Question

The least value of k for which the function \[{{x}^{2}}+kx+1\] is an increasing function in the interval \[1 < x < 2\] is
\[\left( a \right)-4\]
\[\left( b \right)-3\]
\[\left( c \right)-1\]
\[\left( d \right)-2\]

Answer 1

Hint: The question deals with finding the value of k by doing the first derivative test. We use the first derivative test for checking whether the function is increasing or not. And it is given that the function is increasing in interval \[1 < x < 2\].

Complete step-by-step answer:
The given function \[{{x}^{2}}+kx+1\]is an increasing function in the interval \[1 < x < 2\] , so we do the first derivative test.
Let the given function be y, so,
\[y={{x}^{2}}+kx+1\]
Now, for first derivative test, we differentiate y with respect to x,
\[\dfrac{dy}{dx}=2x+k\]
For the given function to be increasing , \[\dfrac{dy}{dx}>0\]
So, \[2x+k>0\]
\[2x>-k\]
\[x>-\dfrac{k}{2}\]
We know that \[\text{1So we consider value of x to be the endpoints of the interval \[\left( 1,2 \right)\],
First taking value of x as 1
\[1>-\dfrac{k}{2}\]
Multiplying both sides by 2, we get,
\[2>-k\]
Multiplying both sides by (-1), so the greater than sign will changes to lesser than
\[\text{-2Next, we take x as 2
\[2>-\dfrac{k}{2}\]
Multiplying both sides by 2, we get,
\[4>-k\]
Multiplying both sides by (-1), so the greater than sign will changes to lesser than
\[-4 < k\] ---------(ii)
From, (i) and (ii) we conclude that \[k=-2\] as \[-4<-2\] so all the numbers which are greater than -2 will always be greater than -4 which implies least value of \[k=-2\] for which the given function \[{{x}^{2}}+kx+1\] is an increasing function in the interval \[1 < x < 2\].
Hence, option (d) is the correct answer.
So, the correct answer is “Option d”.

Note: Given problem deals with doing the first derivative test to find the least value of k for which the given function is increasing in the given interval. The main thing to keep in mind while doing this question is to remember that the first derivative of an increasing function is greater than 0. Also be careful while solving the inequality. Take care while doing the calculations.