Functions Question 155
Question: The function defined by $ f(x)= \begin{cases} & |x-3|,;x\ge 1 \\ & \frac{1}{4}x^{2}-\frac{3}{2}x+\frac{13}{4};,x<1 \\ \end{cases} . $ is
[IIT 1988]
Options:
A) Continuous at $ x=1 $
B) Continuous at $ x=3 $
C) Differentiable at $ x=1 $
D) All the above
Show Answer
Answer:
Correct Answer: D
Solution:
Since $ |x-3|,=x-3, $ if $ x\ge 3 $ $ =-x+3, $ if $ x<3 $
$ \therefore $ The given function can be defined as $ f(x)= \begin{cases} \frac{1}{4}x^{2}-\frac{3}{2}x+\frac{13}{4}, & x<1 \\ 3-x, & 1\le x<3 \\ x-3, & x\ge 3, \\ \end{cases} . $ Now proceed to check the continuity and differentiability at $ x=1. $
BETA
