Question

Find the value of $p$ and $q$ for which $f\left(x\right)=\left\{\begin{array}{c}\frac{1-\sin {}^{3}x}{{\cos }^{2}x},ifx<\pi /2\\ p,ifx=\pi /2\\ \frac{q(1-\sin x)}{(\pi -2x{)}^2},ifx>\pi /2\end{array}\right\}$ is continuous at $x=\pi /2$ .

Answer 1

Limit exists if ${lim}_{x\to a^{-}}f\left(x\right)={lim}_{x\to a^{+}}f\left(x\right)=f\left(a\right)$

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

$={lim}_{x\to {\frac{\pi }{2}}^{-}}\frac{(1-\sin x)(1+{\sin }^{2}x+\sin x)}{3(1-{\sin }^{2}x)}$

$={lim}_{x\to {\frac{\pi }{2}}^{-}}\frac{(1-\sin x)(1+{\sin }^{2}x+\sin x)}{3(1-\sin x)(1+\sin x)}$

$={lim}_{x\to {\frac{\pi }{2}}^{-}}\frac{(1+{\sin }^{2}x+\sin x)}{3(1+\sin x)}$

$=\frac{(1+{\sin }^2\frac{\pi }{2}+\sin \frac{\pi }{2})}{3(1+\sin \frac{\pi }{2})}$

$=\frac{1+1+1}{3(1+1)}$

$=\frac{1}{2}$

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

$={lim}_{x\to {\frac{\pi }{2}}^{+}}\frac{q(1-\cos (\frac{\pi }{2}-x))}{{(\pi -2x)}^2}$

$={lim}_{x\to {\frac{\pi }{2}}^{+}}\frac{q\left(2{\sin }^2(\frac{\pi }{4}-\frac{x}{2})\right)}{16{(\frac{\pi }{4}-\frac{x}{2})}^2}$

$=\frac{2q}{16}=\frac{q}{8}$

Given that the function is continuous

$\Rightarrow \frac{q}{8}=p=\frac{1}{2}$

$\Rightarrow q=\frac{8}{2}=4,p=\frac{1}{2}$

$\therefore q=4,p=\frac{1}{2}$

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