Answer
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Hint: Here we need to find the minimum value of the given variable. The given function is increasing function. So we will first differentiate the given function and form an inequality such that its derivative will be greater than zero. From there, there we will get the relation between the two variables. We will then simplify it further to get the required minimum value of the given variable.
Complete step-by-step answer:
The given function is \[f\left( x \right) = {x^2} + kx + 1\].
It is given that the given function is monotonic increasing in \[\left[ {1,2} \right]\]. So, its derivative will be greater than zero.
Now, we will differentiate the given function with respect to \[x\] .
\[ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} + kx + 1} \right)\]
On differentiating each term, we get
\[ \Rightarrow f'\left( x \right) = 2x + k\]
As the given function is a monotonic increasing function.
Therefore,
\[ \Rightarrow f'\left( x \right) > 0\]
Now, we will substitute the value of \[f'\left( x \right)\] obtained in equation 1 here.
\[ \Rightarrow 2x + k > 0\]
Subtracting the term \[2x\] from both the sides, we get
\[ \Rightarrow 2x + k - 2x > 0 - 2x\]
\[ \Rightarrow k > - 2x\] ………. \[\left( 1 \right)\]
As it is given that:
\[ \Rightarrow 1 \le x \le 2\]
Now, we will multiply \[ - 2\] to the given inequality.
\[ \Rightarrow - 4 \ge - 2x \ge 2\]
On rewriting the inequality, we get
\[ \Rightarrow - 2 \le - 2x \le - 4\] …….. \[\left( 2 \right)\]
Now, we will compare inequality of equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\], we get
\[ \Rightarrow k \ge - 2\]
Hence, the minimum value of \[k\] is equal to \[ - 2\].
Hence, the correct option is option D.
Note: The given function is the monotonic increasing function. A monotonic increasing function is defined as a function whose value will increase when we increase the value of the variable \[x\] , here \[x\] includes the real number. We should remember that the derivative of the monotonic increasing function will always be greater than zero.
Complete step-by-step answer:
The given function is \[f\left( x \right) = {x^2} + kx + 1\].
It is given that the given function is monotonic increasing in \[\left[ {1,2} \right]\]. So, its derivative will be greater than zero.
Now, we will differentiate the given function with respect to \[x\] .
\[ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} + kx + 1} \right)\]
On differentiating each term, we get
\[ \Rightarrow f'\left( x \right) = 2x + k\]
As the given function is a monotonic increasing function.
Therefore,
\[ \Rightarrow f'\left( x \right) > 0\]
Now, we will substitute the value of \[f'\left( x \right)\] obtained in equation 1 here.
\[ \Rightarrow 2x + k > 0\]
Subtracting the term \[2x\] from both the sides, we get
\[ \Rightarrow 2x + k - 2x > 0 - 2x\]
\[ \Rightarrow k > - 2x\] ………. \[\left( 1 \right)\]
As it is given that:
\[ \Rightarrow 1 \le x \le 2\]
Now, we will multiply \[ - 2\] to the given inequality.
\[ \Rightarrow - 4 \ge - 2x \ge 2\]
On rewriting the inequality, we get
\[ \Rightarrow - 2 \le - 2x \le - 4\] …….. \[\left( 2 \right)\]
Now, we will compare inequality of equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\], we get
\[ \Rightarrow k \ge - 2\]
Hence, the minimum value of \[k\] is equal to \[ - 2\].
Hence, the correct option is option D.
Note: The given function is the monotonic increasing function. A monotonic increasing function is defined as a function whose value will increase when we increase the value of the variable \[x\] , here \[x\] includes the real number. We should remember that the derivative of the monotonic increasing function will always be greater than zero.
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