JEE Main 2025 Jan 24 Shift 2, Mathematics Q8: Points and Straight Lines

JEE Main2025Jan 24, Shift 2Mathematics

Let the points $(\frac{11}{2}, α)$ lie on or inside the triangle with sides $x + y = 11$ , $x + 2 y = 16$ and $2 x + 3 y = 29$ . Then the product of the smallest and the largest values of $α$ is equal to :

A
$22$
B
$44$
C
$33$
D
$55$

Solution

The vertical line $x = \frac{11}{2}$ crosses each of the three sides. The portion of this vertical line that lies inside the triangle is bounded above and below by two of the three sides, and the extreme values of $α$ occur at those intersections.

Find $y$ on the line $x = 11/2$ for each side:

On $x + y = 11$ : $y = 11 - 11/2 = 11/2$
On $x + 2 y = 16$ : $2 y = 16 - 11/2 = 21/2 \Rightarrow y = 21/4$
On $2 x + 3 y = 29$ : $3 y = 29 - 11 = 18 \Rightarrow y = 6$

The three candidate ordinates are $11/2 = 5.5$ , $21/4 = 5.25$ , $6$ .

By checking which two of these bound the interior of the triangle (the smallest and largest values that still lie inside), the extremes work out to $α_{m i n} = \frac{11}{2}$ and $α_{m a x} = 6$ .

$α_{m i n} \cdot α_{m a x} = \frac{11}{2} \times 6 = 33$

Correct option: (3)

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JEE Main 2025 Jan 24 Shift 2, Mathematics Q8: Points and Straight Lines

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