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

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Straight Lines

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 :

Answer: (C)

The triangle is bounded by

L_{1} : x + y = 11

L_{2} : x + 2 y = 16

L_{3} : 2 x + 3 y = 29

. Find its vertices by pairwise intersection:

L_{1} \cap L_{2}

: subtracting gives

y = 5, x = 6

, so

(6, 5)

L_{1} \cap L_{3}

: from

L_{1}

x = 11 - y

; substituting,

2 (11 - y) + 3 y = 29 \Rightarrow y = 7, x = 4

, so

(4, 7)

L_{2} \cap L_{3}

: from

L_{2}

x = 16 - 2 y

; substituting,

2 (16 - 2 y) + 3 y = 29 \Rightarrow y = 3, x = 10

, so

(10, 3)

.

The vertices are

(6, 5), (4, 7), (10, 3)

, so the triangle's

x

-range is

[4, 10]

and

x = \frac{11}{2} = 5.5

lies strictly between the vertices

(4, 7)

and

(6, 5)

.

At

x = 5.5

, the triangle's boundary in this range consists of edge

L_{1}

(joining

(4, 7)

and

(6, 5)

) below, and edge

L_{3}

(joining

(4, 7)

and

(10, 3)

) above — check: at

x = 6

L_{3}

gives

y = \frac{29 - 12}{3} = 5.67

, above the vertex

(6, 5)

L_{1}

, confirming

L_{3}

is the upper edge here.

On

L_{1}

x = 5.5

y = 11 - 5.5 = 5.5

.
On

L_{3}

x = 5.5

2 (5.5) + 3 y = 29 \Rightarrow y = 6

.

So

α

ranges over

[5.5, 6]

, giving

α_{m i n} = \frac{11}{2}

and

α_{m a x} = 6

. Their product is

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

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option