Question

Let $f(x)=\{\begin{array}{ll}-2\sin x& \text{if }x\le -\frac{\pi }{2}\\ A\sin x+B& \text{if }-\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos & \text{if }x\le \frac{\pi }{2}\end{array}$
For what values of A and B, the function $f(x)$ is continuous throughout the 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: (B)

Given,
$f(x)=\{\begin{array}{l}-2\sin X,\text{ if }x\le -\frac{\pi }{2}\\ A\sin x+B,\text{ if }-\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos x,\text{ if }x\ge \frac{\pi }{2}\end{array}...(i)$
From above conditions function $f(x)$ is continuous throughout the real line, when function $f(x)$ is continuous at $x=-\frac{\pi }{2}$ and $\frac{\pi }{2}$.
For continuity at $x=-\frac{\pi }{2}$
$\underset{x\to -\frac{{\pi }^{-}}{2}}{\lim }f(x)=\underset{x\to -\frac{{\pi }^{+}}{2}}{\lim }f(x)=f\left(-\frac{\pi }{2}\right)...(ii)$
$\underset{x\to -\frac{{\pi }^{-}}{2}}{\lim }f(x)=2$
$\underset{x\to -\frac{{\pi }^{+}}{2}}{\lim }f(x)=-A+B$
$f\left(-\frac{\pi }{2}\right)=2$
$\therefore$ From Eq. (ii), we get
$-A+B=2...(iii)$
For continuity at $x=\frac{\pi }{2}$
$\underset{x\to \frac{\pi }{2}}{\lim }f(x)=\underset{x\to \frac{{\pi }^{+}}{2}}{\lim }f(x)=f\left(\frac{\pi }{2}\right)...(iv)$
Here, $\underset{x\to \frac{{\pi }^{-}}{2}}{\lim }f(x)=A+B$
$\Rightarrow \underset{x\to \frac{{\pi }^{+}}{2}}{\lim }f(x)=0$
And $f\left(\frac{\pi }{2}\right)=0$
$\therefore$ From Eq. (iv)
$A+B=0...(v)$
$\therefore$ From Eqs. (iii) and (v)
$A=-1,B=1$