JEE PYQ: Quadratic Equation Question 16
Question 16 - 2020 (02 Sep Shift 2)
Let $f(x)$ be a quadratic polynomial such that $f(-1) + f(2) = 0$. If one of the roots of $f(x) = 0$ is 3, then its other root lies in:
(1) $(-1, 0)$
(2) $(1, 3)$
(3) $(-3, -1)$
(4) $(0, 1)$
Show Answer
Answer: (1)
Solution
Let $f(x) = ax^2 + bx + c$
Given: $f(-1) + f(2) = 0$
$a - b + c + 4a + 2b + c = 0$
$\Rightarrow 5a + b + 2c = 0$ …(i)
and $f(3) = 0 \Rightarrow 9a + 3b + c = 0$ …(ii)
From equations (i) and (ii):
$\frac{a}{1-6} = \frac{b}{18-5} = \frac{c}{15-9} \Rightarrow \frac{a}{-5} = \frac{b}{13} = \frac{c}{6}$
Product of roots, $\alpha\beta = \frac{c}{a} = \frac{-6}{5}$ and $\alpha = 3$
$\Rightarrow \beta = \frac{-2}{5} \in (-1, 0)$
BETA
