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

Hint: ‘f’ is a function from real numbers to real numbers as R represents real numbers. The first R is for domain and the second R is for co-domain. Here when x is less than or equal to -1, f has one definition and when x is greater than -1, f has another definition. And at $ x = - 1 $ , these two given definitions of f will be equal. So equate the functions and substitute the value of x as -1 and find the value of k.

Complete step-by-step answer:
We are given that a function $ f:R\to R $ be defined by $ f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}
  {k - 2x,if\;x \leqslant - 1} \\
  {2x + 3,if\;x > - 1}
\end{array}} \right. $ and f has a local minimum at $ x = - 1 $ . We have to find the possible value of k.
When x is less than or equal to -1, f is defined as $ f\left( x \right) = k - 2x $ and when x is greater than -1, f is defined as $ f\left( x \right) = 2x + 3 $ .
And we are also given that the function f has a local minimum at $ x = - 1 $ . This means at $ x = - 1 $ , both the given definitions of f give the same result.
This gives us, $ k - 2x = 2x + 3 $
On substituting $ x = - 1 $ in the above equation, we get
 $ \Rightarrow k - 2\left( { - 1} \right) = 2\left( { - 1} \right) + 3 $
 $ \Rightarrow k + 2 = - 2 + 3 $
 $ \therefore k = 1 - 2 = - 1 $
Therefore, the possible value of k is $-1$.
So, the correct answer is “Option C”.

Note: We can also write the given function using limits and then solve for the value of k.
When x is less than or equal to -1,
$ \mathop {\lim }\limits_{x\Rightarrow {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x\Rightarrow {1^ - }} \left( {k - 2x} \right) $ as x approaches -1 and will never exceed it.
When x is greater than -1,
$ \mathop {\lim }\limits_{x\Rightarrow {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x\Rightarrow {1^ + }} \left( {2x + 3} \right) $ as x will start from -1 and will never fall behind it.
At $ x = - 1 $ ,
$ \mathop {\lim }\limits_{x\Rightarrow {1^ - }} \left( {k - 2x} \right) = \mathop {\lim }\limits_{x\Rightarrow {1^ + }} \left( {2x + 3} \right) $
 $ \Rightarrow k - 2\left( { - 1} \right) = 2\left( { - 1} \right) + 3 $
 $ \Rightarrow k + 2 = 1\Rightarrow k = - 1 $