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.