#### Please solve RD Sharma class 12 chapter Continuity exercise 8.2 question 6 maths textbook solution

$a = \pi /6,\; b=-\pi /12$

Hint:

Put at LHL = RHL at

$x = \pi /4,\; x=\pi /2$

Given:

$f(x)= \begin{cases}x+a \sqrt{2} \sin x & 0 \leq x<\pi / 4 \\ 2 x \cot x+b & \pi / 4 \leq x<\pi / 2 \\ a \cos 2 x-b \sin x & \pi / 2 \leq x<\pi\end{cases}$

Explanation:

At x = $\pi$/4

$L.H.L = R.H.L =f(\frac{\pi }{4}) \qquad ....(A)$

Now,

\begin{aligned} &f\left(\frac{\pi}{4}\right)=2 \cdot \frac{\pi}{4} \cdot \cot \left(\frac{\pi}{4}\right)+b\\ &=\frac{\pi}{2} \cdot 1+b=\frac{\pi}{2}+b \qquad ....(1) \end{aligned}

L.H.L

\begin{aligned} &\lim _{x \rightarrow \pi / 4^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}-h\right)=\lim _{h \rightarrow 0}\left(\frac{\pi}{4}-h\right)+a \sqrt{2} \sin \left(\frac{\pi}{4}-h\right) \\ &=\frac{\pi}{4}+a \sqrt{2} \cdot \frac{1}{\sqrt{2}}=\frac{\pi}{4}+a \end{aligned}

Using (A)

\begin{aligned} &\frac{1}{2}+b=\frac{\pi}{4}+a\\ &a-b=\frac{\pi}{4} \qquad ....(B) \end{aligned}

\begin{aligned} &\text { At } x=\frac{\pi}{2} \\ &\text { L.H.L }=\text { R.H.L }=f\left(\frac{\pi}{2}\right) \quad \ldots(C) \end{aligned}

Now

$f\left(\frac{\pi}{2}\right)=a \cos 2 \cdot \frac{\pi}{2}-b \sin \frac{\pi}{2}=-a-b \qquad ....(2)$

\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{h \rightarrow 0}\left(\frac{\pi}{2}-h\right) \\ &=\lim _{h \rightarrow 0} 2\left(\frac{\pi}{2}-h\right) \cot \left(\frac{\pi}{2}-h\right)+b=b \end{aligned}

Using (C)

$-a-b=b \Rightarrow 2 b=-a \Rightarrow b=\frac{-a}{2}$

From (B)

\begin{aligned} &a+\frac{a}{2}=\frac{\pi}{4} \\ &\frac{3}{2} a=\frac{\pi}{4} \\ &\Rightarrow a=\frac{\pi}{6} \end{aligned}

\begin{aligned} &b=\frac{-a}{2}=-\frac{\pi}{12} \\ \end{aligned}

Hence,

\begin{aligned} &a=\pi / 6, b=-\pi / 12 \end{aligned}