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

Correct answer: A. f(x) is monotonically increasing on [1, ∞) f ^ ′ (x) = 2 e ^ - (x + (1)/(x)) x Which is increasing in [1, ∞) Also, f(x) + f ((1)/(x)) = 0 g(x) = f( 2 ^ x ) = ∫ _ 2 ^ - x ^ 2 ^ x e ^ - (t + (1)/(t)) t dt g(- x) = ∫ _ 2 ^ x ^ 2 ^ - x e ^ - (t + (1)/(t)) t dt = - g(x) Hence, an odd function

JEE Advanced 2014Paper 1

Exam: JEE Advanced
Year: 2014
Paper: Paper 1
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2014 · Paper 1 · Mathematics · Application of Derivatives

Let f: (0, ∞) → R be given by f(x) = ∫ _ (1)/(x) ^ x e ^ - (t +...

Let

f: \left(0, \infty\right) \to R

be given by

f\left(x\right) = \mathop{∫}_{\dfrac{1}{x}}^{x}{e}^{- \left(t + \dfrac{1}{t}\right)}\dfrac{dt}{t},

then

Options

A $f\left(x\right)$ is monotonically increasing on \left[1, \infty\right)
Bf $\left(x\right)$ is monotonically decreasing on (0, 1)
C $f\left(x\right) + f\left(\dfrac{1}{x}\right) = 0,$ for all $x \in \left(0, \infty\right)$
Df $\left({2}^{x}\right)$ is an odd function of $x$ on R

Correct answer

A. $f\left(x\right)$ is monotonically increasing on \left[1, \infty\right)

Step-by-step solution

{f}^{&prime;}\left(x\right) = \dfrac{2{e}^{- \left(x + \dfrac{1}{x}\right)}}

{x}

<br >Which is increasing in \left[1, \infty\right)

Also,

f\left(x\right) + f

\left(\dfrac{1}{x}\right) = 0

<br >

g\left(x\right) = f\left({2}^{x}\right) = \mathop{∫}_{{2}^{- x}}^{{2}^{x}}\dfrac{{e}^{- \left(t + \dfrac{1}{t}\right)}}{t}dt

<br >

g\left(- x\right) = \mathop{∫}_{{2}^{x}}^{{2}^{- x}}\dfrac{{e}^{-

\left(t + \dfrac{1}{t}\right)}}{t}dt = - g\left(x\right)

Hence, an odd function

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