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

Here, we will first find the equation in which the value of the function is less than 0 at point 1 and 3 of the interval given. Then we will solve the obtained equation to get the intervals. Finally, we will write all the values coming in between both the intervals and count them to get our desired answer.

Complete step by step solution:
We have to find the value of $k$ in equation
$5{x^2} + \left( {k + 1} \right)x + k = 0$…….$\left( 1 \right)$
This lies in the interval $\left( {1,3} \right)$.
So, first of all we will let the equation as $f\left( x \right)$.
$f\left( x \right) = 5{x^2} + \left( {k + 1} \right)x + k$
As the interval given is open so,
$f\left( 1 \right) < 0$….$\left( 2 \right)$
Substituting $x = 1$in equation $\left( 1 \right)$, we get
$f\left( 1 \right) = 5 \times {1^2} + \left( {k + 1} \right) \times 1 + k$
Multiplying the terms, we get
$\Rightarrow f\left( 1 \right) = 5 + k + 1 + k$
Adding the like term, we get
$\Rightarrow f\left( 1 \right) = 6 + 2k$
Substituting above value in equation $\left( 2 \right)$, we get
$\begin{array}{l}6 + 2k < 0\\\ \Rightarrow 2k < \- 6\end{array}$
Dividing both side by 2, we get
$\therefore k<-3$…..$\left( 3 \right)$
Also,
$f\left( 3 \right) > 0$ …. $\left( 4 \right)$
Now, substituting $x = 3$ in equation $\left( 1 \right)$, we get
$f\left( 3 \right) = 5 \times {3^2} + \left( {k + 1} \right) \times 3 + k$
Simplifying the equation, we get
$\Rightarrow f\left( 3 \right) = 45 + 3k + 3 + k$
Adding the like terms, we get
$\Rightarrow f\left( 3 \right) = 48 + 4k$
Substituting above value in equation $\left( 4 \right)$, we get
$\begin{array}{l}48 + 4k > 0\\\ \Rightarrow 4k > - 48\end{array}$
Dividing both side by 4, we get
$\therefore k>-12$……$\left( 5 \right)$
Now from equation $\left( 3 \right)$ and $\left( 5 \right)$, we get
$- 12 < k < \- 3$
So the value of $k = - 11, - 10, - 9, - 8, - 7, - 6, - 5, - 4$
So, we get that for 8 integral values the given equation has only one root.

Hence, the option (D) is correct.

Note:
Inequality is used to show the non-equal comparison between two mathematical expressions. Mostly it is used to compare two numbers on a number line by their sizes. An algebraic inequality such as $y > 6$ can be read as “$y$ is greater than 6” in this case $y$ can have many infinite values and its upper case is not given. An integral value is either a numerical value or a function for some interval. It can be two or three dimensional.