CUET UG 2025 — Mathematics Algebra

2025mcqmedium

For the linear programming problem (LPP):

Maximize $Z = x + 1.5y$ , subject to constraints,

$x + 2y \leq 40$ , $2x + y \leq 40$ , $x + y \leq 25$ , $x \geq 0$ , $y \geq 0$ .

Which of the following is NOT correct?

Official previous-year question

Held on 21 May 2025 · Verified 13 Jul 2026.

Options

A
The feasible region is bounded.
B
The corner points of the feasible region are $(0,0)$ , $(20,0)$ , $(15,10)$ , $(10,15)$ and $(0,20)$
C
The optimal value of the objective function is attained at the point $(15, 10)$ .
D
The LPP has a unique optimal solution.

Did you get this right?

Solution

For $f(x)$ to be continuous at $x = \pi/2$ , the left-hand limit must equal the right-hand limit.

From the left $(x \leq \pi/2)$ :

$\lim_{x \to \pi/2^-} f(x) = \sin\left(\frac{\pi}{2}\right) + n$

$= 1 + n$

From the right $(x \geq \pi/2)$ :

$\lim_{x \to \pi/2^+} f(x) = m \cdot \frac{\pi}{2} + 1$

$= \frac{m\pi}{2} + 1$

Setting the two limits equal for continuity:

$1 + n = \frac{m\pi}{2} + 1$

$n = \frac{m\pi}{2}$

Substituting into $\sin 2n$ :

$\sin(2n) = \sin\left(2 \cdot \frac{m\pi}{2}\right)$

$= \sin(m\pi)$

Since $m \in \mathbb{Z}$ , we have $\sin(m\pi) = 0$ for any integer $m$ .

Therefore, $\sin 2n = 0$

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Options

The feasible region is bounded.

The corner points of the feasible region are $(0,0)$ , $(20,0)$ , $(15,10)$ , $(10,15)$ and $(0,20)$

The optimal value of the objective function is attained at the point $(15, 10)$ .

The LPP has a unique optimal solution.

Solution

For $f(x)$ to be continuous at $x = \pi/2$ , the left-hand limit must equal the right-hand limit.

From the left $(x \leq \pi/2)$ :

$\lim_{x \to \pi/2^-} f(x) = \sin\left(\frac{\pi}{2}\right) + n$

$= 1 + n$

From the right $(x \geq \pi/2)$ :

$\lim_{x \to \pi/2^+} f(x) = m \cdot \frac{\pi}{2} + 1$

$= \frac{m\pi}{2} + 1$

Setting the two limits equal for continuity:

$1 + n = \frac{m\pi}{2} + 1$

$n = \frac{m\pi}{2}$

Substituting into $\sin 2n$ :

$\sin(2n) = \sin\left(2 \cdot \frac{m\pi}{2}\right)$

$= \sin(m\pi)$

Since $m \in \mathbb{Z}$ , we have $\sin(m\pi) = 0$ for any integer $m$ .

Therefore, $\sin 2n = 0$

CUET UG 2025 — Mathematics Algebra

Options

Did you get this right?

Solution

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CUET UG Algebra in other years

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Other Mathematics topics for CUET UG

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CUET UG 2025 — Mathematics Algebra

Options

Did you get this right?

Solution

Your note

CUET UG Algebra in other years

More CUET UG Algebra questions

Other Mathematics topics for CUET UG

Calculus

Applied-Mathematics

Statistics & Applications

Geometry

Arithmetic

Coordinate Geometry

Keep practicing Algebra