Question

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{1-\sin x}{{\left(\mathrm{\pi }-2x\right)}^2}.\frac{\log \sin x}{\log \left(1+{\mathrm{\pi }}^2-4\mathrm{\pi x}+4{\mathrm{x}}^2\right)},& x\ne \frac{\mathrm{\pi }}{2}\\ k,& x=\frac{\mathrm{\pi }}{2}\end{array}\right.$
is continuous at x = π/2, then k =
(a) $-\frac{1}{16}$

(b) $-\frac{1}{32}$

(c) $-\frac{1}{64}$

(d) $-\frac{1}{28}$

Answer 1

$\left(c\right)\frac{-1}{64}$

If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{2}$, then
$$\begin{gathered} \underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\lim }f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right) \\ \mathrm{If}\frac{\mathrm{\pi }}{2}-x=t,\mathrm{then} \\ \Rightarrow \underset{t\to 0}{\lim }f\left(\frac{\mathrm{\pi }}{2}-t\right)=f\left(\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \underset{t\to 0}{\lim }\left(\frac{1-\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-t\right)}{4t^2}\times \frac{\log \sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-t\right)}{\log \left(1+{\mathrm{\pi }}^2-4\mathrm{\pi }\left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{t}\right)+4{\left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{t}\right)}^2\right)}\right)=k \\ \Rightarrow \underset{t\to 0}{\lim }\left(\frac{\left(1-\cos t\right)}{4t^2}\times \frac{\log \cos t}{\log \left(1+{\mathrm{\pi }}^2-2{\mathrm{\pi }}^2+4\mathrm{\pi t}+4\left({\displaystyle \frac{{\mathrm{\pi }}^2}{4}}+{\mathrm{t}}^2-\mathrm{\pi t}\right)\right)}\right)=k \\ \Rightarrow \underset{t\to 0}{\lim }\left(\frac{\left(1-\cos t\right)}{4t^2}\times \frac{\log \cos t}{\log \left(1-{\mathrm{\pi }}^2+4\mathrm{\pi t}+\left({\mathrm{\pi }}^2+4{\mathrm{t}}^2-4\mathrm{\pi t}\right)\right)}\right)=k \\ \Rightarrow \underset{t\to 0}{\lim }\left(\frac{\left(1-\cos t\right)}{4t^2}\times \frac{\log \cos t}{\log \left(1+4{\mathrm{t}}^2\right)}\right)=k \\ \Rightarrow \underset{t\to 0}{\lim }\left(\frac{2{\sin }^2{\displaystyle \frac{t}{2}}}{16\times {\displaystyle \frac{t^2}{4}}}\times \frac{\log \cos t}{\log \left(1+4{\mathrm{t}}^2\right)}\right)=k \\ \Rightarrow \frac{2}{16}\underset{t\to 0}{\lim }\left(\frac{{\sin }^2{\displaystyle \frac{t}{2}}}{\left({\displaystyle \frac{t^2}{4}}\right){\displaystyle }}\times \frac{\log \cos t}{\left({\displaystyle \frac{4{\mathrm{t}}^2\log \left(1+4{\mathrm{t}}^2\right)}{4{\mathrm{t}}^2}}\right){\displaystyle }}\right)=k \\ \Rightarrow \frac{1}{8}\underset{t\to 0}{\lim }\left(\frac{{\sin }^2{\displaystyle \frac{t}{2}}}{{\displaystyle {\left(\frac{t}{2}\right)}^2}}\times \frac{\left({\displaystyle \frac{\log \cos t}{4t^2}}\right)}{\left({\displaystyle \frac{\log \left(1+4{\mathrm{t}}^2\right)}{4{\mathrm{t}}^2}}\right)}\right)=k \\ \Rightarrow \frac{1}{8}\underset{t\to 0}{\lim }\left(\frac{{\sin }^2{\displaystyle \frac{t}{2}}}{{\displaystyle {\left(\frac{t}{2}\right)}^2}}\times \frac{\left({\displaystyle \frac{\log \sqrt{1-{\sin }^{2}t}}{4t^2}}\right)}{\left({\displaystyle \frac{\log \left(1+4{\mathrm{t}}^2\right)}{4{\mathrm{t}}^2}}\right)}\right)=k \\ \Rightarrow \frac{1}{8}\underset{t\to 0}{\lim }\left(\frac{{\sin }^2{\displaystyle \frac{t}{2}}}{{\displaystyle {\left(\frac{t}{2}\right)}^2}}\times \frac{\left({\displaystyle \frac{\log \left(1-{\sin }^{2}t\right)}{\left(8t^2\right)}}\right)}{\left({\displaystyle \frac{\log \left(1+4{\mathrm{t}}^2\right)}{4{\mathrm{t}}^2}}\right)}\right)=k \\ \Rightarrow \frac{1}{64}\underset{t\to 0}{\lim }\left(\frac{{\sin }^2{\displaystyle \frac{t}{2}}}{{\displaystyle {\left(\frac{t}{2}\right)}^2}}\times \frac{\left({\displaystyle \frac{\log \left(1-{\sin }^{2}t\right)}{t^2}}\right)}{\left({\displaystyle \frac{\log \left(1+4{\mathrm{t}}^2\right)}{4{\mathrm{t}}^2}}\right)}\right)=k \\ \Rightarrow \frac{1}{64}\left(\underset{t\to 0}{\lim }{\left(\frac{\sin {\displaystyle \frac{t}{2}}}{{\displaystyle \left(\frac{t}{2}\right)}}\right)}^2\times \frac{\underset{t\to 0}{\lim }{\displaystyle \left(\frac{\log \left(1-{\sin }^{2}t\right)}{t^2}\right)}}{\underset{t\to 0}{\lim }{\displaystyle \left(\frac{\log \left(1+4{\mathrm{t}}^2\right)}{4{\mathrm{t}}^2}\right)}}\right)=k \\ \Rightarrow \frac{1}{64}\left(1\times \underset{t\to 0}{\lim }\frac{\left(-{\sin }^{2}t\right)\log \left(1-{\sin }^{2}t\right)}{t^2\left(-{\sin }^{2}t\right)}\right)=k \\ \Rightarrow \frac{-1}{64}\left(\underset{t\to 0}{\lim }\frac{\left({\sin }^{2}t\right)\log \left(1-{\sin }^{2}t\right)}{t^2\left(-{\sin }^{2}t\right)}\right)=k \\ \Rightarrow \frac{-1}{64}\left(\underset{t\to 0}{\lim }{\left(\frac{\sin t}{t}\right)}^2\underset{t\to 0}{\lim }\frac{\log \left(1-{\sin }^{2}t\right)}{\left(-{\sin }^{2}t\right)}\right)=k \\ \Rightarrow \frac{-1}{64}\left(\underset{t\to 0}{\lim }{\left(\frac{\sin t}{t}\right)}^2\underset{t\to 0}{\lim }\frac{\log \left(1-{\sin }^{2}t\right)}{\left(-{\sin }^{2}t\right)}\right)=k \\ \Rightarrow k=\frac{-1}{64}\left[\because \underset{x\to 0}{\lim }\frac{\log \left(1-x\right)}{x}=1\right] \end{gathered}$$