Let $f:(0, \infty) \rightarrow R$ be given by $f(x)=\int\limits_{1 / x}^{x} e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$, then
- $f(x)$ is monotonically increasing on $[1, \infty$ )
- $f(x)$ is monotonically decreasing on $(0,1)$
- $f(x)+f\left(\frac{1}{x}\right)=0$, for all $x \in(0, \infty)$
- $f\left(2^{x}\right)$ is an odd function of $x$ on $R$.
The Correct Option is D
Solution and Explanation
Given that
\(f\left(x\right)=\int\limits_{1/x}^{x} e^{-(t+\frac{1}{t})} d t\)
⇒\( \frac{d}{dx}f(x) = \frac{e^{-x+\frac{1}{x}}}{x}\frac{dx}{dx}-\frac{e^{-(\frac1 x+x)}}{1/x}x \frac{d \frac{1}{x}}{dx}\)
= \(\frac{e^{-(x+\frac{1}{x})}}{x} + xe^{-(x+\frac{1}{x})}x\frac{-1}{x^2}\)
= \(e^{-(x+\frac{1}{x})} + \frac{1}{x}e^{-(x+\frac{1}{x})}\)
= \(2e^{-(x+\frac{1}{x})}= 0\)
Hence, \(f(x)+\)\(f(\frac{1}{x}) = \)
\[\int\limits_{1/x}^{x} \frac{e^{-(t+\frac{1}{t}})} {t}dt+\int\limits_{x}^{1/x} \frac{e^{-(t+\frac{1}{t}})}{t} d t\]\[\int\limits_{1/x}^{1/x} \frac{e^{-(t+\frac{1}{t})}}{t} d t=0\]Now,
\(f(2^x)+f(\frac{1}{2^x})=f(2^x)+f(2^x)=0\)
Therefore, \(f(2^x) \) is an odd function.
Note, Let 2x= μ
log2μ =x
∴ f(2x) = h(x) an odd function
Then h(-x)
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Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C