Question

The set of values of $m$ for which $f\left(x\right)=x^2-(m-3)x+m$ intersects the positive direction of $x-$axis atleast once, is

• A. $(-\infty ,0)\cup (3,\infty )$
• B. $(-\infty ,1]\cup (9,\infty )$
• C. $(-\infty ,0]\cup (3,\infty )$
• D. $(-\infty ,0)\cup [9,\infty )$

Answer 1

The correct option is D $(-\infty ,0)\cup [9,\infty )$

The parabola $f\left(x\right)=x^2-(m-3)x+m=0$ intersects the $x-$axis atleast once, so the roots are real
Now,
Case $1:$ Both the roots are positve,
$\left(i\right)D\ge 0\Rightarrow (m-3{)}^2-4m\ge 0\Rightarrow m^2-10m+9\ge 0\Rightarrow (m-9\left)\right(m-1)\ge 0\Rightarrow x\in (-\infty ,1]\cup [9,\infty \left)\right(ii\left)f\right(0)>0\Rightarrow m>0(iii)-\frac{b}{2a}>0\Rightarrow \frac{m-3}{2}>0\Rightarrow m>3\Rightarrow m\in (3,\infty )\therefore m\in [9,\infty )\cdots (1)$

Case $2:$ One positive and one negative root.
So,
$f\left(0\right)<0$
$\Rightarrow m<0\Rightarrow m\in (-\infty ,0)\cdots \cdots \left(2\right)$

Case $3:$ One roots is $0$ and other is positve.
So,
$\left(i\right)f\left(0\right)=0\Rightarrow m=0\left(ii\right)-\frac{b}{2a}>0\Rightarrow \frac{m-3}{2}>0\Rightarrow m>3$
$\therefore m\in \varphi$

Now, using the above equations $\left(1\right)$ and $\left(2\right)$,
$\therefore m\in (-\infty ,0)\cup [9,\infty )$