The function $$ f(x)= \begin{cases}x+a \sqrt{2} \sin x, & 0 \leq x

Question

The function $$ f(x)= \begin{cases}x+a \sqrt{2} \sin x, & 0 \leq x<\pi / 4 \ 2 x \cot x+b, & \pi / 4 \leq x \leq \pi / 2 \ a \cos 2 x-b \sin x, & \pi / 2<x \leq \pi\end{cases} $$ is continuous for $0 \leq x \leq \pi$. Then a+b=$\pi/n, $ n=?

Professor PC Thomas Classes · Answer

Since $\sin x$ and $\cos x$ are continuous for every $x$ and $\cot x$ is continuous in $[\pi / 4, \pi / 2]$. Therefore $f(x)$ will be continuous in ( $0, \pi$ ) provided it is continuous at $\pi / 4$ and $\pi / 2$. For $f(x)$ to be continuous at $x=\pi / 4$\ $\lim _{x ightarrow(\pi / 4)^{-}} f(x)=\lim _{x ightarrow(\pi / 4)^{+}} f(x)$\ $\Rightarrow \lim _{x ightarrow(\pi / 4)^{-}}(x+a \sqrt{2} \sin x)=\lim _{x ightarrow(\pi / 4)^{+}}(2 x \cot x+b)$\ $\Rightarrow \pi / 4+a \sqrt{2} 	imes 1 / \sqrt{2}=2 \pi / 4+b$\ $\Rightarrow \quad(\pi / 4)+a=(\pi / 2)+b$\ For $f(x)$ to be continuous at $x=\pi / 2$\ $\lim _{x ightarrow(\pi / 2)^{-}} f(x)=\lim _{x ightarrow(\pi / 2)^{+}} f(x)$\ $\Rightarrow \quad \lim _{x ightarrow(\pi / 2)^{-}}(2 x \cot x+b)$\ $=\lim _{x ightarrow(\pi / 2)^{+}}(a \cos 2 x-b \sin x)$\ $\Rightarrow \quad 2 	imes \pi / 2 	imes 0+b=a(-1)-b$\ $\Rightarrow \quad b=-a-b$\ Solving (1) & (2),\ $a=\pi / 6$ and $b=-\pi / 12$\