
Let $f:R \to R$ be a function defined by $f(x) = \left\{ k - 2x,{\text{ if }}x \leqslant - 1 \\
2x + 3,{\text{ if }}x > - 1 \\
\right.$ , of $f$has a local minimum at $x = - 1$, then possible values of $k$is.
A. $1$
B. $0$
C. $ - \dfrac{1}{2}$
D. $ - 1$
Answer
232.8k+ views
Hint: Here, we have to find out the possible values of $k$. In this question we, have to take limits to find out the value of $k$. Firstly, we take the function $f(x) > - 1$ and take the limit of this function and find out the $f(x)$.
Secondly, we take the function $f(x) \leqslant - 1$and take a limit of the function and calculate the value of $f(x)$in terms of $k$.We are given the local minimum at $x = - 1$.Now these three conditions and find the value of .
Complete step by step solution:
We have;
$f(x) = \left\{
k - 2x,{\text{ if }}x \leqslant - 1 \\
2x + 3,{\text{ if }}x > - 1 \\
\right.$
Let us take the limit of a function $f(x) > - 1$ we get;
$
\mathop {\lim }\limits_{x \to - {1^ + }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} (2x + 3) \\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = 2( - 1) + 3 \\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = - 2 + 3 \\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = + 1 \\
$
Similarly, take the function with limit $f(x) \leqslant - 1$ we have,
$
\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ - }} (k - 2x) \\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ - }} f(x) = k - 2( - 1) \\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ - }} f(x) = k + 2 \\
$
Here we are having the local minimum of function $f(x)$at $x = - 1$.
Therefore, we can write the function as follows:
$f( - {1^ + }) \geqslant f( - 1) \geqslant f( - {1^ - })$
Putting all the values we get;
$1 \geqslant k + 2$
Subtract from both sides we get;
$
1 - 2 \geqslant k + 2 - 2 \\
\Rightarrow - 1 \geqslant k \\
\Rightarrow k \leqslant - 1 \\
$
Therefore, according to the local minimum of the function, we can take the possible value of $k$ is taken as$ - 1$.
Option ‘D’ is correct
Note: In this type of question, we must consider the conditions for each function we have given, and we must determine the limit of the given variable based on the conditions. Always keep an eye out for signs of inequity in the questions, such as
(less than equal to),
(greater than equal to),
(less than),
and (greater than). If we took 'x' conditions in a function. Always solve inequalities by adding, subtracting, multiplying, or dividing both sides until the variable is left alone. Do not attempt to solve it directly.
Secondly, we take the function $f(x) \leqslant - 1$and take a limit of the function and calculate the value of $f(x)$in terms of $k$.We are given the local minimum at $x = - 1$.Now these three conditions and find the value of .
Complete step by step solution:
We have;
$f(x) = \left\{
k - 2x,{\text{ if }}x \leqslant - 1 \\
2x + 3,{\text{ if }}x > - 1 \\
\right.$
Let us take the limit of a function $f(x) > - 1$ we get;
$
\mathop {\lim }\limits_{x \to - {1^ + }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} (2x + 3) \\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = 2( - 1) + 3 \\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = - 2 + 3 \\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = + 1 \\
$
Similarly, take the function with limit $f(x) \leqslant - 1$ we have,
$
\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ - }} (k - 2x) \\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ - }} f(x) = k - 2( - 1) \\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ - }} f(x) = k + 2 \\
$
Here we are having the local minimum of function $f(x)$at $x = - 1$.
Therefore, we can write the function as follows:
$f( - {1^ + }) \geqslant f( - 1) \geqslant f( - {1^ - })$
Putting all the values we get;
$1 \geqslant k + 2$
Subtract from both sides we get;
$
1 - 2 \geqslant k + 2 - 2 \\
\Rightarrow - 1 \geqslant k \\
\Rightarrow k \leqslant - 1 \\
$
Therefore, according to the local minimum of the function, we can take the possible value of $k$ is taken as$ - 1$.
Option ‘D’ is correct
Note: In this type of question, we must consider the conditions for each function we have given, and we must determine the limit of the given variable based on the conditions. Always keep an eye out for signs of inequity in the questions, such as
(less than equal to),
(greater than equal to),
(less than),
and (greater than). If we took 'x' conditions in a function. Always solve inequalities by adding, subtracting, multiplying, or dividing both sides until the variable is left alone. Do not attempt to solve it directly.Recently Updated Pages
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