JEE Main 2025 Apr 7 Shift 1, Mathematics Q14: Roots Lie In An Interval

JEE Main2025Apr 7, Shift 1Mathematics

Let the set of all values of $p \in R$ , for which both the roots of the equation $x^{2} - (p + 2) x + (2 p + 9) = 0$ are negative real numbers, be the interval $(α, β]$ . Then $β - 2 α$ is equal to

A
$0$
B
$9$
C
$5$
D
$20$

Solution

Apply the location-of-roots conditions for both roots negative:

(i) Discriminant $\geq 0$ :

$(p + 2)^{2} - 4 (2 p + 9) \geq 0$

$p^{2} + 4 p + 4 - 8 p - 36 \geq 0 \Rightarrow p^{2} - 4 p - 32 \geq 0 \Rightarrow (p - 8) (p + 4) \geq 0$

$\Rightarrow p \leq - 4$ or $p \geq 8$ .

(ii) Sum of roots $< 0$ : Sum $= p + 2 < 0 \Rightarrow p < - 2$ .

(iii) Product of roots $> 0$ : Product $= 2 p + 9 > 0 \Rightarrow p > - \frac{9}{2}$ .

Intersection of the three:

From (i): $p \leq - 4$ or $p \geq 8$
From (ii): $p < - 2$
From (iii): $p > - 9/2$

Combining: $- \frac{9}{2} < p \leq - 4$ , i.e. $p \in (- \frac{9}{2}, - 4]$ .

So $α = - \frac{9}{2}$ and $β = - 4$ .

$β - 2 α = - 4 - 2 (- \frac{9}{2}) = - 4 + 9 = 5$

$∴$ Option (3).

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JEE Main 2025 Apr 7 Shift 1, Mathematics Q14: Roots Lie In An Interval

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