Question

Let $f\left(x\right)=\{\begin{array}{cc}k+x^2,& x<2\\ (x-2{)}^2+3,& x\ge 2\end{array}.$ Then which of the following statements is/are CORRECT?

• A. If $f\left(x\right)$ is continuous at $x=2,$ then $k=-1$
• B. If $f\left(x\right)$ is differentiable at $x=2,$ then $k=-1$
• C. $f\left(x\right)$ is not differentiable at $x=2,$ for any real value of $k$
• D. If $k>3,$ then the least value of $f\left(x\right)$ occurs at $x=2$

Answer 1

Answer: (A), (C), (D)

If $k>3,$ then the least value of $f\left(x\right)$ occurs at $x=2$

$f\left(x\right)=\{\begin{array}{cc}k+x^2,& x<2\\ (x-2{)}^2+3,& x\ge 2\end{array}$

$\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}(k+x^2)=k+4\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}\left(\right(x-2{)}^2+3)=3$
For being continuous at $x=2,$
$3=k+4$
$\Rightarrow k=-1$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}2x,& x<2\\ 2(x-2),& x\ge 2\end{array}$

$f^{\prime }\left(2^{-}\right)=4f^{\prime }\left(2^{+}\right)=0$
Clearly, $f$ is not differentiable for any real of value of $k.$


From above graph, clearly for $k>3,$ least value of $f\left(x\right)$ occurs at $x=2$