Functions Question 237
Question: Let $ f(x),= \begin{cases} x, & when & x<1 \\ 2-x, & when & 1 \le x\ge 2 \\ −2+3x−x^2, & when & x\gt 2 \\ \end{cases}, $
then $ {f}(x)$ is :
[Karnataka CET 2002]
Options:
A) differentiable at x=1
B) differentiable at x=2
C) differentiable at x=1 and x=2
D) not differentiable at x=10
Show Answer
Answer:
Correct Answer: B
Solution:
Checking Differentiability at $ x = 1 $
-
Left-hand derivative at $ x = 1 $: $ f’(1^-) = \lim_{{h \to 0^-}} \frac{{f(1 + h) - f(1)}}{h} = \lim_{{h \to 0^-}} \frac{{(1 + h) - 1}}{h} = \lim_{{h \to 0^-}} \frac{h}{h} = 1 $
-
Right-hand derivative at $ x = 1 $: $ f’(1^+) = \lim_{{h \to 0^+}} \frac{{f(1 + h) - f(1)}}{h} = \lim_{{h \to 0^+}} \frac{{(2 - (1 + h)) - 1}}{h} = \lim_{{h \to 0^+}} \frac{{1 - h - 1}}{h} = \lim_{{h \to 0^+}} \frac{-h}{h} = -1 $
Since $ f’(1^-) \neq f’(1^+) $, $ f(x) $ is not differentiable at $ x = 1 $.
Checking Differentiability at $ x = 2 $
-
Left-hand derivative at $ x = 2 $: $ f’(2^-) = \lim_{{h \to 0^-}} \frac{{f(2 + h) - f(2)}}{h} = \lim_{{h \to 0^-}} \frac{{(2 - (2 + h)) - 0}}{h} = \lim_{{h \to 0^-}} \frac{{-h}}{h} = -1 $
-
Right-hand derivative at $ x = 2 $:
$ f’(2^+) = \lim_{{h \to 0^+}} \frac{{f(2 + h) - f(2)}}{h} = \lim_{{h \to 0^+}} \frac{{(-2 + 3(2 + h) - (2 + h)^2) - 0}}{h} $
$ = \lim_{{h \to 0^+}} \frac{{(-2 + 6 + 3h - 4 - 4h - h^2) - 0}}{h} = \lim_{{h \to 0^+}} \frac{{-h^2 - h}}{h} = -1 $
Since $ f’(2^-) = f’(2^+) $, $ f(x) $ is differentiable at $ x = 2 $.
BETA
