A) \[-\frac{1}{16}\]
B) \[-\frac{1}{32}\]
C) \[-\frac{1}{64}\]
D) \[-\frac{1}{28}\]
Correct Answer: C
Solution :
For \[f(x)\] to be continuous at \[x=\frac{\pi }{2},\] we must have \[\underset{x\to \pi /2}{\mathop{\lim }}\,f(x)=f(\pi /2)\] \[\Rightarrow \]\[\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{1-\sin x}{(\pi =2{{x}^{2}})}.\frac{\log (\sin x)}{\log (1+{{\pi }^{2}}-4\pi x+4{{x}^{2}})}=k\] \[\Rightarrow \] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{1-\cosh }{4{{h}^{2}}}\times \frac{\log \cosh }{\log (1+4{{h}^{2}})}=k\] \[\Rightarrow \] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{1-\cosh }{4{{h}^{2}}}\times \frac{\log \{1+\cosh -1\}}{\cosh -1}\] \[\times \frac{4{{h}^{2}}}{\log (1+4{{h}^{2}})}\times \frac{\cosh -1}{4{{h}^{2}}}=k\] \[\Rightarrow \] \[\underset{h\to 0}{\mathop{\lim }}\,{{\left( \frac{1-\cosh }{4{{h}^{2}}} \right)}^{2}}\frac{\log (1+(\cosh -1))}{\cosh -1}\] \[\times \frac{4{{h}^{2}}}{\log (1+4{{h}^{2}})}=k\] \[\Rightarrow \] \[-\underset{h\to 0}{\mathop{\lim }}\,{{\left( \frac{{{\sin }^{2}}h/2}{2{{h}^{2}}} \right)}^{4}}\frac{\log (1+(\cosh -1))}{\cosh -1}\] \[\times \frac{4{{h}^{2}}}{\log (1+4{{h}^{2}})}=k\] \[\Rightarrow \] \[-\frac{1}{64}\underset{h\to 0}{\mathop{\lim }}\,{{\left( \frac{\sinh /2}{h/2} \right)}^{4}}\frac{\log (1+(\cosh -1))}{\cosh -1}\] \[\times \frac{4{{h}^{2}}}{\log (1+4{{h}^{2}})}=k\] \[\Rightarrow \] \[-\frac{1}{64}=k\]
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