Question

For the quadratic equation ${{x}^{2}}-\left( m-3 \right)x+m=0\left( m\in R \right)$, find the value of m for which at least one root lies in the interval (1,2).
(a) (0,10)
(b) (10, ∞)
(c) (-∞, 10)
(d) None of these

Answer 1

Hint: First, by using the determinant method, we form one equation for values of m. By using the interval given in the problem statement, we obtain another relation for values of m. Now, we combine both the obtained relations to get the final possible range. By using this methodology, we can easily solve the question.

Complete step-by-step answer:
According to the problem statement, ${{x}^{2}}-\left( m-3 \right)x+m=0$ must have at least one root lying between the interval (1, 2). This confirms that the determinant is greater than 0.
$\therefore D\ge 0$
For a quadratic expression, the determinant may be expressed as:
$D=\sqrt{{{b}^{2}}-4ac}$ for any equation $a{{x}^{2}}+bx+c=0$.
Now, applying it for ${{x}^{2}}-\left( m-3 \right)x+m=0$, we get
$\begin{align}
  & {{\left( m-3 \right)}^{2}}-4m\ge 0 \\
 & {{m}^{2}}-6m+9-4m\ge 0 \\
 & {{m}^{2}}-10m+9\ge 0 \\
\end{align}$
Now, by using the middle term splitting method, we factorize the above expression as:
$\begin{align}
  & {{m}^{2}}-m-9m+9\ge 0 \\
 & m(m-1)-9(m-1)\ge 0 \\
 & (m-9)(m-1)\ge 0 \\
\end{align}$
So, the possible values of m lie in the interval $m\in (-\infty ,1)\cup (9,\infty )\ldots (1)$.
Since, one root lie between the interval (1,2), it must satisfy$f(1)f(2)<0$.
Putting x = 1 in ${{x}^{2}}-\left( m-3 \right)x+m=0$ to evaluate the value of $f(1)$.
$\begin{align}
  & f(1)=1-(m-3)+m \\
 & f(1)=4 \\
\end{align}$
Putting x = 2 in ${{x}^{2}}-\left( m-3 \right)x+m=0$ to evaluate the value of $f(2)$.
$\begin{align}
  & f(2)=4-(m-3)(2)+m \\
 & f(2)=4-2m+6+m \\
 & f(2)=10-m \\
\end{align}$
Putting the obtained values in the obtained condition:
$\begin{align}
  & f(1)f(2)<0 \\
 & \Rightarrow 4\cdot (10-m)<0 \\
 & \Rightarrow m>10 \\
 & \Rightarrow m\in (10,\infty )\ldots (2) \\
\end{align}$
From equation (1) and (2), we obtain the possible values of m by taking the common values from both the equations.
\[\begin{align}
  & m\in (-\infty ,1)\cup (9,\infty )\text{ and }m\in (10,\infty ) \\
 & \therefore m\in (10,\infty ). \\
\end{align}\]
Therefore, option (b) is correct.

Note: The key concept involved in solving this problem is the knowledge of roots of quadratic equations under certain conditions. Students must be very careful while analysing the problem statement to form the correct relation. In the last step, students must take the intersection of both the obtained equations to find possible values of m.