A) 0
B) 1/4
C) ?1/4
D) None of these
Correct Answer: B
Solution :
\[f(x)={{\left[ {{x}^{2}}+{{e}^{\frac{1}{2-x}}} \right]}^{-1}}\] and \[f(2)=k\] If \[f(x)\] is continuous from right at \[x=2\] then \[\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f(x)=f(2)=k\] Þ \[\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,{{\left[ {{x}^{2}}+{{e}^{\frac{1}{2-x}}} \right]}^{-1}}=k\]Þ \[k=\underset{h\to 0}{\mathop{\lim }}\,f(2+h)\] Þ \[k=\underset{h\to 0}{\mathop{\lim }}\,\,{{\left[ {{(2+h)}^{2}}+{{e}^{\frac{1}{2-(2+h)}}} \right]}^{\,-1}}\] Þ \[k=\underset{h\to 0}{\mathop{\lim }}\,\,{{\left[ \,4+{{h}^{2}}+4h+{{e}^{-1/h}}\, \right]}^{\,-1}}\] Þ \[k={{[4+0+0+{{e}^{-\infty }}]}^{\,-1}}\]Þ \[k=\frac{1}{4}\].
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