Question

Consider the function
$f\left(x\right)=\{\begin{array}{ccc}-2\sin x& if& x\le -\frac{\pi }{2}\\ A\sin x+B& if& -\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos x& if& x\ge \frac{\pi }{2}\end{array}$
which is continuous everywhere.
The value of B is

• A. 1
• B. 0
• C. -1
• D. -2

Answer 1

Answer: (A)

1

We have,

$f\left(x\right)=\{\begin{array}{ccc}-2\sin x& if& x\le -\frac{\pi }{2}\\ A\sin x+B& if& -\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos x& if& x\ge \frac{\pi }{2}\end{array}$

Case-I, Continuity at $x=\frac{\pi }{2},$

$\underset{x\to {\left(\pi /2\right)}^{-}}{lim}f\left(x\right)=$$\underset{x\to {\left(\pi /2\right)}^{+}}{lim}f\left(x\right)=$$f\left(\frac{\pi }{2}\right)$

or, $\underset{h\to 0}{lim}f(\frac{\pi }{2}-h)$$=\underset{h\to 0}{lim}f(\frac{\pi }{2}+h)$$=f\left(\frac{\pi }{2}\right)$

or, $\underset{h\to 0}{lim}A\sin (\frac{\pi }{2}-h)+B$$=\underset{h\to 0}{lim}\cos (\frac{\pi }{2}+h)$$=\cos \left(\frac{\pi }{2}\right)$

or, $A\sin \frac{\pi }{2}+B=\cos \frac{\pi }{2}=0$

or, $A+B=\mathrm{0......}\left(i\right)$

Case-II, for continuity at $x=-\frac{\pi }{2},$

$\underset{x\to -{\left(\pi /2\right)}^{-}}{lim}f\left(x\right)=$$\underset{x\to -{\left(\pi /2\right)}^{+}}{lim}f\left(x\right)=$$f(-\frac{\pi }{2})$

or, $\underset{h\to 0}{lim}f(-\frac{\pi }{2}-h)$$=\underset{h\to 0}{lim}f(-\frac{\pi }{2}+h)$$=f(-\frac{\pi }{2})$

or, $\underset{h\to 0}{lim}-2\sin (-\frac{\pi }{2}-h)=$$\underset{h\to 0}{lim}A\sin (-\frac{\pi }{2}-h)+B$$=-2\sin \left(-\frac{\pi }{2}\right)$

or, $2=-A+B=2$

or, $-A+B=\mathrm{2............}\left(ii\right)$

On solving $\left(i\right)$ & $\left(ii\right),$ we get

$B=1$ and $A=-1$

Hence, A is the correct option.