APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems. We started to develop ways to enhance students IQ. We started to leave an indelible mark on the students who have undergone APEX training. That is why APEX INSTITUTE is very well known of its quality of education

1. Mission IIT –2013
sec2 x
2
f (t )dt
Q1. lim equals
x
 2
4 x – 2
16
8 2 2 1
(a) f(2) (b) f(2) (c) f  (d) 4 f(2)
   2
t 2 f(x) – x 2 f (t )
Q2. Let f(x) be differentiable on the interval (0, ) such f(1) = 1, and lim = 1 for
t x t–x
each x > 0. Then f(x) is
1 2x 2 –1 4x 2 –1 2 1
(a) + (b) + (c) + 2 (d)
3x 3 3x 3 x x x
The value of lim   sin x   1  x   , where x > 0 is
1/ x sin x
Q3.
x 0
(a) 0 (b) –1 (c) 1 (d) 2
Q4. If f(x) is continuous and differentiable function and f(1/n) = 0  n  1 and n  I, then
(a) f(x) = 0, x  (0, 1] (b) f(0) = 0, f’(0) = 0
(c) f(0) = 0 = f’(0), x  (0, 1] (d) f(0) = 0 and f’(0) need not to be zero
Q5. The function given by y = | | x | – 1| is differentiable for all real numbers except the points
(a) {0, 1, –1} (b) 1 (c) 1 (d) –1
f (x 2 ) – f (x)
Q6. If (x) is differentiable and strictly increasing function, then the value of lim is
x  0 f (x) – f (0)
(a) 1 (b) 0 (c) –1 (d) 2
f (2h + 2 + h 2 ) – f (2)
Q7. lim , given that f’(2) = 6 and f’(1) = 4
h  0 f (h – h 2  1) – f (1)
(a) does not exist (b) is equal to –3/2 (c) is equal to 3/2 (d) is equal to 3
 (a – n)nx – tan x  sin nx
Q8. If lim = 0, where n is non-zero real number, then a is equal to
x 0 x2

2. Mission IIT –2013
n 1 1
(a) 0 (b) (c) n (d) n+
n n
 f 1  x  
1/ x
Q9. Let f : R R be such that f(1) = 3 and f’(1) = 6. Then lim   equals
x 0
 f (1) 
(a) 1 (b) e1/2 (c) e2 (d) e3
(cos x –1)(cos x – e x )
Q10. The integer n for which lim is a finite non-zero number is
x 0 xn
(a) 1 (b) 2 (c) 3 (d) 4
 tan –1 x if |x|  1

Q11. The domain of the derivative of the function  1 is
 | x | –1 if x  1
2
(a) R – {0} (b) R – {1} (c) R – {–1} (d) R – {–1, 1}
Q12. Which of the following functions is differentiable at x = 0?
(a) cos(|x|) + |x| (b) cos(|x|) – |x| (c) sin(|x|) + |x| (d) sin(|x|) – |x|
Q13. Let f : R R be a function defined by f(x) = max {x, x3}. The set of all points where f(x) is
NOT differentiable is
(a) {–1, 1} (b) {–1, 0} (c) {0, 1} (d) {–1, 0, 1}
Q14. The left-hand derivative of f(x) = [x] sin( x) at x = k, k an integer, is
(a) (–1)k (k – 1) (b) (–1)k – 1(k – 1) (c) (–1)k k (d) (–1)k – 1 k
sin( cos 2 x)
Q15. lim equals
x 0 x2
(a) – (b)  (c) /2 (d) 1
x
 x – 3
Q16. For x  R, lim   =
x 0  x + 2 
(a) e (b) e–1 (c) e–5 (d) e5

3. Mission IIT –2013
x tan 2x – 2x tan x
Q17. lim
x 0 (1 – cos 2x) 2
(a) 2 (b) –2 (c) 1/2 (d) –1/2
Q18. The function f(x) = (x2 – 1) |x2 – 3x + 2| + cos(| x |) is NOT differentiable at
(a) –1 (b) 0 (c) 1 (d) 2
Q19. The function f(x) = [x]2 – [x2] (where [y] is the greatest integer less than or equal to y), is
discontinuous at
(a) all integers (b) all integers except 0 and 1
(c) all integers except 0 (d) all integers except 1
1 2n r
Q20. lim 1 2 2 equals
x  n
r= n r
(a) 1+ 5 (b) –1 + 5 (c) –1 + 2 (d) 1+ 2