Notice Board: Due to unavoidable circumstances, Ctanujit Classes will be closed till 31st May, 2018. Application for ISI 2019 Entrance exam study material & Online classes will start from 1st June, 2018

IIT JEE SOLVED PAPERS ON MONOTONOCITY OF FUNCTIONS

This is IIT-JEE Previous Year Solution on Monotonicity of functions


Q. [IIT-JEE 2001] If  f(x) = xex(1-x), then f(x) is
(a) increasing on [-½,1]                                 (b) decreasing on R     (c) increasing on R                                        (d) decreasing on [-½,1]
Solution : (a) Given f(x) = xex(1-x), differentiating we get                                    f′(x)= -ex(1-x)(2x+1)(x-1) . Now, we see f′(x)≥ 0 for -½≤ x ≤1.
Q. [IIT-JEE 2000] Let f(x)=∫ex(x-1)(x-2)dx. Then f decreases in the interval
(a) (-∞,-2)    (b) (-2,-1)     (c) (1,2)    (d)(2,∞)
Solution :(c) Given f(x)=∫ex(x-1)(x-2)dx
Now,  f(x)=ex(x-1)(x-2) . Now f′(x)< 0 for 1< x <2.
Q. [IIT-JEE 1990] Let f(x) be a quadratic expression which is positive for all real values of x. If g(x) = f(x) + f′(x) + f″(x), then for any real x
(a) g(x)<0      (b) g(x)>0     (c) g(x)=0      d) g(x)≥ 0
Solution :(b) Let  f(x) = ax­2 + bx + c
Since f(x) is positive for all x in R.
  D = b2 – 4ac < 0  and a > 0
Now  g(x) = ax2 + (2a+b)x + (a+b+c)
D1= Discriminant of g(x) = (2a+b)2 – 4a(a+b+c) = b2 – 4ac < 0  and a > 0
∴g(x) > 0 for all x in R.