JEE Mathematics 2025 Question with Solution

Given: $x(x+2)(12-k) = 2$ has equal roots.

Find: The distance of the point $\left(k, \frac{k}{2}\right)$ from the line $3x + 4y + 5 = 0$.

Rewrite the equation as a quadratic in $x$:

$$(12-k)x^2 + 2(12-k)x - 2 = 0$$

So the coefficients are:

$$a = 12-k, \qquad b = 2(12-k), \qquad c = -2$$

For equal roots, the discriminant must be zero:

$$\Delta = b^2 - 4ac = 0$$

Substituting the coefficients:

$$(2(12-k))^2 - 4(12-k)(-2) = 0$$ $$4(12-k)^2 + 8(12-k) = 0$$ $$4(12-k)(14-k) = 0$$

Thus,

$$12-k = 0 \quad \text{or} \quad 14-k = 0$$

So $k = 12$ or $k = 14$.

From the second approach, the quadratic form is taken for $k \ne 12$, so the valid value used is:

$$k = 14$$

Hence the point is:

$$\left(k, \frac{k}{2}\right) = \left(14, 7\right)$$

Now use the distance formula from point $\left(x_1,y_1\right)$ to line $Ax+By+C=0$:

$$D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Substitute $A=3$, $B=4$, $C=5$, $x_1=14$, $y_1=7$:

$$D = \frac{|3\cdot 14 + 4\cdot 7 + 5|}{\sqrt{3^2 + 4^2}}$$ $$D = \frac{|42 + 28 + 5|}{\sqrt{25}} = \frac{75}{5} = 15$$

Therefore, the distance is $15$ and the correct option is A.

Given: $x(x+2)(12-k) = 2$ has equal roots.

Find: The distance of $\left(k, \frac{k}{2}\right)$ from $3x + 4y + 5 = 0$.

Let

$$\lambda = 12-k$$

Then the equation becomes:

$$(x^2+2x)\lambda = 2$$ $$\lambda x^2 + 2\lambda x - 2 = 0$$

For equal roots, discriminant must be zero:

$$D = (2\lambda)^2 - 4(\lambda)(-2) = 0$$ $$4\lambda^2 + 8\lambda = 0$$ $$4\lambda(\lambda+2) = 0$$

So,

$$\lambda = 0 \quad \text{or} \quad \lambda = -2$$

Using the valid quadratic case from the extracted solution, $\lambda = -2$, hence:

$$12-k = -2$$ $$k = 14$$

Therefore,

$$P = \left(k, \frac{k}{2}\right) = \left(14, 7\right)$$

Distance from $3x+4y+5=0$ is:

$$D = \frac{|3(14)+4(7)+5|}{\sqrt{3^2+4^2}}$$ $$D = \frac{|42+28+5|}{5} = \frac{75}{5} = 15$$

Therefore, the correct option is A.

Note: The first extracted solution contains an internal inconsistency when discussing the point corresponding to $k=14$, but it still concludes with distance $15$, which matches option A.