CUET · MATHS · PYQ PAPER 2023

The area enclosed between the curves \(y^2=x\) and \(y=|x|\) is :

A 1 sq. unit
B \(\frac{2}{9}\) sq. units
C \(\frac{1}{6}\) sq. units
D \(\frac{2}{3}\) sq. units

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{6}\) sq. units

Step-by-step Solution

Detailed explanation

Intersection points: \(y^2=x\), \(y=|x|\) Since \(x=y^2 \ge 0\), we use \(y=x\). \(y^2=y \implies y(y-1)=0 \implies y=0, 1\) Intersection points are \((0,0)\) and \((1,1)\). Area \(A = \int_{0}^{1} (y - y^2) dy\) \(A = \left[ \frac{y^2}{2} - \frac{y^3}{3} \right]_{0}^{1}\)…

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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Match the Matrix in List - I with its Adjoint's Determinant in List - II.
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(A) \(\left[\begin{array}{ll}3 & 1 \\ 4 & 2\end{array}\right]\)	(I) 9
(B) \(\left[\begin{array}{cc}5 & -1 \\ 4 & 2\end{array}\right]\)	(II) 8
(C) \(\left[\begin{array}{cc}6 & -1 \\ 2 & 1\end{array}\right]\)	(III) 14
(D) \(\left[\begin{array}{ll}4 & 1 \\ 3 & 3\end{array}\right]\)	(IV) 2

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The corner points of the bounded feasible region determined by the system of linear constraints are \((0,0), (5,0), (6,5)\), \((6,8),(4,10), (0,8)\). Let \(Z=3 x-4 y\) be the objective function. The minimum value of \(Z\) occurs at

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Match List-l with List-ll.Find the derivatives from List-I.

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X	2	3	4	5
P(X)	5/k	7/k	9/k	11/k