Let the points (frac112,alpha)$(\frac{11}{2},\alpha)$ lie on or inside the \triangle with sides x+y=11$x+y=11$, x+2y=16$x+2y=16$ and 2x+3y=29$2x+3y=29$ Then the product of the smallest and the largest values of a is equal to: [cite: 3284, 3285, 3286, 3287, 3288, 3289]
A.22$22$
B.44$44$
C.33$33$
D.55$55$
Solution & Explanation
### Related Formula
For a vertical line x = x_0$x = x_0$ crossing a bounded region, the valid coordinates of y$y$ sit between the boundary lines intersecting that specific line vertical plane.
### Core Logic
The point given is fixed at x = frac112 = 5.5$x = \frac{11}{2} = 5.5$[cite: 3285, 3925]. We evaluate the values of y$y$ along this vertical line segment across each boundary edge.
Linear Programming region graph for Q58 - JEE Main 2025 Evening
### Step 1: Evaluate Intersections
Substitute x = frac112$x = \frac{11}{2}$ into the three linear constraints:
1. From x + y = 11$x + y = 11$:
frac112 + y = 11 Rightarrow y = 11 - 5.5 = 5.5$$\frac{11}{2} + y = 11 \Rightarrow y = 11 - 5.5 = 5.5$$
2. From x + 2y = 16$x + 2y = 16$:
frac112 + 2y = 16 Rightarrow 2y = 16 - 5.5 = 10.5 Rightarrow y = 5.25$$\frac{11}{2} + 2y = 16 \Rightarrow 2y = 16 - 5.5 = 10.5 \Rightarrow y = 5.25$$
3. From 2x + 3y = 29$2x + 3y = 29$:
2left(frac112right) + 3y = 29 Rightarrow 11 + 3y = 29 Rightarrow 3y = 18 Rightarrow y = 6$$2\left(\frac{11}{2}\right) + 3y = 29 \Rightarrow 11 + 3y = 29 \Rightarrow 3y = 18 \Rightarrow y = 6$$
### Step 2: Define Extrema and Multiply
Checking the internal region of the \triangle bounded by these lines [cite: 3286, 3287], the valid range for alpha$\alpha$ along the section line is delimited by y = 5.5$y = 5.5$ and y = 6$y = 6$:
alpha_min = frac112 = 5.5$$\alpha_{\min} = \frac{11}{2} = 5.5$$alpha_max = 6$$\alpha_{\max} = 6$$
The product of the limits is :
alpha_min cdot alpha_max = frac112 times 6 = 33$$\alpha_{\min} \cdot \alpha_{\max} = \frac{11}{2} \times 6 = 33$$
### Pattern Recognition
Instead of drawing full coordinate diagrams or computing all three vertex points, evaluating values directly at the fixed coordinate constraint x = 5.5$x = 5.5$ saves time during multi-line area problems.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Linear Inequalities
Class 11 Mathematics: Straight Lines
Keywords:#\triangle boundary inequality values#JEE Main 2025 Evening Q58#linear programming section line check#straight lines region optimization
More Linear Inequalities Previous-Year Questions — Page 10
Q6jee_main_2024_31_jan_morningComposition of Functions
If f(x) = frac4x + 36x - 4, x neq frac23$f(x) = \frac{4x + 3}{6x - 4}, x \neq \frac{2}{3}$ and (fof)(x) = g(x)$(fof)(x) = g(x)$, where g : mathbbR - left\frac23right\ to mathbbR - left\frac23right\$g : \mathbb{R} - \left\{\frac{2}{3}\right\} \to \mathbb{R} - \left\{\frac{2}{3}\right\}$, then (gogog)(4)$(gogog)(4)$ is equal to
Let A = \1, 2, 3, 4\$A = \{1, 2, 3, 4\}$ and R = \(1, 2), (2, 3), (1, 4)\$R = \{(1, 2), (2, 3), (1, 4)\}$ be a relation on A$A$. Let S$S$ be the equivalence relation on A$A$ such that R subset S$R \subset S$ and the number of elements in S$S$ is n$n$. Then, the minimum value of n$n$ is
Numerical Answer.Answer: 16 to 16
Solution
### Core Logic
S$S$ must be reflexive, symmetric, and transitive, containing (1,2)$(1,2)$, (2,3)$(2,3)$, and (1,4)$(1,4)$.
Symmetric property forces (2,1), (3,2), (4,1) in S$(2,1), (3,2), (4,1) \in S$.
Transitive property: (1,2)$(1,2)$ and (2,3) implies (1,3) in S$(2,3) \implies (1,3) \in S$. Symmetric implies (3,1) in S$\implies (3,1) \in S$.
(4,1)$(4,1)$ and (1,2) implies (4,2) in S$(1,2) \implies (4,2) \in S$. Symmetric implies (2,4) in S$\implies (2,4) \in S$.
(4,1)$(4,1)$ and (1,3) implies (4,3) in S$(1,3) \implies (4,3) \in S$. Symmetric implies (3,4) in S$\implies (3,4) \in S$.
### Step 1: Universal Relation
Since 1 is related to 2, 3, 4 and the relation is an equivalence relation (which creates partitions), all elements 1, 2, 3, and 4 must fall into the same single equivalence class.
Thus, S$S$ must contain all possible ordered pairs in A times A$A \times A$.
### Step 2: Final Count
Number of elements in A times A = 4 times 4 = 16$A \times A = 4 \times 4 = 16$.
Minimum value of n$n$ is 16$16$.
### Pattern Recognition
If a relation connects all elements in a set to each other through a chain, its equivalence closure is the universal relation A times A$A \times A$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Relations and Functions
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