CUET · MATHS · PYQ PAPER 2025

The area (in sq.units) of the region bounded by the curves \(y = x\) and y = \(x^3\) is:

A 1 sq. units
B \(\frac{3}{4}\)sq. units
C \(\frac{1}{2}\) sq. units
D \(\frac{1}{4}\) sq. units

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{2}\) sq. units

Step-by-step Solution

Detailed explanation

Intersection points: \(x^3 = x \implies x^3 - x = 0 \implies x(x^2-1) = 0 \implies x = -1, 0, 1\). Area \(A = \int_{-1}^{0} (x^3 - x) dx + \int_{0}^{1} (x - x^3) dx\)…

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
Let \(\theta\) be the angle between the vector \(\vec{a}\) and \(\vec{b}\).

List-I	List-II
(A) \(\vec{a} \cdot \vec{b}\)	(I) \(\frac{1}{10} \log \left\|\frac{5+x}{5-x}\right\|+C\)
(B) \(\vec{a} \times \vec{b}\)	(II) \(\frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|^2} \vec{b}\)
(C) Projection vector of \(\vec{a}\) on \(\vec{b}(\neq 0)\)	(III) \(\|\vec{a}\|\|\vec{b}\| \sin \theta \hat{n}\) where \(\hat{n}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\)
(D) \(\vec{a}\) and \(\vec{b}\) are orthogonal vectors	(IV) \(\|\vec{a}\|\|\vec{b}\| \cos \theta\)

Choose the correct answer from the options given below:

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