Let f :0, ∞→R be given by fx= ∫1/xxe^(-t+1t)dt/t, then

Correct answer: (A) fx is monotonically increasing on 1, ∞; (C) fx+f1/x=0, for all x ∈0, ∞; (D) f2^x is an odd function of x on R. f^(′)x=(2e^(-x+1x))/x Which is increasing in 1, ∞ Also, fx+f1/x=0 gx=f2^x= ∫2^(-x)2^x(e^(-t+1t))/t dt g-x= ∫2^x2^(-x)(e^(-t+1t))/t dt= -gx Hence, an odd function

Q21 / 60 · JEE Advanced 2014 (Paper 1) · Mathematics

Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{\frac{1}{x}}^{x} e^{- (t + \frac{1}{t})} \frac{d t}{t},$ then

Correct answer
A
$f (x)$ is monotonically increasing on $[1, \infty)$
B
$f (x)$ is monotonically decreasing on (0, 1)
Correct answer
C
$f (x) + f (\frac{1}{x}) = 0,$ for all $x \in (0, \infty)$
Correct answer
D
$f (2^{x})$ is an odd function of $x$ on R

Solution

f^{'} (x) = \frac{2 e^{- (x + \frac{1}{x})}}{x}

Which is increasing in

[1, \infty)

Also,

f (x) + f (\frac{1}{x}) = 0

g (x) = f (2^{x}) = \int_{2^{- x}}^{2^{x}} \frac{e^{- (t + \frac{1}{t})}}{t} d t

g (- x) = \int_{2^{x}}^{2^{- x}} \frac{e^{- (t + \frac{1}{t})}}{t} d t = - g (x)

Hence, an odd function

Hence, the correct answer is options: 'A', 'C', 'D'.

Let f :0, ∞→R be given by fx= ∫1xxe-t+1tdtt, then

Solution

Relevant Questions

Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{\frac{1}{x}}^{x} e^{- (t + \frac{1}{t})} \frac{d t}{t},$ then