JEE Mathematics 2025 Question with Solution

Given: The region is defined by $1 + x^2 \leq y \leq \min(x + 7, 11 - 3x)$.

Find: The value of $3A$, where $A$ is the area of the region.

First identify which line gives the upper boundary. The two lines intersect at

so

and hence

Therefore, for $x < 1$ the minimum is $x + 7$, and for $x > 1$ the minimum is $11 - 3x$.

Now find where the parabola meets each line in the relevant interval.

With $y = x + 7$,

so the roots are $x = 3$ and $x = -2$. Since this part uses $x < 1$, the valid point is $x = -2$.

With $y = 11 - 3x$,

so the roots are $x = -5$ and $x = 2$. Since this part uses $x > 1$, the valid point is $x = 2$.

Hence the area must be split into two parts:

where

and

For the first part,

Now,

and

Therefore,

For the second part,

Now,

and

Therefore,

So the total area is

Hence,

Therefore, the correct option is A.

The first solution shown in the source contains an arithmetic inconsistency; the second solution correctly gives $3A = 50$.

Conclusion: The value of $3A$ is $50$.

Given: $y$ lies above the parabola $y = 1 + x^2$ and below the function $\min(x + 7, 11 - 3x)$.

Find: The required area expression and then compute $3A$.

Because the upper boundary is a minimum of two lines, we must use the lower of the two lines at each value of $x$. Their intersection occurs at $x = 1$, so the upper boundary changes there. This is why the area is not found using a single integral over the whole interval.

The admissible region starts where the parabola first meets the active upper boundary on the left, namely at $x = -2$, and ends where it meets the active upper boundary on the right, namely at $x = 2$. Thus the region exists over

but with a change of upper curve at

Therefore,

Evaluating gives

Hence,

So the correct option is A.