Page 13 - Engineering Mathematics Workbook_Final

Calculus

1  1 z  − 2    y 
 
  

)

5
f
1. The integral     dx dy dz = 4. Let ( , x y = x y 2 tan − 1       . Then
0  0   0     x 
___________ f  f 
x + y = _______
1  1 y  − 2    x  y
  
 


(a)     dx dz dy
0  0   0    (a) 2f (b) 3f
1  1 y  − 1   (c) 5f (d) 7f
  
 


(b)     dx dz dy
0  0   0    [JAM 2006]
2  2  1 z   cos x t − 2
−
  
 


f
(c)     dx dz dy 5. Let ( ) x =  e dt . Then
0  0   0    sin x
1
f  )
( / 4 equals _______
2  2  1 y  
−
 

  

(d)     dx dz dy
0  0   0    (a) 1 e (b) − 2 e
[JAM 2005]
(c) 2 e (d) − 1 e
2 n + 1 + 3 n + 1
2. Lt
n
n→  2 + 3 n [JAM 2006]
(a) 3 (b) 2  1 cx 1 x 
+
6. If Lt       = 4 then
−
(c) 0 (d) 1 x→ 0  1 cx 
+
[JAM 2006]  1 2cx 1 x 
Lt   −     is _________
17
f x =
3. Let ( ) (x − 2 ) (x + ) 5 24 then x→ 0  1 2cx 
_______ (a) 2 (b) 4
(a) f does not have a critical point at (c) 16 (d) 64
[IISC 2008]
(b) f has maxima at 2
2
y
(c) f has minima at 2 7. Evaluate   xe dx dy where R is
R
(d) f has neither minima nor maxima at 2 the region bounded by the lines x = 0, y
2
[JAM 2006] = 1 and the parabola y = x .

[JAM 2006]

11