A) \[[.]\]
B) \[a=2,\,b=4\]
C) \[a=2,\,\]any \[b\]
D) Any \[a,\,\,\,b=4\]
Correct Answer: C
Solution :
\[\because f\] is continuous at \[x=0\], \\[f({{0}^{-}})=f({{0}^{+}})=f(0)=-1\] Also \[L{f}'(0)=R{f}'(0)\] Þ \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}\] Þ \[\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{{{e}^{-2h}}-1+1}{-h} \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{ah+\frac{b{{h}^{2}}}{2}-1+1}{h} \right)\] Þ \[\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{-2{{e}^{-2h}}}{-1} \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( a+\frac{bh}{2} \right)\] Þ \[2=a+0\] Þ \[a=2,\,\,b\] any number.
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