JEE Mathematics 2025 Question with Solution

Given: The region is defined by

$$2y \leq x^2 + 3$$ $$y + |x| \leq 3$$ $$y \geq |x-1|$$

Find: The value of $6A$, where $A$ is the area of the region.

From the inequalities,

$$y \leq \frac{x^2}{2} + \frac{3}{2}, \qquad y \leq 3 - |x|, \qquad y \geq |x-1|$$

The solution notes that the parabola and the lines $y = x-1$ and $y = 1-x$ do not intersect in real points, since

$$\frac{x^2}{2} + \frac{3}{2} = x-1$$

gives discriminant

$$\Delta = 4 - 10 = -6$$

and

$$\frac{x^2}{2} + \frac{3}{2} = 1-x$$

gives

$$\Delta = 1 - 10 = -9$$

So the bounded region is determined effectively by the linear constraints together with the stated inequalities.

Using the split form of the absolute values,

$$y + |x| \leq 3$$

becomes

$$y \leq 3-x \quad (x \geq 0), \qquad y \leq 3+x \quad (x < 0)$$

and

$$y \geq |x-1|$$

means the region lies above

$$y = x-1 \quad (x \geq 1), \qquad y = 1-x \quad (x < 1)$$

The extracted solution states that the area is computed by splitting the region into suitable $x$-intervals and evaluating the bounded part using the upper curve

$$y = \min\left(\frac{x^2}{2}+\frac{3}{2}, 3-|x|\right)$$

and the lower curve

$$y = |x-1|$$

According to the solution, the simplified integral gives

$$A = 2$$

Hence,

$$6A = 12$$

Therefore, the correct option is D.