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If $f(x)=\left\{\begin{array}{cl}2 \sin x & ;-\pi \leq x \leq \frac{-\pi}{2} \\ a \sin x+b & ;-\frac{\pi}{2} < x < \frac{\pi}{2} \quad \text { and it is } \\ \cos x & ; \frac{\pi}{2} \leq x \leq \pi\end{array}\right.$ continuous on $[-\pi, \pi]$, then
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Verified Answer
The correct answer is:
$a=1$ and $b=-1$
$\begin{aligned}
& \text { At, } x=\frac{\pi}{2} \\
& \text { LHL }=\lim _{x \rightarrow \pi / 2^{-}}(a \sin x+b)=a+b \\
& \text { RHL }=\lim _{x \rightarrow \pi / 2^{+}}(\cos x)=0
\end{aligned}$
Since, $f(x)$ is continuous at $x=\pi / 2$
$\begin{aligned}
& \text { At, } x=-\frac{\pi}{2} \\
& \text { LHL }=\lim _{x \rightarrow-\pi / 2^{-}}(2 \sin x)=-2 \\
& \text { RHL }=\lim _{x \rightarrow-\pi / 2^{+}}(a \sin x+b)=-a+b
\end{aligned}$
Since, $f(x)$ is continuous at $x=-\pi / 2$
On solving Eqs. (i) and (ii), we get $a=1$ and $h=-1$
& \text { At, } x=\frac{\pi}{2} \\
& \text { LHL }=\lim _{x \rightarrow \pi / 2^{-}}(a \sin x+b)=a+b \\
& \text { RHL }=\lim _{x \rightarrow \pi / 2^{+}}(\cos x)=0
\end{aligned}$
Since, $f(x)$ is continuous at $x=\pi / 2$
$\begin{aligned}
& \text { At, } x=-\frac{\pi}{2} \\
& \text { LHL }=\lim _{x \rightarrow-\pi / 2^{-}}(2 \sin x)=-2 \\
& \text { RHL }=\lim _{x \rightarrow-\pi / 2^{+}}(a \sin x+b)=-a+b
\end{aligned}$
Since, $f(x)$ is continuous at $x=-\pi / 2$
On solving Eqs. (i) and (ii), we get $a=1$ and $h=-1$
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