JEE Advanced 2014 · Paper 1 · Mathematics Q43

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MathematicsMultiple correct+3Question 43 of 60

Let $f:(0,\infty)\to\mathbb R$ be given by
$f(x)=\int_{1/x}^{x}e^{-\left(t+\frac1t\right)}\frac{dt}t.$
Then

f(x)

is monotonically increasing on

[1,\infty)

f(x)

is monotonically decreasing on

(0,1)

f(x)+f\left(\frac1x\right)=0

, for all

x\in(0,\infty)

f(2^x)

is an odd function of

x

\mathbb R

Answer and worked solution

Official answer: A, C, D

Derivative (Leibniz rule):
$f'(x)=\frac{e^{-(x+1/x)}}{x}-\frac{e^{-(1/x+x)}}{1/x}\cdot\left(-\frac1{x^2}\right)=\frac{2e^{-(x+1/x)}}{x}>0\quad(x>0).$

(A) $f$ is increasing on all of $(0,\infty)$ , so on $[1,\infty)$ . True.
(B) It is increasing on $(0,1)$ too. False.
(C) $f(1/x)=\displaystyle\int_x^{1/x}=-\int_{1/x}^x=-f(x)$ . True.
(D) $f(2^{-x})=f\!\left(\frac1{2^x}\right)=-f(2^x)$ , so $f(2^x)$ is odd. True.

(A), (C), (D)

Topics:Leibniz rule Monotonicity Types of functions

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JEE Advanced 2014 · Paper 1, Mathematics question 43