Question

Let $f\left(x\right)=\{\begin{array}{c}-2\sin xifx\le -\frac{\pi }{2}\\ A\sin x+Bif-\frac{\pi }{2}<x<\frac{\pi }{2};\\ \cos xifx\ge \frac{\pi }{2}\end{array}$
For what values of $A$ and $B$, the function $f\left(x\right)$ is continuous throughout real line?

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

Answer 1

Answer: (A)

$A=-1,B=1$

Given
$f\left(x\right)=\{\begin{array}{c}-2\sin xifx\le -\frac{\pi }{2}\\ A\sin x+Bif-\frac{\pi }{2}<x<\frac{\pi }{2};\\ \cos xifx\ge \frac{\pi }{2}\end{array}$
From above conditions function $f\left(x\right)$ is continuous throughout the real line, when function $f\left(x\right)$ is continuous at $x=-\frac{\pi }{2}$ and $\frac{\pi }{2}$
For continuity at $x=-\frac{\pi }{2}$
$\underset{x\to \frac{{\pi }^{-}}{2}}{lim}f\left(x\right)=\underset{x\to \frac{{\pi }^{+}}{2}}{lim}f\left(x\right)=f(-\frac{\pi }{2})...\left(ii\right)$
$\underset{x\to \frac{{\pi }^{-}}{2}}{lim}f\left(x\right)=2$
$\underset{x\to \frac{{\pi }^{+}}{2}}{lim}f\left(x\right)=-A+B$
$f(-\frac{\pi }{2})=2$
$\therefore$ From Eq (ii) we get
$-A+B=\mathrm{2....}\left(iii\right)$
For continuity at $x=\frac{\pi }{2}$
$\underset{x\to \frac{{\pi }^{-}}{2}}{lim}f\left(x\right)=\underset{x\to \frac{{\pi }^{+}}{2}}{lim}f\left(x\right)=f\left(\frac{\pi }{2}\right)...\left(iv\right)$
Here $\underset{x\to \frac{{\pi }^{-}}{2}}{lim}f\left(x\right)=A+B$

$\Rightarrow \underset{x\to \frac{{\pi }^{+}}{2}}{lim}f\left(x\right)=0$
And $f\left(\frac{\pi }{2}\right)=0$
$\therefore$ from Eq(iv)
$A+B=\mathrm{0....}\left(v\right)$
From Eqs. (iii) and (iv)
$A=-1,B=1$