CUET · MATHS · PYQ PAPER 2025

Match List - I with List - II
List - I List - II
(A) If vectors \(\vec{a}\) and \(\vec{b}\) are such that \(\vec{a}\) = \(\lambda \vec{b}\) and \(|\vec{a}|=|\vec{b}|\), then (I) \(\vec{a}\) and \(\vec{b}\) are orthogonal
(B) Projection vector of \(\vec{a}\) on \(\vec{b}\) (II) [0, 12]
(C) If \(\vec{a}\) and \(\vec{b}\) are non-zero vectors such that \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\), then (III) \(\vec{a}= \pm \vec{b}\)
(D) If \(|\vec{a}|=4\) and \(-3 \leq \lambda \leq 2\), then the range of \(|\lambda \vec{a}|\) (IV) \(\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}\)
Choose the correct answer from the options given below:

A (А) - (II), (В) - (III), (C) - (IV), (D) - (I)
B (A) - (I), (B) - (IV), (C) - (II), (D) - (III)
C (А) - (III), (В) - (IV), (C) - (I), (D) - (II)
D (A) - (IV), (B) - (III), (С) - (II), (D) - (I)

Verified Solution

Answer & Solution

Correct Answer

(C) (А) - (III), (В) - (IV), (C) - (I), (D) - (II)

Step-by-step Solution

Detailed explanation

For (A): \(\vec{a}=\lambda\vec{b}, |\vec{a}|=|\vec{b}|\) \(|\vec{a}|=|\lambda||\vec{b}| \Rightarrow |\vec{b}|=|\lambda||\vec{b}| \Rightarrow |\lambda|=1 \Rightarrow \vec{a}=\pm\vec{b}\) Match: (A) - (III) For (B): Projection vector of \(\vec{a}\) on \(\vec{b}\):…

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. \(\left( \frac{dy}{dx} \right)^2 + \frac{dy}{dx} + 5 = 0\)	I. order 3, degree 1
B. \(xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0\)	II. order 2, degree 2
C. \(\left( \frac{d^2y}{dx^2} \right)^2 + \frac{dy}{dx} = 0\)	III. order 2, degree 1
D. \(\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0\)	IV. order 1, degree 2

Choose the correct answer from the options given below:

CUET 2023 Hard

If A is a square matrix and I is the identity matrix of same order such that \(A^2=I\), then \((A-I)^3+(A+I)^3-3 A\) is equal to:

CUET 2025 Easy

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List - I	List - II
(A) If vectors \(\vec{a}\) and \(\vec{b}\) are such that \(\vec{a}\) = \(\lambda \vec{b}\) and \(\|\vec{a}\|=\|\vec{b}\|\), then	(I) \(\vec{a}\) and \(\vec{b}\) are orthogonal
(B) Projection vector of \(\vec{a}\) on \(\vec{b}\)	(II) [0, 12]
(C) If \(\vec{a}\) and \(\vec{b}\) are non-zero vectors such that \(\|\vec{a}+\vec{b}\|=\|\vec{a}-\vec{b}\|\), then	(III) \(\vec{a}= \pm \vec{b}\)
(D) If \(\|\vec{a}\|=4\) and \(-3 \leq \lambda \leq 2\), then the range of \(\|\lambda \vec{a}\|\)	(IV) \(\left(\frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|^2}\right) \vec{b}\)