Question

Determine $a$ and $b$ if the function defined as under be continuous at $x=\pi /2.$ $f\left(x\right)=\{\begin{array}{cc}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x},& x<\pi /2\\ a,& x=\pi /2\\ \frac{b(1-\sin x)}{{(\pi -2x)}^2},& x>\pi /2\end{array}$

• A. $a=\frac{1}{2}$
• B. $b=4$
• C. $b=\frac{1}{2}$
• D. $a=\frac{1}{4}$

Answer 1

Answer: (A), (B)

The correct options are
A $a=\frac{1}{2}$
B $b=4$

$\underset{x\to {\frac{\pi }{2}}^{-}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x}\frac{0}{0}$form

$=\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{-3{\sin }^{2}x\cos x}{6\cos x(-\sin x)}$

$=\frac{1}{2}\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\sin x=\frac{1}{2}$

$f\left(\frac{\pi }{2}\right)=a$

$\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{b(1-\sin x)}{{(\pi -2x)}^2}$

$=\frac{b}{4}\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{(1-\cos (\frac{\pi }{2}-x))}{{(\frac{\pi }{2}-x)}^2}$

$x\to {\frac{\pi }{2}}^{+}$

$\frac{\pi }{2}-x\to 0^{+}$$=\frac{b}{4}\times \frac{1}{2}=\frac{b}{8}$

Since $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$

RHL $=$ LHL $=f\left(\frac{\pi }{2}\right)$

$\Rightarrow \frac{1}{2}=\frac{b}{8}=a$

$\Rightarrow b=4,a=\frac{1}{2}$