Question

Set of values of $a$ for which both the roots of the quadratic polynomial $f\left(x\right)=a{x}^2+(a-3)x+1$ lie on one side of the $y-$axis is

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

Answer 1

The correct option is D $(0,1]\cup [9,\infty )$

For the given polynomial, $a\ne 0$
$\because$ Roots are only on one side,
$\therefore$ Both roots either be $+ve$ or $-ve$

For both the roots to be real,
$D\ge 0\Rightarrow (a-3{)}^2-4a\ge 0\Rightarrow a^2-10a+9\ge 0\Rightarrow a\in (-\infty ,1]\cup [9,\infty )-\{0\}\cdots (1)$

Now,
Case $1:$ Both roots are positive
Product and sum of the roots both will be $>0$
$\therefore -\frac{a-3}{a}>0,\frac{1}{a}>0\Rightarrow \frac{a-3}{a}<0,a\in (0,\infty )\Rightarrow a\in (0,3),a\in (0,\infty )\Rightarrow a\in (0,3)\cdots \left(2\right)$

Case $2:$ Both roots are negative
Product of the root will be positive and sum will be negative.
$\therefore \frac{1}{a}>0,-\frac{a-3}{a}<0\Rightarrow a\in (0,\infty ),\frac{a-3}{a}>0\Rightarrow a\in (3,\infty )\cdots \left(3\right)$

From above equations, $a\in (0,1]\cup [9,\infty )$