Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

Let $f: R \rightarrow R$ be defined by

$f(x)=\left\{\begin{array}{l}
k-2 x, \text { if } x \leq-1 \\
2 x+3, \text { if } x>-1
\end{array}\right.$

If f has a local minimum at $x=-1$, then a possible value of $k$, is

Options:

-1/2

-1

1

0

Correct Answer:

-1

Explanation:

If $f(x)$ has a local minimum at $x=-1$, then

$\lim\limits_{x \rightarrow-1^{+}} f(x)=\lim\limits_{x \rightarrow-1^{-}} f(x)$

$\Rightarrow \lim\limits_{x \rightarrow-1^{+}} 2 x+3=\lim\limits_{x \rightarrow-1^{-}} k-2 x$

$\Rightarrow -2+3=k+2 \Rightarrow k=-1$