Find the values of a and b so that the function f(x) defined by


Find the values of a and b so that the function f(x) defined by

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

becomes continuous on [0, π].


Given: $f$ is continuous on $[0, \pi]$.

$\therefore f$ is continuous at $x=\frac{\pi}{4}$ and $\frac{\pi}{2}$

At $x=\frac{\pi}{4}$, we have

$\lim _{x \rightarrow \frac{\pi}{4}^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}-h\right)=\lim _{h \rightarrow 0}\left[\left(\frac{\pi}{4}-h\right)+a \sqrt{2} \sin \left(\frac{\pi}{4}-h\right)\right]=\left[\frac{\pi}{4}+a \sqrt{2} \sin \left(\frac{\pi}{4}\right)\right]=\left[\frac{\pi}{4}+a\right]$

$\lim _{x \rightarrow \frac{\pi}{4}^{+}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}+h\right)=\lim _{h \rightarrow 0}\left[2\left(\frac{\pi}{4}+h\right) \cot \left(\frac{\pi}{4}+h\right)+b\right]=\left[\frac{\pi}{2} \cot \left(\frac{\pi}{4}\right)+b\right]=\left[\frac{\pi}{2}+b\right]$

$\lim _{x \rightarrow \frac{\pi}{2}^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right)=\lim _{h \rightarrow 0}\left[2\left(\frac{\pi}{2}-h\right) \cot \left(\frac{\pi}{2}-h\right)+b\right]=b$

$\lim _{x \rightarrow \frac{\pi}{2}^{+}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right)=\lim _{h \rightarrow 0}\left[a \cos 2\left(\frac{\pi}{2}+h\right)-b \sin \left(\frac{\pi}{2}+h\right)\right]=-a-b$

Since $f$ is continuous at $x=\frac{\pi}{4}$ and $x=\frac{\pi}{2}$, we get

$\lim _{x \rightarrow \frac{\pi}{2}^{-}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}^{+}} f(x)$ and $\lim _{x \rightarrow \frac{\pi}{4}^{-}} f(x)=\lim _{x \rightarrow \frac{\pi}{4}^{+}} f(x)$

$\Rightarrow-b-a=b$ and $\frac{\pi}{4}+a=\frac{\pi}{2}+b$

$\Rightarrow b=\frac{-a}{2} \quad \ldots(1)$ and $\frac{-\pi}{4}=b-a \quad \ldots(2)$

$\Rightarrow \frac{-\pi}{4}=\frac{-3 a}{2}$      [Substituting the value of $b$ in eq. (2)]

$\Rightarrow a=\frac{\pi}{6}$

$\Rightarrow b=\frac{-\pi}{12}$            [From eq. (1)]

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