Limits Continuity And Differentiability Question 76
If $ f(x)= \begin{cases} {e^{x^{2}+x}} & x>0 \ ax+b & x\le 0 \ \end{cases} . $ is differentiable at x=0, then
Options:
A) $ a=1,b=-1 $
B) $ a=-1,b=1 $
C) $ a=1,b=1 $
D) $ a=-1,b=-1 $
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Answer:
Correct Answer: C
Solution:
For $ f(x) $ to be continuous at x=0, we have $ f({0^{-}})=f({0^{+}}) $
or $ a(0)+b=1 $ or b=1 $ f({0^{+}})=\underset{h\to 0}{\mathop{\lim }}\frac{f(h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\frac{{e^{h^{2}+h}}-b}{h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{{e^{h^{2}+h}}-1}{h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{{e^{h^{2}+h}}-1}{h(h+1)}(h+1)=1 $
$ \therefore f’({0^{-}})=\alpha $
Hence, a=1.
BETA
