• # question_answer Let f(x)=\left\{ \begin{align} & \frac{1+\cos x}{{{(\pi /x)}^{2}}}.\frac{{{\sin }^{2}}x}{\log (1+{{\pi }^{2}}-2\pi x+{{x}^{2}})},x\ne \pi \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x=\pi \\ \end{align} \right.. If f (x) is continuous function at $x=\pi ,$then k is equal to A)  $\frac{1}{4}$                                   B)  $\frac{1}{2}$ C)  $\frac{-1}{2}$                                 D)  $-\frac{1}{4}$

Idea If f(x) is continuous at x = a, then $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=f(a)$ The given function is $f(x)=\left\{ \begin{matrix} \frac{1+\cos x}{{{(\pi -x)}^{2}}}.\frac{{{\sin }^{2}}x}{\log (1+{{\pi }^{2}}-2\pi x+{{x}^{2}})}, & x\ne \pi \\ k & ,x=\pi \\ \end{matrix} \right\}$Since, f(x) is continuous at $x=\pi$ $\therefore$  $\underset{x\to \pi }{\mathop{\lim }}\,f(x)=f(\pi )$ $\Rightarrow$$\underset{h\to 0}{\mathop{\lim }}\,f(\pi -h)=k\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\frac{1+\cos (\pi -h)}{{{(\pi -\pi +h)}^{2}}}.$ $\Rightarrow$$\frac{{{\sin }^{2}}(\pi -h)}{\log {{[1+{{\pi }^{2}}-2\pi (\pi -h)+(\pi -h)]}^{2}}}=k$ $\Rightarrow$$\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos \,h}{{{\pi }^{2}}}\times \frac{{{\sin }^{2}}\,h}{\log (1+{{h}^{2}})}=k$ $\Rightarrow$$\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{2}{{\left( \frac{\sin \,h/2}{h/2} \right)}^{2}}.\frac{{{h}^{2}}\,}{\log \,(1+{{h}^{2}})}.{{\left( \frac{\sin \,h}{h} \right)}^{2}}$ $\Rightarrow$               $\frac{1}{2}=k$ TEST Edge   Continuity   over   an interval discontinuity related questions are asked. To solve such types of questions, students are advised to understand the concept of continuity.