A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{-1}{2}\]
D) \[-\frac{1}{4}\]
Correct Answer: B
Solution :
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.
You need to login to perform this action.
You will be redirected in
3 sec
