#### Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 33 Maths Textbook Solution.

$f(\pi)=\frac{49}{10}$

Hint:

$f(x)$ must be defined. The limit of the $f(x)$ approaches the value $x$ must exist.

Given:

$f(x)=\frac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}, x=\pi$

Solution:

$f(x)=\frac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}, x=\pi$

If $f\left ( x \right )$ is continuous at $x=\pi$ , then

\begin{aligned} &\lim _{x \rightarrow \pi} f(x)=f(\pi) \\ &\lim _{x \rightarrow \pi} \frac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}=f(\pi) \end{aligned}

$\frac{2}{5} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(x-\pi)}{2}\right)}{(x-\pi)^{2}}=f(\pi)$                                                 $\left[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}\right]$

$\frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(x-\pi)}{2}\right)}{\frac{49}{4}(x-\pi)^{2}}=f(\pi)$                           [Multiplying and dividing by $\frac{49}{4}$ ]

$\frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(x-\pi)}{2}\right)}{\left(\frac{7}{2}(x-\pi)\right)^{2}}=f(\pi)$

$\frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi}\left[\frac{\sin \left(\frac{7(x-\pi)}{2}\right)}{\left(\frac{7}{2}(x-\pi)\right)}\right]^{2}=f(\pi)$

$\frac{2}{5} \times \frac{49}{4} \times 1=f(\pi)$                                                                               $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

\begin{aligned} &\frac{1}{5} \times \frac{49}{2} \times 1=f(\pi) \\ &f(\pi)=\frac{49}{10} \end{aligned}