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CUET Previous Year Questions (PYQs)

CUET Rectangular Cartesian Coordinates PYQ

CUET PYQ
List I List II
A. Kailash Satyarthi I. Chemistry
B. Abhijit Banerjee II. Peace
C. Vinkatraman Ramakrishnan III. Physics
D. Subrahmanyan Chandrasekhar IV. Economics






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CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

List I List II
A. Kailash Satyarthi II. Peace
B. Abhijit Banerjee IV. Economics
C. Venkatraman Ramakrishnan I. Chemistry
D. Subrahmanyan Chandrasekhar III. Physics
CUET PYQ
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: If $a \ne b$ then $(ab) \ne (b,a)$.
Reason R: $(4,-3)$ lies in quadrant IV.

In the light of the above statements, choose the correct answer from the options given below:





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CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

Assertion A is true because ordered pairs depend on order.
Reason R is also true since $(+,-)$ lies in quadrant IV.
But R does not explain A.
CUET PYQ
If the vertices of a triangle are O(0,0), A(a,0) and B(0,a). Then, the distance between its circumcenter and orthocenter is:





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CUET Previous Year PYQ CUET CUET 2022 PYQ

Solution

CUET PYQ
The points $(K,2-2K)$, $(-K+1,2K)$ and $(-4-K,6-2K)$ are collinear if:

(A) $K=\frac{1}{2}$

(B) $K=\frac{-1}{2}$

(C) $K=\frac{3}{2}$

(D) $K=-1$

(E) $K=1$

Choose the correct answer from the options given below:






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CUET Previous Year PYQ CUET CUET 2025 PYQ

Solution

Given points: \[ P_1(K,2-2K), \quad P_2(1-K,2K), \quad P_3(-4-K,6-2K) \] These three points are collinear if the area of triangle formed by them is zero: \[ \Delta = \begin{vmatrix} K & 2-2K & 1\\ 1-K & 2K & 1\\ -4-K & 6-2K & 1 \end{vmatrix} = 0 \] Expanding determinant: \[ 8K^2 + 4K - 4 = 0 \;\;\Rightarrow\;\; 2K^2 + K - 1 = 0 \] Solving quadratic: \[ K = \frac{-1 \pm 3}{4} \;\;\Rightarrow\;\; K=\tfrac{1}{2},\; -1 \]
CUET PYQ
If the vertices of a triangle are (1, 2), (2, 5) and (4, 3) then the area of the triangle is:





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CUET Previous Year PYQ CUET CUET 2024 PYQ

Solution


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