Question

Let ${\displaystyle g\left(x\right)=\ln x}$ and $f\left(x\right)=\left(\frac{1-x\cos x}{1+x\cos x}\right)$ then ${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{dx}$ is equal to

• A. $\ln \left(1\right)$
• B. $\ln \left(2\right)$
• C. $\ln \left(\mathrm{e}\right)$
• D. $\ln \left(4\right)$

Answer 1

The correct option is A

$\ln \left(1\right)$

Determine the value of ${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{dx}$

Step 1: Check if the function is odd or even

Since the limit of integral is given from $-\frac{\mathrm{\pi }}{4}\mathrm{to}\frac{\mathrm{\pi }}{4}$, We will check if the given function is odd or even.

We know that the function is odd when $\mathrm{f}(-\mathrm{x})=-\mathrm{f}\left(\mathrm{x}\right)$, and the function is even when $\mathrm{f}(-\mathrm{x})=\mathrm{f}\left(\mathrm{x}\right)$.

We have,

$$\begin{gathered} \mathrm{h}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right) \\ =\ln \left(\mathrm{f}\right(\mathrm{x}\left)\right)[\because g(x)=\ln (x),\mathrm{Given}] \\ =\ln \left(\frac{1-\mathrm{xcos}\left(\mathrm{x}\right)}{1+\mathrm{xcos}\left(\mathrm{x}\right)}\right)\left[\because f\left(x\right)=\left(\frac{1-x\cos x}{1+x\cos x}\right),\mathrm{Given}\right] \end{gathered}$$

Now replace $\mathrm{x}\mathrm{by}-\mathrm{x}$ we get,

$$\begin{gathered} \Rightarrow \mathrm{h}(-\mathrm{x})=\ln \left(\frac{1-(-\mathrm{x})\cos \left(-\mathrm{x}\right)}{1+(-\mathrm{x})\cos \left(-\mathrm{x}\right)}\right) \\ \Rightarrow \mathrm{h}(-\mathrm{x})=\ln \left(\frac{1+\mathrm{xcos}\left(\mathrm{x}\right)}{1-\mathrm{xcos}\left(\mathrm{x}\right)}\right) \\ \Rightarrow \mathrm{h}(-\mathrm{x})=\ln {\left(\frac{1-\mathrm{xcos}\left(\mathrm{x}\right)}{1+\mathrm{xcos}\left(\mathrm{x}\right)}\right)}^{-1} \\ \Rightarrow \mathrm{h}(-\mathrm{x})=-\ln \left(\frac{1-\mathrm{xcos}\left(\mathrm{x}\right)}{1+\mathrm{xcos}\left(\mathrm{x}\right)}\right) \\ \Rightarrow \mathrm{h}(-\mathrm{x})=-\mathrm{h}\left(\mathrm{x}\right) \end{gathered}$$

Therefore, the given function is odd and its value under the given integral will become equal to zero, i,e ${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{dx}={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\ln \left(\frac{1-\mathrm{xcos}\left(\mathrm{x}\right)}{1+\mathrm{xcos}\left(\mathrm{x}\right)}\right)d\mathrm{x}=0$

We know that the value of $\ln \left(1\right)=0$, hence the correct option is (A)