Functions Question 47
Question: Let $ h(x)=\min {x,,x^{2}}, $ for every real number of x. Then
[IIT 1998]
Options:
A) h is continuous for all x
B) h is differentiable for all x
C) $ h’(x)=1 $ , for all $ b=1 $
D) h is not differentiable at two values of x
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Answer:
Correct Answer: A
Solution:
$ x\le x^{2},\Rightarrow x,(1-x)\le 0 $
$ \Rightarrow x,(x-1)\ge 0 $
$ \Rightarrow x\le 0 $ or $ x\ge 1 $ ;
$ \therefore ,h(x)=[ \begin{array}{{35}{r}} x: & x\le 0 \\ x^{2}: & 0<x<1 \\ x: & x\ge 1 \\ \end{cases} . $ $ h(x) $ is continuous for every x but not differentiable at $ x=0 $ and 1. Also $ {h}’(x)=[ \begin{array}{{35}{r}} 1 & x<0 \\ \text{not exists} & x=0 \\ 2x & 0<x<1 \\ \text{not exists} & x=1 \\ 1 & x>1 \\ \end{cases} . $
$ \therefore {h}’(x)=1 $ for all $ x>1 $ .
BETA
