Question

Let $f:R\to R$ be defined by $f\left(x\right)=\left\{\begin{array}{l}k-2x,ifx\le -1\\ 2x+3,ifx>-1\end{array}\right.$. If f has a local minimum at $x=-1$, then a possible value of $k$ is

• A. $1$
• B. $0$
• C. $-1/2$
• D. $-1$

Answer 1

The correct option is D

$-1$

Explanation for the correct option:

Taking limit of $f\left(x\right)$:

$f\left(x\right)$ has local minima at $x=-1$

Therefore, the right-hand limit at this point is greater than or equal to $f(-1)$ and left-hand limit is less than or equal to $f(-1)$.

$$\begin{gathered} \therefore \underset{x=-1^{+}}{\lim }f\left(x\right)=2(-1)+3 \\ =1 \end{gathered}$$

$$\begin{gathered} \therefore \underset{x=-1^{-}}{\lim }f\left(x\right)=k-2(-1) \\ =k+2 \end{gathered}$$

And,

$$\begin{gathered} f(-1)=k-2(-1) \\ =k+2 \end{gathered}$$

$$\begin{gathered} \Rightarrow \underset{x=-1^{+}}{\lim }f\left(x\right)\ge f(-1)\ge \underset{x=-1^{-}}{\lim }f\left(x\right) \\ \Rightarrow 1\ge k+2 \\ \Rightarrow -1\ge k \end{gathered}$$

Therefore, the correct answer is option (D).