#### Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 31

The correct option is (a)

Hint:

f(x) is continuous at x = 0

$(\text { LHL of } f(x) \text { at } x=0)=(\text { RHL of } f(x) \text { at } x=0)=f(0)$

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)\: \: \: \: \: \: \: \: \: \: .....(i)$

\begin{aligned} \text { Now, }&\lim _{x \rightarrow 0^{-}} f(x) \\ &=\lim _{x \rightarrow 0} a \sin \frac{\pi}{2}(x+1) \end{aligned}

\begin{aligned} &=\lim _{x \rightarrow 0} a \sin \left(\frac{\pi}{2}+\frac{\pi}{2} x\right) \\ &=\lim _{x \rightarrow 0} a \cos \frac{\pi x}{2}=a \end{aligned}                            \begin{aligned} &{\left[\therefore f(x)=a \sin \frac{\pi}{2}(x+1), \text { if } x \leq 0\right]} \\ &{[\cos 0={1}]} \end{aligned}

\begin{aligned} \text { Again, }&\lim _{x \rightarrow 0^{+}} f(x) \end{aligned}

$=\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}} \quad\left[\because f(x)=\frac{\tan x-\sin x}{x^{3}} \text { if } x>0\right]$

\begin{aligned} &=\lim _{x \rightarrow 0} \frac{\frac{\sin x}{\cos x}-\sin x}{x^{3}} \\ &=\lim _{x \rightarrow 0} \frac{\sin x-\sin x \cos x}{\cos x \times x^{3}}=\lim _{x \rightarrow 0} \frac{\sin x(1-\cos x)}{\cos x \times x^{3}} \end{aligned}

$=\lim _{x \rightarrow 0} \frac{1}{\cos x} \lim _{x \rightarrow 0} \frac{\sin x}{x} \times \frac{2 \sin ^{2} \frac{x}{2}}{\frac{x^{2}}{4} \times 4} \quad\left[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}\right]$

$=\frac{1}{1} \times 1 \times \frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}$

$=\frac{1}{2}\left[\lim _{\frac{x}{2} \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}\right]=\frac{1}{2} \times 1=\frac{1}{2}$

$\begin{gathered} \text { Also, } f(0)=a \sin \frac{\pi}{2}(0+1) \\ \qquad=a \sin \frac{\pi}{2}=a \end{gathered}$

Putting above values in (i) we get,

$\begin{gathered} a=\frac{1}{2} \end{gathered}$

So, the correct option is (a)