Question

Let $f(x) = \begin{cases} - 2 \sin x & \quad \text{if } x \leq - \frac{\pi}{2}\\ A \ \sin x + B & \quad \text{if } - \frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos & \quad \text{if } x \leq \frac{\pi}{2} \end{cases} $
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,
$f(x)=\begin{cases}-2 \sin X, \text { if } x \leq-\frac{\pi}{2} \\ A \sin x +B, \text { if }-\frac{\pi}{2}<\,x<\,\frac{\pi}{2} \\ \cos x, \text { if } x \geq \frac{\pi}{2}\end{cases} \,...(i)$
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 $x=-\frac{\pi}{2}$
$\displaystyle\lim _{x \rightarrow-\frac{\pi^{-}}{2}} f(x)=\displaystyle\lim _{x \rightarrow-\frac{\pi^{+}}{2}} f(x)=f\left(-\frac{\pi}{2}\right) \,...(ii)$
$\displaystyle\lim _{x \rightarrow-\frac{\pi^{-}}{2}} f(x)=2$
$\displaystyle\lim _{x \rightarrow-\frac{\pi^{+}}{2}} f(x) =-A+B $
$f\left(-\frac{\pi}{2}\right)=2$
$\therefore $ From Eq. (ii), we get
$-A+B=2\,...(iii)$
For continuity at $x=\frac{\pi}{2}$
$\displaystyle\lim _{x \rightarrow \frac{\pi}{2}} f(x)=\displaystyle\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=f\left(\frac{\pi}{2}\right)\,...(iv)$
Here, $ \displaystyle\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=A+B$
$\Rightarrow \displaystyle\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=0$
And $f\left(\frac{\pi}{2}\right)=0$
$\therefore $ From Eq. (iv)
$A+B=0\,...(v)$
$\therefore $ From Eqs. (iii) and (v)
$A=-1, B=1$