Functions Question 169
Question: If $ f(x)= \begin{cases} e^{x}+ax, & x<0 \\ b{{(x-1)}^{2}}, & x\ge 0 \\ \end{cases} . $ is differentiable at $ x=0, $ then $ (a,,b) $ is
[MP PET 2000]
Options:
A) $ (-3,,-1) $
B) $ (-3,1) $
C) $ (3,1) $
D) $ (3,,-1) $
Show Answer
Answer:
Correct Answer: B
Solution:
Given $ f(x) $ is differentiable at $ x=0 $ . Hence, $ f(x) $ will be continuous at $ x=0 $ . \ $ \underset{x\to {0^{-}}}{\mathop{lim}},(e^{x}+ax)=\underset{x\to {0^{+}}}{\mathop{lim}},b{{(x-1)}^{2}} $
Þ $ e^{0}+a\times 0=b{{(0-1)}^{2}} $
Þ $ b=1 $ ?.. (i) But $ f(x) $ is differentiable at $ x=0 $ , then $ L{f}’(x)=R{f}’(x) $
Þ $ \frac{d}{dx}(e^{x}+ax)=\frac{d}{dx}b{{(x-1)}^{2}} $
Þ $ e^{x}+a=2b(x-1) $ At $ x=0, $ $ e^{0}+a=-2b $
Þ $ a+1=-2b $
Þ $ a=-3 $
Þ $ (a,b)=(-3,1) $ .
BETA
