Question

# If $f\left(x\right)=\left\{\begin{array}{cc}\frac{1-{\mathrm{sin}}^{2}x}{3{\mathrm{cos}}^{2}x},& x<\frac{\mathrm{\pi }}{2}\\ a,& x=\frac{\mathrm{\pi }}{2}\\ \frac{b\left(1-\mathrm{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

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## (b) $a=\frac{1}{3},b=\frac{8}{3}$ Given: $f\left(x\right)=\left\{\begin{array}{c}\frac{1-{\mathrm{sin}}^{2}x}{3{\mathrm{cos}}^{2}x},\mathrm{if}\mathrm{x}<\frac{\mathrm{\pi }}{2}\\ a,\mathrm{if}x=\frac{\mathrm{\pi }}{2}\\ \frac{b\left(1-\mathrm{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 = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{\mathrm{\pi }}{2}-h\right)$ $=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1-{\mathrm{sin}}^{2}\left(\frac{\mathrm{\pi }}{2}-h\right)}{3{\mathrm{cos}}^{2}\left(\frac{\mathrm{\pi }}{2}-h\right)}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1-{\mathrm{cos}}^{2}h}{3{\mathrm{sin}}^{2}h}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\mathrm{sin}}^{2}h}{{\mathrm{sin}}^{2}h}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$ (RHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{\mathrm{\pi }}{2}+h\right)$ $=\underset{h\to 0}{\mathrm{lim}}\left(\frac{b\left[1-\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}+h\right)\right]}{{\left[\mathrm{\pi }-2\left(\frac{\mathrm{\pi }}{2}+h\right)\right]}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{b\left(1-\mathrm{cos}h\right)}{{\left[-2h\right]}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{2b{\mathrm{sin}}^{2}\frac{h}{2}}{4{h}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{2b{\mathrm{sin}}^{2}\frac{h}{2}}{16\frac{{h}^{2}}{4}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{b}{8}\underset{h\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{sin}\frac{h}{2}}{\frac{h}{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{b}{8}×1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{b}{8}$ Also, $f\left(\frac{\mathrm{\pi }}{2}\right)=a$ If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$, then ​$\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$ $⇒\frac{1}{3}=\frac{b}{8}=a$ $⇒a=\frac{1}{3}\mathrm{and}b=\frac{8}{3}$

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