Question

If $f\left(x\right)=(x-1{)}^2,$ then which of the following is/are true

• A. $f$ is continuous at $x=1$
• B. $f$ is not derivable at $x=1$
• C. $f$ is discontinuous at $x=1$
• D. $f$ is derivable at $x=1$

Answer 1

Answer: (A), (D)

$f$ is derivable at $x=1$

For continuity,
$L.H.L.=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(x-1{)}^2$
$=\underset{h\to 0}{lim}(1-h-1{)}^2=\underset{h\to 0}{lim}{h}^2$
$=0$

$R.H.L.=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(x-1{)}^2$
$=\underset{h\to 0}{lim}(1+h-1{)}^2=\underset{h\to 0}{lim}{h}^2$
$=0$
$\Rightarrow L.H.L.=R.H.L.=f\left(1\right)$
So, $f\left(x\right)$ is continuous at $x=1$

For differentiablity,
$R.H.D.$
$f^{\prime }\left(1^{+}\right)=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}$
$=\underset{h\to 0}{lim}\frac{h^2-0}{h}$
$=\underset{h\to 0}{lim}h$
$\Rightarrow f^{\prime }\left(1^{+}\right)=0$

$L.H.D.$
$f^{\prime }\left(1^{-}\right)=\underset{h\to 0}{lim}\frac{f(1-h)-f\left(1\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{h^2-0}{-h}$
$=-\underset{h\to 0}{lim}(-h)$
$\Rightarrow f^{\prime }\left(1^{-}\right)=0$
$\Rightarrow L.H.D.=R.H.D.$
Hence, $f\left(x\right)$ is derivable at $x=1$