SATHEE: Application of Derivatives 2 Question 18

Application of Derivatives 2 Question 18

####18. If $f : (0, \infty) \to R$ be given by

(2014 Adv.)

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

Then,

(a) $f (x)$ is monotonically increasing on $[1, \infty)$

(b) $f (x)$ is monotonically decreasing on $[0, 1)$

(d) $f (2^{x})$ is an odd function of $x$ on $R$

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Answer:

Correct Answer: 18. (c)

Solution:

Given, $f (x) = \int_{\frac{1}{x}}^{x} \frac{e^{- t + \frac{1}{t}}}{t} d t$

$\begin{aligned} f^{'} (x) & = 1 \cdot \frac{e^{- x + \frac{1}{x}}}{x} - \frac{- 1}{x^{2}} \frac{e^{- \frac{1}{x} + x}}{1 / x} \\ = \frac{e^{- x + \frac{1}{x}}}{x} + \frac{e^{- x + \frac{1}{x}}}{x} = \frac{2 e^{- x + \frac{1}{x}}}{x} \end{aligned}$

As $f^{'} (x) > 0, \forall x \in (0, \infty)$

$∴ f (x)$ is monotonically increasing on $(0, \infty)$ .

$\Rightarrow$ Option (a) is correct and option (b) is wrong.

Now, $f (x) + f \frac{1}{x} = \int_{1 / x}^{x} \frac{e^{- t + \frac{1}{t}}}{t} d t + \int_{x}^{1 / x} \frac{e^{- t + \frac{1}{t}}}{t} d t$ $= 0, \forall x \in (0, \infty)$

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

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

$∴ f (2^{x})$ is an odd function.

Application of Derivatives 2 Question 18

Answer:

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Application of Derivatives 2 Question 18

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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