Page 39 - Engineering Mathematics Workbook_Final

Calculus

3
2
) ( ) t dt ,
209. Let ( ) x = f 0  x ( x − 2 t g 213. If 0  x f ( ) t dt = x 2 sin x + x . Then
where g is real valued continuous f     
function on R. Then f’(x) is equal to   2     is

( )
3
(a) 0 (b) x g x    2    3 3 2




 x x (a)   2   +   2   (b)  + 4
(c) g ( ) t dt (d) 2x 0  g ( ) t dt    
0 3 2
(c)  − (d) 0
[JAM MA 2008] 4
210. Let a be non-zero real number. Then e 1 +
1 214. The value of 1  2 e (1 ln ) x dx is
lim 2  x sin t 2 x
( ) dt equals
2
→
x a x − a a
(a) 1 (b) 1/e
1 1 (c) e (d) 0
( )
( )
(a) sin a 2 (b) cos a 2
2a 2a [JAM CA 2005]
1 1
( )
( )
(c) − sin a 2 (d) − cos a 2
2a 2a  dx
215. The integral 1  2 x )
+
[JAM MA 2009] x (1 e
→
211. Let : f R R be defined as (a) converges and has value < 1
(b) converges and has value equal to 1
 tant


f ( ) t =  t , t  0 (c) converges and has value > 1
 (d) diverges
 1, t = 0

0
216. For   , the value of the integral
1 3 0   −  x 2
Then the value of lim 2  x 2 f ( ) t dt e dx equals
x→ 0 x x

1  
(a) is equal to (-1) (b) is equal to 0 (a) (b)
(c) is equal to 1 (d) does not exist 2  2
2 
[JAM MS 2006] (c) (d) 2
 
d sin x 2
t
212.  e dt is equal to [JAM CA 2007]
dx 0
)
x
2
2
(a) e sin x cos x (b) e sin x 217. The integral 0   /2 min (sin ,cos x dx
)
2
(c) (2sin x e sin x (d) e 2sin x equals

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