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

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Let $\mathrm{f}(\mathrm{x}) \begin{cases}=\mathrm{x}+\mathrm{a} \sqrt{2} \sin \mathrm{x}, & 0 \leq \mathrm{x} < \frac{\pi}{4} \ =2 \mathrm{x} \cot \mathrm{x}+\mathrm{b}, & \frac{\pi}{4} \leq \mathrm{x} < \frac{\pi}{2} \ =\mathrm{a} \cos 2 \mathrm{x}-\mathrm{b} \sin \mathrm{x}, & \frac{\pi}{2} \leq \mathrm{x} \leq \pi\end{cases}$ If $\mathrm{f}(\mathrm{x})$ is continuous for $0 \leq \mathrm{x} \leq \pi$, then

Aicharya · Accepted Answer

$\lim _{x ightarrow \frac{\pi^{-}}{4}} f(x)=\lim _{x ightarrow \frac{\pi^{-}}{4}} x+a \sqrt{2} \sin x$ $\lim _{x ightarrow \frac{\pi^{+}}{4}} f(x)=\lim _{x ightarrow \frac{\pi^{+}}{4}} 2 x \cot x+b$ $\lim _{x ightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{x ightarrow \frac{\pi^{-}}{2}} 2 x \cot x+b$ $\begin{aligned} & \lim _{x ightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x ightarrow \frac{\pi^{+}}{2}} a \cos 2 x-b \sin x \ & =a \cos 2\left(\frac{\pi}{2}ight)-b \sin \left(\frac{\pi}{2}ight)\end{aligned}$ Since $f(x)$ is continuous at $\frac{\pi}{4}$ and $\frac{\pi}{2}$, we write $\frac{\pi}{4}+\mathrm{a}=\frac{\pi}{2}+\mathrm{b} \quad \ldots[$ From (1) and (2) $]$ $\mathrm{b}=-\mathrm{a}-\mathrm{b} \quad \ldots[$ From (3) and (4) $]$

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

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