Functions Question 332
Question: The value of m for which the function $ f(x)= \begin{cases} & mx^{2},x\le 1 \\ & 2x,x>1 \\ \end{cases} $ is differentiable at $ x=1 $ is
[MP PET 1998]
Options:
A) 0
B) 1
C) 2
D) Does not exist
Show Answer
Answer:
Correct Answer: D
Solution:
$ L{f}’(1)=\underset{h\to 0}{\mathop{\lim }}\frac{f(1-h)-f(1)}{-h} $ $ =\underset{h\to 0}{\mathop{\lim }}\frac{m{{(1-h)}^{2}}-m}{-h} $ $ =\underset{h\to 0}{\mathop{\lim }}\frac{m[1+h^{2}-2h-1]}{-h} $ $ =\underset{h\to 0}{\mathop{\lim }}m(2-h)=2m $ and $ R{f}’(1)=\underset{h\to 0}{\mathop{\lim }}\frac{f(1+h)-f(1)}{h} $ $ =\underset{h\to 0}{\mathop{\lim }}\frac{2(1+h)-m}{h} $ .
For differentiability, $ L{f}’(1)=R{f}’(1) $ .
But for any value of $ m,R{f}’(1)=L{f}’(1) $ not possible.
BETA
