Question

$limi{t}_{x\to \frac{\mathrm{\pi }}{4}}\left(\frac{{\int }_2^{se{c}^{2}x}f\left(t\right)dt}{{\displaystyle x^2-\frac{{\mathrm{\pi }}^2}{16}}}\right)=$

• A. $\frac{8}{\mathrm{\pi }}f\left(2\right)$
• B. $\frac{2}{\mathrm{\pi }}f\left(2\right)$
• C. $\frac{2}{\mathrm{\pi }}f\left(\frac{1}{2}\right)$
• D. $4f\left(2\right)$

Answer 1

The correct option is A

$\frac{8}{\mathrm{\pi }}f\left(2\right)$

Explanation for Correct Answer:

Finding the value:

$$\begin{gathered} limi{t}_{x\to \frac{\mathrm{\pi }}{4}}\left(\frac{{\int }_2^{se{c}^{2}x}f\left(t\right)dt}{{\displaystyle x^2-\frac{{\mathrm{\pi }}^2}{16}}}\right)=\frac{{\int }_2^{se{c}^2\frac{\mathrm{\pi }}{4}}f\left(t\right)dt}{{\displaystyle {\left(\frac{\mathrm{\pi }}{4}\right)}^2-\frac{{\mathrm{\pi }}^2}{16}}} \\ =\frac{{\int }_2^{{\left(\sqrt{2}\right)}^2}f\left(t\right)dt}{{\displaystyle \frac{{\mathrm{\pi }}^2}{16}-\frac{{\mathrm{\pi }}^2}{16}}}\left[\because sec\frac{\mathrm{\pi }}{4}=\sqrt{2}\right] \\ =\frac{{\int }_2^{2}f\left(t\right)dt}{{\displaystyle 0}} \\ =\frac{0}{0}\left[\because {\int }_a^{a}f\left(t\right)dt=0\right] \end{gathered}$$

$\frac{0}{0}$Indeterminant form using L Hospitals Rule.

$limi{t}_{x\to \frac{\mathrm{\pi }}{4}}\left(\frac{{\int }_2^{se{c}^{2}x}f\left(t\right)dt}{{\displaystyle x^2-\frac{{\mathrm{\pi }}^2}{16}}}\right)\left(\because usingLeibnitzrule\right)$

$$\begin{gathered} =limi{t}_{x\to \frac{\mathrm{\pi }}{4}}\left(\frac{f\left({\sec }^2\left(x\right)\right).2\sec \left(x\right)\sec \left(x\right)\tan \left(x\right)}{2x}\right)\left[\because \frac{d}{dx}sec\left(x\right)=sec\left(x\right)tan\left(x\right)\text{applying chain rule}\right] \\ =\left(\frac{f\left({\sec }^2\left(\frac{\mathrm{\pi }}{4}\right)\right).2\sec \left(\frac{\mathrm{\pi }}{4}\right)\sec \left(\frac{\mathrm{\pi }}{4}\right)\tan \left(\frac{\mathrm{\pi }}{4}\right)}{2\frac{\mathrm{\pi }}{4}}\right) \\ =\left(\frac{f\left({\left(\sqrt{2}\right)}^2\right).2.\sqrt{2}.\sqrt{2}.1}{2\frac{\mathrm{\pi }}{4}}\right)\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{e}\mathbf{c}\left(\frac{\pi }{4}\right)\mathbf{=}\sqrt{\mathbf{2}}\mathbf{;}\mathbf{t}\mathbf{a}\mathbf{n}\left(\frac{\pi }{4}\right)\mathbf{=}\mathbf{1}\mathbf{]} \\ =4\left(\frac{f\left(2\right)}{\frac{\mathrm{\pi }}{2}}\right) \\ =4\left(\frac{2}{\mathrm{\pi }}\right)f\left(2\right) \\ =\frac{8}{\mathrm{\pi }}f\left(2\right) \end{gathered}$$

Hence, the correct answer is option (A).