Question

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{1-{\sin }^{2}x}{3{\cos }^{2}x},& x<\frac{\mathrm{\pi }}{2}\\ a,& x=\frac{\mathrm{\pi }}{2}\\ \frac{b\left(1-\sin x\right)}{{\left(\mathrm{\pi }-2\mathrm{x}\right)}^2},& x>\frac{\mathrm{\pi }}{2}\end{array}\right.$. Then, f (x) is continuous at $x=\frac{\mathrm{\pi }}{2}$, if
(a) $a=\frac{1}{3},$ b = 2

(b) $a=\frac{1}{3},b=\frac{8}{3}$

(c) $a=\frac{2}{3},b=\frac{8}{3}$

(d) none of these

Answer 1

(b) $a=\frac{1}{3},b=\frac{8}{3}$


Given: $f\left(x\right)=\left\{\begin{array}{c}\frac{1-{\sin }^{2}x}{3{\cos }^{2}x},\mathrm{if}\mathrm{x}<\frac{\mathrm{\pi }}{2}\\ a,\mathrm{if}x=\frac{\mathrm{\pi }}{2}\\ \frac{b\left(1-\sin x\right)}{{\left(\mathrm{\pi }-2\mathrm{x}\right)}^2},\mathrm{if}x>\frac{\mathrm{\pi }}{2}\end{array}\right.$

We have
(LHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\frac{\mathrm{\pi }}{2}-h\right)$

$$\begin{gathered} =\underset{h\to 0}{\lim }\left(\frac{1-{\sin }^2\left(\frac{\mathrm{\pi }}{2}-h\right)}{3{\cos }^2\left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{1-{\cos }^{2}h}{3{\sin }^{2}h}\right) \\ =\frac{1}{3}\underset{h\to 0}{\lim }\left(\frac{{\sin }^{2}h}{{\sin }^{2}h}\right) \\ =\frac{1}{3} \end{gathered}$$

(RHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\frac{\mathrm{\pi }}{2}+h\right)$

$$\begin{gathered} =\underset{h\to 0}{\lim }\left(\frac{b\left[1-\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}+h\right)\right]}{{\left[\mathrm{\pi }-2\left({\displaystyle \frac{\mathrm{\pi }}{2}+}h\right)\right]}^2}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{b\left(1-\cos h\right)}{{\left[-2h\right]}^2}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{2b{\sin }^2{\displaystyle \frac{h}{2}}}{4{\displaystyle h^2}}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{2b{\sin }^2{\displaystyle \frac{h}{2}}}{16{\displaystyle \frac{{\displaystyle h^2}}{4}}}\right) \\ =\frac{b}{8}\underset{h\to 0}{\lim }{\left(\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\right)}^2 \\ =\frac{b}{8}\times 1 \\ =\frac{b}{8} \end{gathered}$$

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

If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$, then

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

$\Rightarrow \frac{1}{3}=\frac{b}{8}=a$

$\Rightarrow a=\frac{1}{3}\mathrm{and}b=\frac{8}{3}$