Question

Number of integral values of $ k $ for which exactly one root of the equation $ 5{{x}^{2}}+\left( k+1 \right)x+k=0 $ lies in the interval $ \left( 1,3 \right) $ is
(a) 5
(b) 6
(c) 7
(d) 8

Answer 1

Hint:
 We will understand the meaning of the root of the equation lying in a certain interval. The value of the equation at the endpoints will be less than zero and greater than zero. So we will find the value of the function represented by the equation at the endpoints of this interval. This will give us a range of values for $ k $. We will count these integral values to obtain the answer.

Complete step by step answer:
The given equation is $ f\left( x \right)=5{{x}^{2}}+\left( k+1 \right)x+k=0 $ . Exactly one root of the given equation lies in the interval $ \left( 1,3 \right) $ . This means that the value of the function is equal to zero for some $ x\in \left( 1,3 \right) $ . Therefore, the value of the function at $ x=1 $ will be less than zero and the value of the function at $ x=3 $ will be greater than zero.
So, we have $ f\left( 1 \right)<0 $ and $ f\left( 3 \right)>0 $ . Let us find the value of the function at the end points of the given interval. At $ x=1 $ , we have the following,
 $ \begin{align}
  & f\left( 1 \right)=5{{\left( 1 \right)}^{2}}+\left( k+1 \right)\cdot 1+k \\
 & \Rightarrow f\left( 1 \right)=5+k+1+k \\
 & \therefore f\left( 1 \right)=6+2k \\
\end{align} $
So, we get the following bound for the value of $ k $ ,
 $ \begin{align}
  & 6+2k<0 \\
 & \Rightarrow 2k<-6 \\
 & \therefore k<-3 \\
\end{align} $
Next, we will find the value of the function at $ x=3 $ in the following manner,
 $ \begin{align}
  & f\left( 3 \right)=5{{\left( 3 \right)}^{2}}+\left( k+1 \right)\cdot 3+k \\
 & \Rightarrow f\left( 3 \right)=5\times 9+3k+3+k \\
 & \therefore f\left( 3 \right)=48+4k \\
\end{align} $
Therefore, we get the following bound for the value of $ k $ ,
 $ \begin{align}
  & 48+4k>0 \\
 & \Rightarrow 4k>-48 \\
 & \therefore k>-12 \\
\end{align} $
This implies that the values of $ k $ are in the range $ -12 < k < -3 $ . There are 8 integral values that $ k $ can take. Hence, the correct option is (d).

Note:
It is important to understand the meaning of a root of an equation. Since exactly one root of the given equation was in the given interval, we could compare the values of the function at the endpoints of the interval to zero. Another fact that supports this assumption is that we are given a quadratic equation. We know that the graph of a quadratic equation is always a parabola.