ be a continuous function.
 1 ln x 
+
(a) ln tan       + c 0  x f ( )dt = x sin ( ) x
2t
 2  If  for all
 1 ln x  
−
(b) ln tan       + c x R , then f(2) is equal to
 2 
(a) -1 (b) 0
(c) 1 (d) 2

[JAM MA 2007]

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