CUET · MATHS · PYQ PAPER 2023

\(y=\log _e\left(\frac{1-x^2}{1+x^2}\right)\), then \(\frac{d y}{d x}\) is equal to:

A \(\frac{4 x^3}{1-x^4}\)
B \(\frac{-4 x}{1-x^4}\)
C \(\frac{1}{4-x^4}\)
D \(\frac{-4 x^3}{1-x^4}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{-4 x}{1-x^4}\)

Step-by-step Solution

Detailed explanation

\(\frac{d y}{d x} = \frac{1}{1-x^2}(-2x) - \frac{1}{1+x^2}(2x)\) \(\frac{d y}{d x} = -2x \left( \frac{1}{1-x^2} + \frac{1}{1+x^2} \right)\) \(\frac{d y}{d x} = -2x \left( \frac{1+x^2+1-x^2}{(1-x^2)(1+x^2)} \right)\) \(\frac{d y}{d x} = \frac{-4x}{1-x^4}\)

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. For any vector \(\vec{a}, \vec{a} \times \vec{a}\)	\(I\). \(\|\vec{a}\|^2\)
B. For any vector \(\vec{a}, \vec{a} \cdot \vec{a}\)	\(II.\) \(0\)
C. If either \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\), then \(\vec{a} \cdot \vec{b}\) is	\(III.\) \(\|\vec{a}\|\|\vec{b}\|\)
D. If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\theta=0\), then \(\vec{a} \cdot \vec{b}\) is	\(IV.\) \(\overrightarrow{0}\)

Choose the correct answer from the options given below:

CUET 2023 Medium

A die is tossed 6 times and getting 1 or 5 is considered a success.
The probability of getting at least one success in six tosses is:

CUET 2025 Medium

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