Q45 / 75 · JEE Main 2025 (3 Apr Shift 2) · Mathematics

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

  • Correct answer
    A
    $15$
  • B
    $5 \sqrt{3}$
  • C
    $15 \sqrt{5}$
  • D
    $12$

Solution

$\begin{aligned}<br/>& \left(\mathrm{x}^2+2 \mathrm{x}\right)(12-\mathrm{k})=2 \\ & \lambda \mathrm{x}^2+2 \lambda \mathrm{x}-2=0 \quad \mathrm{k} \neq 12 \text { Let } 12-\mathrm{k}=\lambda \\ & \mathrm{D}=0 \\ & 4 \lambda^2+8 \lambda=0 \\ & \lambda=0 \text { or } \lambda=-2 \\ & \Rightarrow 12-\mathrm{k}=-2 \\ & \mathrm{k}=14<br/>\end{aligned}$
$\text { So } P\left(k, \frac{k}{2}\right)=(14,7)$
$\mathrm{d}=\left|\frac{3 \times 14+4 \times 7+5}{5}\right|=15$
option (1)

Hence, the correct answer is option: 'A'.

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