Question

Consider the function $f\left(x\right)=\left(x - 2\right)\left|x^{2} - 3 x + 2\right|$ , then the incorrect statement is

• A. $f\left(x\right)$ is continuous at $x=1$
• B. $f\left(x\right)$ is continuous at $x=2$
• C. $f\left(x\right)$ is differentiable at $x=1$
• D. $f\left(x\right)$ is differentiable at $x=2$

Answer 1

Answer: (C)

$\because x-2 \&\left|x^{2}-3 x+2\right|$ both are continuous $\forall x \in R$
$\therefore f(x)$ is continuous $\forall x \in R$
Now, $\left|x^{2}-3 x+2\right|=\mid(x-1)(x-2)$
$\therefore f(x)=\left\{\begin{array}{lll}(x-2)(x-1)(x-2) & : & x \in(-\infty, 1) \cup(2, \infty) \\ -(x-2)(x-1)(x-2) & : & x \in[1,2]\end{array}\right.$
$f(x)=\left\{\begin{array}{lll}(x-1)(x-2)^{2} & : & x \in(-\infty, 1) \cup(2, \infty) \\ -(x-1)(x-2)^{2} & : & x \in[1,2]\end{array}\right.$
$f^{\prime}(x)$
$=\left\{\begin{array}{lll}(x-1) 2(x-2)+(x-2)^{2} & : & x \in(-\infty, 1) \cup(2, \infty) \\ -(x-1) 2(x-2)-(x-2)^{2} & : & x \in(1,2)\end{array}\right.$
Clearly, $f^{\prime}\left(2^{-}\right)=f^{\prime}\left(2^{+}\right)=0 \Rightarrow f(x)$ is differentiable at $x=2$ and $f^{\prime}\left(1^{-}\right)=1, f^{\prime}\left(1^{+}\right)=-1 \Rightarrow f(x)$ is non-differentiable at $x=1$