Question

If $I={\int }_0^1\frac{dx}{1+x^{{\displaystyle \frac{\mathrm{\pi }}{2}}}}$, then

• A. ${\log }_{e}2<I<\frac{\pi }{4}$
• B. ${\log }_{e}2>I$
• C. $I=\frac{\mathrm{\pi }}{4}$
• D. $I={\log }_{e}2$

Answer 1

The correct option is A

${\log }_{e}2<I<\frac{\pi }{4}$

Explanation for correct option:

Find the relation:

Since, $x\in (0,1)$, then

$$\begin{gathered} \Rightarrow x^2<x^{\frac{\mathrm{\pi }}{2}}<x \\ \Rightarrow 1+x^2<1+x^{\frac{\mathrm{\pi }}{2}}<1+x\mathbf{[}\mathbf{}\mathbf{add}\mathbf{}\mathbf{1}\mathbf{}\mathbf{and}\mathbf{}\mathbf{take}\mathbf{}\mathbf{reciprocal}\mathbf{]} \\ \Rightarrow \frac{1}{1+x^2}>\frac{1}{1+x^{{\displaystyle \frac{\mathrm{\pi }}{2}}}}>\frac{1}{1+x} \\ \Rightarrow {\int }_0^1\frac{1}{1+x^2}dx>{\int }_0^1\frac{1}{1+x^{{\displaystyle \frac{\mathrm{\pi }}{2}}}}dx>{\int }_0^1\frac{1}{1+x}dx \\ \Rightarrow {\left[{\tan }^{-1}x\right]}_0^1>I>{[In|1+x\left|\right]}_0^1 \\ \Rightarrow \frac{\mathrm{\pi }}{4}>I>In2-In1 \\ \Rightarrow \frac{\mathrm{\pi }}{4}>I>In2 \end{gathered}$$

Hence, the correct option is A.