JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 6)
Let $$f:\left( {0,\infty } \right) \to R$$ be given by $$f\left( x \right) $$= $$\int\limits_{{1 \over x}}^x {{{{e^{ - \left( {t + {1 \over t}} \right)}}} \over t}} dt$$. Then
$$f(x)$$ is monotonically increasing on $$\left[ {1,\infty } \right)$$
$$f(x)$$ is monotonically decreasing on $$(0,1)$$
$$f(x)$$ $$ + f\left( {{1 \over x}} \right) = 0$$, for all $$x \in \left( {0,\infty } \right)$$
$$f\left( {{2^x}} \right)$$ is an odd function of $$x$$ on $$R$$
Explanation
Given, $$f\left( x \right) $$= $$\int\limits_{{1 \over x}}^x {{{{e^{ - \left( {t + {1 \over t}} \right)}}} \over t}} dt$$
Therefore, $\frac{d}{d x} f(x)=\frac{e^{-\left(x+\frac{1}{x}\right)}}{x} \frac{d}{d x} (x)-\frac{e^{-\left(\frac{1}{x}+x\right)}}{1 / x} \times \frac{d}{d x}\left(\frac{1}{x}\right)$
$$ \begin{aligned} & =\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}+x e^{-\left(x+\frac{1}{x}\right)} \times\left(-\frac{1}{x^2}\right) \\\\ & =\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}+\frac{1}{x} e^{-\left(x+\frac{1}{x}\right)} \\\\ & =\frac{2 e^{-\left(x+\frac{1}{x}\right)}}{x}>0 \end{aligned} $$
As, $ f^{\prime}(x)>0, \forall x \in(0, \infty)$
$\therefore f(x)$ is monotonically increasing on $(0, \infty)$.
$\Rightarrow$ options (a) is correct and (b) is wrong.
Now,
$\begin{aligned} f(x)+f\left(\frac{1}{x}\right) & =\int\limits_{1 / x}^x \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t+\int\limits_x^{\frac{1}{x}} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t \\\\ & =\int\limits_{1 / x}^{1 / x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t=0\end{aligned}$
Now, let
$$ \begin{aligned} g(x) & =f\left(2^x\right)=\int\limits_{2^{-x}}^{2^x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t \\\\ g(-x) & =f\left(2^{-x}\right) \\\\ & =\int\limits_{2^x}^{2^{-x}} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t \\\\ & =-g(x) \end{aligned} $$
$\therefore f\left(2^x\right)$ is an odd function.
Therefore, $\frac{d}{d x} f(x)=\frac{e^{-\left(x+\frac{1}{x}\right)}}{x} \frac{d}{d x} (x)-\frac{e^{-\left(\frac{1}{x}+x\right)}}{1 / x} \times \frac{d}{d x}\left(\frac{1}{x}\right)$
$$ \begin{aligned} & =\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}+x e^{-\left(x+\frac{1}{x}\right)} \times\left(-\frac{1}{x^2}\right) \\\\ & =\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}+\frac{1}{x} e^{-\left(x+\frac{1}{x}\right)} \\\\ & =\frac{2 e^{-\left(x+\frac{1}{x}\right)}}{x}>0 \end{aligned} $$
As, $ f^{\prime}(x)>0, \forall x \in(0, \infty)$
$\therefore f(x)$ is monotonically increasing on $(0, \infty)$.
$\Rightarrow$ options (a) is correct and (b) is wrong.
Now,
$\begin{aligned} f(x)+f\left(\frac{1}{x}\right) & =\int\limits_{1 / x}^x \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t+\int\limits_x^{\frac{1}{x}} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t \\\\ & =\int\limits_{1 / x}^{1 / x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t=0\end{aligned}$
Now, let
$$ \begin{aligned} g(x) & =f\left(2^x\right)=\int\limits_{2^{-x}}^{2^x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t \\\\ g(-x) & =f\left(2^{-x}\right) \\\\ & =\int\limits_{2^x}^{2^{-x}} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t \\\\ & =-g(x) \end{aligned} $$
$\therefore f\left(2^x\right)$ is an odd function.
Comments (0)
