Let $f:R \to R$ be defined by $f(x) = \left\{ {\begin{array}{*{20}{c}}
  {k - 2x} \\
  {2x + 3}
\end{array}} \right.$ if $\begin{array}{*{20}{c}}
  {x \leqslant - 1} \\
  {x > - 1}
\end{array}$ if $f$ has a local minimum at $x = - 1$ , then a possible value of $k$ is:
A. $1$
B. $0$
C. $- \dfrac{1}{2}$
D. $- 1$

Hint: To solve these questions first find the value of the function at the given critical point. If the function has a local minimum at that point then the value of the function at that critical point would be the least. Then, try finding the value of the function by taking any other value other than the critical point and solve for the unknown variable.

Complete step-by-step solution:
For the given function in the question, $f(x) = \left\{ {\begin{array}{*{20}{c}}
  {k - 2x} \\
  {2x + 3}
\end{array}} \right.$ if $\begin{array}{*{20}{c}}
  {x \leqslant - 1} \\
  {x > - 1}
\end{array}$ , where the range and domain of the function $f$ is $f:R \to R$ , the value of the function when $x$ tends to the positive side of the critical point $- 1$ will be,
$\mathop {\lim }\limits_{x \to 1 + } f\left( x \right) = 1$
When the value of $x$ will be equal to $- 1$ , then the function will be $k - 2x$ and on substituting the value of $x$ we will get the function as,
As $f\left( { - 1} \right) = k + 2$
For local minima of a function, we can write that,
$f(a) \leqslant f(x)$ , for all $x$ in the interval and where $a$ is the point of local minima.
Now, $f$has a local minimum at $x = - 1$ and $f(x) = \left\{ {\begin{array}{*{20}{c}}
  {k - 2x} \\
  {2x + 3}
\end{array}} \right.$ if $\begin{array}{*{20}{c}}
  {x \leqslant - 1} \\
  {x > - 1}
\end{array}$
Taking limits on both sides of the critical point we get,
${\lim _{x \to - {1^2}}}k - 2x = {\lim _{x \to - {1^ + }}}2x + 3$
Substituting $x = - 1$ , we get,
$\Rightarrow$$k + 2 = - 2 + 3$
Adding $2$ on both sides of the equation to get,
$\Rightarrow$$k + 2 + 2 = - 2 + 3 + 2$
$\Rightarrow k + 4 = 3$
Simplifying further we get,
$\Rightarrow$$k = - 1$
Therefore, a possible value of $k$ for the function $f:R \to R$ defined by $f(x) = \left\{ {\begin{array}{*{20}{c}}
  {k - 2x} \\
  {2x + 3}
\end{array}} \right.$ if $\begin{array}{*{20}{c}}
  {x \leqslant - 1} \\
  {x > - 1}
\end{array}$
Will be $k = - 1$ .

Hence, the correct option will be D.

Note: The question can also be solved with the help of linear inequalities. The condition for the minima of the function can be expressed as, $f\left( {{1^ + }} \right) \geqslant f\left( { - 1} \right) \geqslant f\left( { - {1^ - }} \right)$ as $f( - 1) = k + 2$ . The inequality can further be simplified using basic rules of linear inequalities to get,
$\Rightarrow k + 2 \leqslant 1$
$\Rightarrow k \leqslant - 1$ . Therefore, again we get the possible answer as $k = - 1$ .