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Question: Let \( f:R\to R \) be defined by \( f\left( x \right) = \left\\{ {\begin{array}{*{20}{l}} {k - 2...

Let f:R→Rf: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=−1x = - 1 , then a possible value of k is
A. 0
B. −12- \dfrac{1}{2}
C. -1
D. 1

Explanation

Solution

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=−1x = - 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→Rf: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=−1x = - 1 . We have to find the possible value of k.
When x is less than or equal to -1, f is defined as f(x)=k−2xf\left( x \right) = k - 2x and when x is greater than -1, f is defined as f(x)=2x+3f\left( x \right) = 2x + 3 .
And we are also given that the function f has a local minimum at x=−1x = - 1 . This means at x=−1x = - 1 , both the given definitions of f give the same result.
This gives us, k−2x=2x+3k - 2x = 2x + 3
On substituting x=−1x = - 1 in the above equation, we get
⇒k−2(−1)=2(−1)+3\Rightarrow k - 2\left( { - 1} \right) = 2\left( { - 1} \right) + 3
⇒k+2=−2+3\Rightarrow k + 2 = - 2 + 3
∴k=1−2=−1\therefore k = 1 - 2 = - 1
Therefore, the possible value of k is −1-1.
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,
lim⁡x⇒1−f(x)=lim⁡x⇒1−(k−2x)\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,
lim⁡x⇒1+f(x)=lim⁡x⇒1+(2x+3)\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=−1x = - 1 ,
lim⁡x⇒1−(k−2x)=lim⁡x⇒1+(2x+3)\mathop {\lim }\limits_{x\Rightarrow {1^ - }} \left( {k - 2x} \right) = \mathop {\lim }\limits_{x\Rightarrow {1^ + }} \left( {2x + 3} \right)
⇒k−2(−1)=2(−1)+3\Rightarrow k - 2\left( { - 1} \right) = 2\left( { - 1} \right) + 3
⇒k+2=1⇒k=−1\Rightarrow k + 2 = 1\Rightarrow k = - 1