Question

Let $f\left(x\right)=\{\begin{array}{c}-2\sin x,-\pi \le x\le -\frac{\pi }{2}\\ a\sin x+b,-\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos x,\frac{\pi }{2}\le x\le \pi \end{array}$
If $f\left(x\right)$ is continuous on $[-\pi ,\pi ]$, then

• A. $a=1,b=1$
• B. $a=-1,b=-1$
• C. $a=-1b=1$
• D. $a=1,b=-1$

Answer 1

The correct option is C $a=-1b=1$

Given $f\left(x\right)=\{\begin{array}{c}-2\sin x,-\pi \le x\le -\frac{\pi }{2}\\ a\sin x+b,-\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos x,\frac{\pi }{2}\le x\le \pi \end{array}$

Since $f\left(x\right)$ is continuous on $[-\pi ,\pi ]$, then

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

$\Rightarrow \underset{x\to -\frac{\pi }{2}}{lim}-2\sin x=\underset{x\to -\frac{\pi }{2}}{lim}a\sin x+b$

$\Rightarrow -2\sin (-\frac{\pi }{2})=a\sin (-\frac{\pi }{2})+b$

$\Rightarrow 2=-a+b...\left(i\right)$

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

$\Rightarrow {lim}_{x\to \frac{\pi }{2}}a\sin x+b={lim}_{x\to \frac{\pi }{2}}\cos x$

$\Rightarrow a\sin \frac{\pi }{2}+b=\cos \frac{\pi }{2}$

$\Rightarrow a+b=\mathrm{0...}\left(ii\right)$

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

$a=-1$ and $b=1$