The values of A and B so that function f (x) defined by
$f(x)=\left\{\begin{array}{l}x+A \sqrt{2} \sin x, & 0 \leq x<\frac{\pi}{4} \\ 2 x \cot x+B, & \frac{\pi}{4} \leq x<\frac{\pi}{2} \\ A \cos 2 x-B \sin x, & \frac{\pi}{2} \leq x \leq \pi\end{array}\right.$
become continuous, respectively are
Answer & explanation
Correct answer: option 2
$f(x)=\left\{\begin{array}{l} x+A \sqrt{2} \sin x & 0 \leq x<\pi / 4 \\ 2 x \cot x+B & \pi / 4 \leq x<\pi / 2 \\ A \cos 2 x-B \sin x & \pi / 2 \leq x \leq \pi \end{array}\right.$
L.H. limit at $x<\frac{\pi}{4}$
$=\lim\limits_{x \rightarrow \pi / 4^{-}} x+A \sqrt{2} \sin x=\frac{\pi}{4}+A \sqrt{2} \sin \frac{\pi}{4}=\frac{\pi}{4}+A \sqrt{2} \times \frac{1}{\sqrt{2}}=A+\frac{\pi}{4}$
R.H. limit $=\lim\limits_{x \rightarrow \pi / 4^{+}} 2 x \cot x+B=\frac{2 \pi}{4} \cdot \cot \frac{\pi}{4}+B=\frac{\pi}{2}+B$
$A+\frac{\pi}{4}=B+\frac{\pi}{2} \Rightarrow A-B=\frac{\pi}{4}$ ..........(1)
L.H. limit at $x<\frac{\pi}{2}$
$=\lim\limits_{x \rightarrow \pi / 2}-(2 x \cot x+B)=2 \times \frac{\pi}{2} \cot \frac{\pi}{2}+B=B$
RH limit = $x \lim\limits_{x \rightarrow \pi / 2}+A \cos 2 x-B \sin x=A \cos \pi-B \sin \frac{\pi}{2}$
= − A − B
− A − B = B ⇒ a = −2B ..........(2)
$-3 B=\frac{\pi}{4} \Rightarrow B=\frac{-\pi}{12}, A=+\frac{\pi}{6}=\frac{\pi}{6}$
Hence (2) is the correct answer.