JEE Mathematics 2025 Question with Solution

Given: The point is $\left( \frac{11}{2}, \alpha \right)$ and the triangle is formed by the lines $x + y = 11$, $x + 2y = 16$, and $2x + 3y = 29$.

Find: The product of the smallest and largest values of $\alpha$ for which the point lies on or inside the triangle.

First find the vertices of the triangle.

Intersection of $x + y = 11$ and $x + 2y = 16$:

$$\begin{aligned} (x+y) - (x+2y) &= 11 - 16 \\ -y &= -5 \\ y &= 5 \end{aligned}$$

Then

$$x = 11 - 5 = 6$$

So one vertex is $A(6,5)$.

Intersection of $x + y = 11$ and $2x + 3y = 29$:

$$y = 11 - x$$

Substitute into the third line:

$$\begin{aligned} 2x + 3(11-x) &= 29 \\ 2x + 33 - 3x &= 29 \\ -x &= -4 \\ x &= 4 \end{aligned}$$

Hence

$$y = 11 - 4 = 7$$

So another vertex is $B(4,7)$.

Intersection of $x + 2y = 16$ and $2x + 3y = 29$:

$$x = 16 - 2y$$

Substitute into the third line:

$$\begin{aligned} 2(16-2y) + 3y &= 29 \\ 32 - 4y + 3y &= 29 \\ -y &= -3 \\ y &= 3 \end{aligned}$$

Then

$$x = 16 - 2(3) = 10$$

So the third vertex is $C(10,3)$.

Now consider the vertical line

$$x = \frac{11}{2}$$

At this $x$-value, the corresponding $y$-values on the three sides are:

$$\text{On } x+y=11, \quad y = 11 - \frac{11}{2} = \frac{11}{2}$$ $$\text{On } x+2y=16, \quad y = \frac{16 - \frac{11}{2}}{2} = \frac{21}{4}$$ $$\text{On } 2x+3y=29, \quad y = \frac{29 - 11}{3} = 6$$

The line $x = \frac{11}{2}$ cuts the triangle between the sides $x+y=11$ and $2x+3y=29$, so $\alpha$ ranges from

$$\frac{11}{2} \le \alpha \le 6$$

Therefore, the smallest value of $\alpha$ is $\frac{11}{2}$ and the largest value is $6$. Their product is

$$\frac{11}{2} \times 6 = 33$$

Therefore, the correct option is B.

For the fixed point $\left( \frac{11}{2}, \alpha \right)$, only the vertical line $x = \frac{11}{2}$ matters.

A point on or inside the triangle must lie between the two sides of the triangle intersected by this vertical line. Evaluating the side equations at $x = \frac{11}{2}$ gives:

$$y = 11 - x = \frac{11}{2}$$ $$y = \frac{16-x}{2} = \frac{21}{4}$$ $$y = \frac{29-2x}{3} = 6$$

Among these, the actual segment inside the triangle is from $y = \frac{11}{2}$ to $y = 6$. Thus

$$\alpha_{\min} = \frac{11}{2}, \qquad \alpha_{\max} = 6$$

Hence

$$\alpha_{\min} \alpha_{\max} = \frac{11}{2} \cdot 6 = 33$$

So the answer is B.