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CUET · MATHS · PYQ PAPER 2025

Match List-I with List-II
List-IList-II
(A) Maximum value of \(f(x)=\sin ^2 x-\cos ^2 x \forall x \in(\pi, 2 \pi)\)(I) \(0\)
(B) Minimum value of \(f(x)=\sin x \cdot \cos x\)(II) 1
(C) Point of Minima of \(f(x)=x^x(x>0)\)(III) \(-\frac{1}{2}\)
(D) Maximum value of \(f(x)=-x^{2026}\)(IV) \(\frac{1}{e}\)
Choose the correct answer from the options given below :

  1. A (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
  2. B (A) - (III), (B) - (II), (C) - (IV), (D) - (I)
  3. C (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
  4. D (A) - (II), (B) - (III), (C) - (I), (D) - (IV)
Verified Solution

Answer & Solution

Correct Answer

(C) (A) - (II), (B) - (III), (C) - (IV), (D) - (I)

Step-by-step Solution

Detailed explanation

(A) \(f(x) = \sin^2 x - \cos^2 x = -\cos(2x)\) For \(x \in (\pi, 2\pi)\), \(2x \in (2\pi, 4\pi)\). Max \(-\cos(2x)\) is \(1\). (B) \(f(x) = \sin x \cos x = \frac{1}{2}\sin(2x)\) Min \(\frac{1}{2}\sin(2x)\) is \(-\frac{1}{2}\). (C)…
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