Question

If exactly one root of the quadratic equation $x^2-(a+1)x+2a=0$ lies in the interval $(0,3)$, then the set of values $a$ is given by

• A. $(-\infty ,0)\cup (6,\infty )$
• B. $(-\infty ,0]\cup (6,\infty )$
• C. $(-\infty ,0]\cup [6,\infty )$
• D. $(0,6)$

Answer 1

The correct option is B $(-\infty ,0]\cup (6,\infty )$

For the equation $x^2-(a+1)x+2a=0$ to have exactly one root lying between the interval $(0,3)$, the following conditions should be satisfied,
(i)$f\left(0\right)\times f\left(3\right)<0$
$\Rightarrow \left(2a\right)(9-3(a+1)+2a)<0$
$\Rightarrow a(6-a)<0$
$\Rightarrow a(a-6)>0$
$\Rightarrow a<0$ and $a>6$
(ii) Discriminant, $D\ge 0$
$\Rightarrow (a+1{)}^2-8a\ge 0$
$\Rightarrow a^2-6a+1\ge 0$
$\Rightarrow (a-3{)}^2-8\ge 0$
$\Rightarrow (a-(3+2\sqrt{2}\left)\right)(a-(3-2\sqrt{2}\left)\right)\ge 0$
$\Rightarrow a\ge (3+2\sqrt{2})$ and $a\le (3-2\sqrt{2})$
The part of intersection for both of these conditions gives the set of values of $a$
i.e. $a<0$ and $a>6$
Also we have to check for the end points.
For $a=0$ we get the roots to be $0$ and $1$. Hence $0$ is also possible. But for $a=6$ we get the roots to be $3$ and $4$, which does not lies in $(0,3)$.
Hence, option 'B' is correct.