Q54 / 150 · MHT CET 2023 (12 May Shift 1) · Mathematics
The values of $a$ and $b$, so that the function
$$
f(x)=\left\{\begin{array}{l}
x+\mathrm{a} \sqrt{2} \sin x, 0 \leq x \leq \frac{\pi}{4} \\
2 x \cot x+\mathrm{b}, \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\
\operatorname{acos} 2 x-\mathrm{b} \sin x, \frac{\pi}{2} < x \leq \pi
\end{array}\right.
$$
is continuous for $0 \leq x \leq \pi$, are respectively given by
- A$-\frac{\pi}{12}, \frac{\pi}{6}$
- B$-\frac{\pi}{6},-\frac{\pi}{12}$
- C$\frac{\pi}{6}, \frac{\pi}{12}$
- Correct answerD$\frac{\pi}{6},-\frac{\pi}{12}$
Solution
As the given function is continuous at $x=\frac{\pi}{4}$ and $\frac{\pi}{2}$, we get
$$
\begin{array}{ll}
& \lim _{x \rightarrow \frac{\pi^{-}}{4}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{4}} \mathrm{f}(x) \\
\therefore \quad & \lim _{x \rightarrow \frac{\pi}{4}}(x+\mathrm{a} \sqrt{2} \sin x)=\lim _{x \rightarrow \frac{\pi}{4}}(2 x \cot x+\mathrm{b}) \\
\therefore \quad & \frac{\pi}{4}+\mathrm{a}=\frac{2 \pi}{4}+\mathrm{b} \\
\therefore \quad & \mathrm{a}-\mathrm{b}=\frac{\pi}{4} \\
& \text { Also, } \lim _{x \rightarrow \frac{\pi^{-}}{2}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \mathrm{f}(x) \\
& \lim _{x \rightarrow \frac{\pi}{2}} 2 x \cot x+\mathrm{b}=\lim _{x \rightarrow \frac{\pi}{2}} \mathrm{a} \cos 2 x-\mathrm{b} \sin x \\
\therefore \quad & 0+\mathrm{b}=-\mathrm{a}-\mathrm{b} . \\
\therefore \quad & \mathrm{a}+2 \mathrm{~b}=0
\end{array}
$$
Solving equations (i) and (ii), we get $a=\frac{\pi}{6}$ and $b=\frac{-\pi}{12}$
Hence, the correct answer is option: 'D'.