JEE Main 2025 Apr 3 Shift 2, Mathematics Q7: Quadratic Equations

JEE Main2025Apr 3, Shift 2Mathematics

Let the equation $x (x + 2) (12 - k) = 2$ have equal roots. Then the distance of the point $(k, \frac{k}{2})$ from the line $3 x + 4 y + 5 = 0$ is

Solution

Let $λ = 12 - k$ (assume $k \neq = 12$ ). The equation becomes

$λ (x^{2} + 2 x) = 2 \Rightarrow λ x^{2} + 2 λ x - 2 = 0$ .

For equal roots, discriminant is zero:

$(2 λ)^{2} - 4 \cdot λ \cdot (- 2) = 0 \Rightarrow 4 λ^{2} + 8 λ = 0 \Rightarrow 4 λ (λ + 2) = 0$ .

So $λ = 0$ or $λ = - 2$ . The choice $λ = 0$ degenerates the equation, so take $λ = - 2$ , i.e. $12 - k = - 2 \Rightarrow k = 14$ .

Then the point is $(14, 7)$ . Distance from $3 x + 4 y + 5 = 0$ :

$d = \frac{∣3 \cdot 14 + 4 \cdot 7 + 5∣}{9 + 16} = \frac{∣42 + 28 + 5∣}{5} = \frac{75}{5} = 15$ .

Concept behind this question

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