Functions Question 278
Question: sThe function $ f(x)= \begin{cases} e^{2x}-1 & , & x\le 0 \\ ax+\frac{bx^{2}}{2}-1 & , & x>0 \\ \end{cases} . $ is continuous and differentiable for
[AMU 2002]
Options:
A) $ [.] $
B) $ a=2,,b=4 $
C) $ a=2,, $ any $ b $
D) Any $ a,,b=4 $
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Answer:
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 }},( \frac{{e^{-2h}}-1+1}{-h} )=\underset{h\to 0}{\mathop{\lim }},( \frac{ah+\frac{bh^{2}}{2}-1+1}{h} ) $
Þ $ \underset{h\to 0}{\mathop{\lim }},( \frac{-2{e^{-2h}}}{-1} )=\underset{h\to 0}{\mathop{\lim }},( a+\frac{bh}{2} ) $
Þ $ 2=a+0 $
Þ $ a=2,b $ any number.
BETA
