Question

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{\sin \left(\cos x\right)-\cos x}{{\left(\mathrm{\pi }-2x\right)}^2},& x\ne \frac{\mathrm{\pi }}{2}\\ k,& x=\frac{\mathrm{\pi }}{2}\end{array}\right.$is continuous at x = π/2, then k is equal to
(a) 0
(b) $\frac{1}{2}$
(c) 1
(d) −1

Answer 1

(a) 0

Given: $f\left(x\right)=\left\{\begin{array}{l}\frac{\sin \left(\cos x\right)-\cos x}{{\left(\mathrm{\pi }-2\mathrm{x}\right)}^2},x\ne \frac{\mathrm{\pi }}{2}\\ k,x=\frac{\mathrm{\pi }}{2}\end{array}\right.$

If f(x) is continuous at $x=\frac{\mathrm{\pi }}{2}$, then
$\underset{x\mathit{\to }\frac{\mathrm{\pi }}{2}}{\lim }f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$

$\Rightarrow \underset{x\mathit{\to }\frac{\mathrm{\pi }}{2}}{\lim }\frac{\sin \left(\cos x\right)-\cos x}{{\left(\mathrm{\pi }-2\mathrm{x}\right)}^2}=k$


Now,
$\frac{\mathrm{\pi }}{2}-x=y$
$\Rightarrow \mathrm{\pi }-2x=2y$

Also, $x\to \frac{\mathrm{\pi }}{2},y\to 0$

$\Rightarrow \underset{y\mathit{\to }0}{\lim }\frac{\sin \left(\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-y\right)\right)-\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-y\right)}{4y^2}=k$

$\Rightarrow \underset{y\mathit{\to }0}{\lim }\frac{\sin \left(\sin y\right)-\sin \left(y\right)}{4{\mathrm{y}}^2}=k$

$$\begin{gathered} \Rightarrow \underset{y\mathit{\to }0}{\lim }\frac{2\sin \left({\displaystyle \frac{\sin y-y}{2}}\right)\cos \left({\displaystyle \frac{\sin y+y}{2}}\right)}{4y^2}=k\left[\because \sin C-\sin D=2s\mathrm{in}\left(\frac{C-D}{2}\right)\cos \left(\frac{C+D}{2}\right)\right] \\ \Rightarrow \frac{1}{2}\underset{y\mathit{\to }0}{\lim }\frac{\sin \left({\displaystyle \frac{\sin y-y}{2}}\right)}{y}\frac{\cos \left({\displaystyle \frac{\sin y+y}{2}}\right)}{y}=k \\ \Rightarrow \frac{1}{2}\underset{y\mathit{\to }0}{\lim }\frac{\left({\displaystyle \frac{\sin y-y}{2}}\right)\sin \left({\displaystyle \frac{\sin y-y}{2}}\right)}{y\left({\displaystyle \frac{\sin y-y}{2}}\right)}\frac{\cos \left({\displaystyle \frac{\sin y+y}{2}}\right)}{y}=k \\ \Rightarrow \frac{1}{2}\underset{y\mathit{\to }0}{\lim }\left(\frac{\left({\displaystyle \frac{\sin y-y}{2}}\right)}{y}\right)\left(\frac{\sin \left({\displaystyle \frac{\sin y-y}{2}}\right)}{\left({\displaystyle \frac{\sin y-y}{2}}\right)}\right)\left(\frac{\cos \left({\displaystyle \frac{\sin y+y}{2}}\right)}{y}\right)=k \\ \Rightarrow \frac{1}{2}\underset{y\mathit{\to }0}{\lim }\left(\frac{\left({\displaystyle \frac{\sin y-y}{2}}\right)}{y}\right)\underset{y\mathit{\to }0}{\lim }\left(\frac{\sin \left({\displaystyle \frac{\sin y-y}{2}}\right)}{\left({\displaystyle \frac{\sin y-y}{2}}\right)}\right)\underset{y\mathit{\to }0}{\lim }\left(\frac{\cos \left({\displaystyle \frac{\sin y+y}{2}}\right)}{y}\right)=k \\ \Rightarrow \frac{1}{4}\underset{y\mathit{\to }0}{\lim }\left(\frac{\sin y}{y}-1\right)\underset{y\mathit{\to }0}{\lim }\left(\frac{\sin \left({\displaystyle \frac{\sin y-y}{2}}\right)}{\left({\displaystyle \frac{\sin y-y}{2}}\right)}\right)\underset{y\mathit{\to }0}{\lim }\left(\frac{\cos \left({\displaystyle \frac{\sin y+y}{2}}\right)}{y}\right)=k \\ \Rightarrow \frac{1}{4}\times 0\times 1\times \underset{y\mathit{\to }0}{\lim }\left(\frac{\cos \left({\displaystyle \frac{\sin y+y}{2}}\right)}{y}\right)=k \\ \Rightarrow 0=k \end{gathered}$$​