CUET · MATHS · PYQ PAPER 2025

The value of \(\left|\begin{array}{lll}2^x & 1 & 6^x \\ 4^x & 1 & 3^x \\ 2^x & 1 & 6^x\end{array}\right|\), where \(x \neq 0 ;\) is:

A 0
B \(2^x+3^x\)
C \(3^x+4^x\)
D \(4^x+6^x\)

Verified Solution

Answer & Solution

Correct Answer

(A) 0

Step-by-step Solution

Detailed explanation

Given determinant: \( \left|\begin{array}{lll}2^x & 1 & 6^x \\ 4^x & 1 & 3^x \\ 2^x & 1 & 6^x\end{array}\right| \) Since the first row (\(R_1\)) and the third row (\(R_3\)) are identical, \(R_1 = R_3\). Thus, the value of the determinant is \(0\).

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.

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From CUET

Match List - I with List - II.

LIST I Differential equation	LIST II Integrating Factor (I.F)
A. \(\frac{d y}{d x}-\frac{y}{x}=2 x\)	I. \(x^2-1\)
B. \(\frac{d y}{d x}+\left(\frac{2 x}{x^2-1}\right) y=\frac{2}{\left(x^2-1\right)^2}\)	II. \(\sqrt{1-x^2}\)
C. \(\frac{d y}{d x}-\left(\frac{x}{1-x^2}\right) y=\frac{1}{1-x^2}\)	III. \(\frac{1}{x}\)
D. \(\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{\cot x}{1+x^2}\)	IV. \(1+x^2\)

Choose the correct answer from the options given below:

CUET 2023 Hard

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X	0	1	2
\(P ( X )\)	\(1 / 4\)	\(1 / 2\)	\(1 / 4\)

x	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)
\(P(X)\)	\[3k\]	\(3k\)	\(2k\)	\(2k^2\)	\(3k^2\)	\(4k^2\)