Question

Consider the function $f\left(x\right)=\left(x-2\right)\left|x^2-3x+2\right|$ , 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)

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