CUET · MATHS · PYQ PAPER 2025

For a linear programming problem, the feasible region is shown in the figure by shaded portion, then linear constraints are

A \(3 x+4 y \leq 24, x+2 y \leq 10, x \geq 0, y \geq 0\)
B \(3 x+4 y \geq 24, x+2 y \geq 10, x \geq 0, y \geq 0\)
C \(3 x+4 y \geq 24, x+2 y \leq 10, x \geq 0, y \geq 0\)
D \(3 x+4 y \leq 24, x+2 y \geq 10, x \geq 0, y \geq 0\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(3 x+4 y \leq 24, x+2 y \leq 10, x \geq 0, y \geq 0\)

Step-by-step Solution

Detailed explanation

\(3x + 4y \leq 24\) \(x + 2y \geq 10\) \(x \geq 0, y \geq 0\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match List I with List II

LIST I	LIST II
A. If \(A^{-1}=\left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right]\) then \(A\) is	I. \(\left[\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right]\)
B. If \(A=\left[\begin{array}{cc}2 & -1 \\ 3 & 1\end{array}\right]\) then \(A A^T\) is	II. \(\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\)
C. If \(A^{-1}=\left[\begin{array}{ll}5 & 7 \\ 2 & 3\end{array}\right]\) then \(\left(A^T\right)^{-1}\) is	III. \(\left[\begin{array}{cc}-3 & 2 \\ \frac{5}{2} & -\frac{3}{2}\end{array}\right]\)
D. If \(A=\left[a_{i j}\right]_{2 \times 2}\) where \(a_{i j}=(i-j)^2\), then \(A\) is	IV. \(\left[\begin{array}{cc}5 & 5 \\ 5 & 10\end{array}\right]\)

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