JEE PYQ: Quadratic Equation Question 49
Question 49 - 2019 (12 Jan Shift 1)
The number of integral values of $m$ for which the quadratic expression, $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$, $x \in R$, is always positive, is:
(1) 3
(2) 8
(3) 7
(4) 6
Show Answer
Answer: (3)
Solution
Let the given quadratic expression $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$ is positive for all $x \in R$, then
$1 + 2m > 0$ …(1)
$D < 0$
$\Rightarrow 4(1 + 3m)^2 - 4(1 + 2m) \cdot 4(1 + m) < 0$
$\Rightarrow 1 + 9m^2 + 6m - 4[1 + 2m^2 + 3m] < 0$
$\Rightarrow m^2 - 6m - 3 < 0$
$\Rightarrow m \in (3 - 2\sqrt{3}, 3 + 2\sqrt{3})$
From (1): $m > -\frac{1}{2}$
$m \in (3 - 2\sqrt{3}, 3 + 2\sqrt{3})$
Then, integral values of $m = {0, 1, 2, 3, 4, 5, 6}$
Hence, number of integral values of $m = 7$
BETA
