#### Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 44

$a=\frac{1}{2}, b=4$

Hint:

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

Given:

$f(x)=\left\{\begin{array}{l} \frac{1-\sin ^{3} x}{3 \cos ^{2} x}, \text { if } x<\frac{\pi}{2} \\\\ a \quad, \text { if } x=\frac{\pi}{2} \\\\ \left(\frac{6(1-\sin x)}{(\pi-2 x)^{2}}\right), \text { if } x>\frac{\pi}{2} \end{array}\right.$

Solution:

$f(x)=\left\{\begin{array}{l} \frac{1-\sin ^{3} x}{3 \cos ^{2} x}, \text { if } x<\frac{\pi}{2} \\\\ a \quad, \text { if } x=\frac{\pi}{2} \\\\ \left(\frac{6(1-\sin x)}{(\pi-2 x)^{2}}\right), \text { if } x>\frac{\pi}{2} \end{array}\right.$

We have

(LHL at  $x=\frac{\pi}{2}$ )

$\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right)=\lim _{h \rightarrow 0}\left(\frac{1-\sin ^{3}\left(\frac{\pi}{2}-h\right)}{3 \cos ^{2}\left(\frac{\pi}{2}-h\right)}\right)$

$=\lim _{h \rightarrow 0}\left(\frac{1-\cos ^{3} h}{3 \sin ^{2} h}\right)=\frac{1}{3} \lim _{h \rightarrow 0}\left(\frac{(1-\cosh )\left(1+\cos ^{2} h+\cosh \right)}{(1-\cosh )(1+\cosh )}\right)$                                                                                                                                                                                                                                    $\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$

$=\frac{1}{3} \lim _{h \rightarrow 0}\left(\frac{\left(1+\cos ^{2} h+\cosh \right)}{(1+\cosh )}\right)=\frac{1}{3}\left(\frac{1+1+1}{1+1}\right)=\frac{1}{2}$

(RHL at   $x=\frac{\pi}{2}$  )

$\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right)=\lim _{h \rightarrow 0}\left[\frac{b\left[1-\sin \left(\frac{\pi}{2}+h\right)\right]}{\left[\pi-2\left(\frac{\pi}{2}+h\right)\right]^{2}}\right]=\lim _{h \rightarrow 0}\left[\frac{b[1-\cosh ]}{[-2 h]^{2}}\right]$

$[\because \sin (90+\theta)=\cos \theta]$

$=\lim _{h \rightarrow 0}\left[\frac{2 b \sin ^{2} \frac{h}{2}}{4 h^{2}}\right]=\lim _{h \rightarrow 0}\left[\frac{2 b \sin ^{2} \frac{h}{2}}{16 \frac{h^{2}}{4}}\right]=\frac{b}{8} \lim _{h \rightarrow 0}\left[\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right]^{2}$                                              $\left[\begin{array}{l} \because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \because 1-\cos x=2 \sin ^{2} \frac{x}{2} \end{array}\right]$

\begin{aligned} &=\frac{b}{8} \times 1 \\ &=\frac{b}{8} \end{aligned}

Also

$f\left(\frac{\pi}{2}\right)=a$

If  $f\left ( x \right )$  is continuous at $x=\frac{\pi}{2}$ , then

\begin{aligned} &\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\\\ &\frac{1}{2}=\frac{b}{8}=a \\\\ &a=\frac{1}{2} \text { and } b=4 \end{aligned}