Question

The values of $A$ and $B$ such that the function
$f\left(x\right)=\{\begin{array}{c}-2\sin x,x\le -\frac{\pi }{2}\\ A\sin x+B,-\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos x,x\ge \frac{\pi }{2}\end{array}$

is continuous everywhere, are-

• A. $A=0,B=1$
• B. $A=1,B=1$
• C. $A=-1,B=1$
• D. $A=-1,B=0$

Answer 1

The correct option is C $A=-1,B=1$

Given $f\left(x\right)=\{\begin{array}{c}-2\sin x,x\le -\frac{\pi }{2}\\ A\sin x+B,-\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos x,x\ge \frac{\pi }{2}\end{array}$
Since, f(x) is continuous everywhere so f(x) is continuous at $x=\frac{\pi }{2}$
$LHL=RHL=f\left(\frac{\pi }{2}\right)$ ...(1)
$f\left(\frac{\pi }{2}\right)=0$
$LHL=\underset{x\to {\frac{\pi }{2}}^{-}}{lim}=\underset{h\to 0}{lim}f(\frac{\pi }{2}-h)$
$=\underset{h\to 0}{lim}A\sin (\frac{\pi }{2}-h)+B$
$=\underset{h\to 0}{lim}A\cos h+B$
$LHL=A+B$
$\Rightarrow A+B=\mathrm{0....}\left(2\right)$ (by (1))
f(x) is also continuous at $x=\frac{-\pi }{2}$
$LHL=RHL=f\left(\frac{-\pi }{2}\right)$ ...(3)
Now, $f(-\frac{\pi }{2})=-2\sin (-\frac{\pi }{2})=2$
$RHL=\underset{x\to {\frac{-\pi }{2}}^{+}}{lim}=\underset{h\to 0}{lim}f(\frac{-\pi }{2}+h)$
$\underset{h\to 0}{lim}f(-\frac{\pi }{2}+h)=\underset{h\to 0}{lim}-A\sin (\frac{\pi }{2}-h)+B$
$=\underset{h\to 0}{lim}-A\cos h+B$
$\Rightarrow RHL=-A+B$
By eqn (3),
$\Rightarrow -A+B=2$ ....(4)
Solving equations (2) and (4), we get
$B=1$, $A=-1$