Let A = [8(1 sin⁡α 1 -sin⁡α 1 sin⁡α -1 -sin⁡α 1)], where 0 ≤ α ≤ 2π, then:

(a) |A|= 0        (b) |A| ∈ (2,∞)

(c) |A| ∈ (2,4)  (d) |A| ∈ [2, 4]

 

This question is inspired from Misc 19 (MCQ) - Chapter 4 Class 12 - Determinants

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Question 45 Let A = [■8(1&sin⁡𝛼&1@−sin⁡𝛼&1&sin⁡𝛼@−1&−sin⁡𝛼&1)], where 0 ≤ α ≤ 2π, then: (a) |A|= 0 (b) |A| ∈ (2,∞) (c) |A| ∈ (2,4) (d) |A| ∈ [2, 4] |A| = |■8(1&sin⁡θ&1@−sin⁡θ&1&sin⁡θ@−1&〖−sin〗⁡θ&1)| = 1 |■8(1&sin⁡θ@−sin⁡θ&1)| – sin θ |■8(−sin⁡θ&sin⁡θ@−1&1)| + 1 |■8(−sin⁡θ&1@−1&〖−sin〗⁡θ )| = 1 (1 + sin2 θ) – sin θ (–sin θ + sin θ) + 1 (sin2 θ + 1) = (1 + sin2 θ) – sin θ × 0 + (1 + sin2 θ) = 2 (1 + sin2 θ) Thus, |A| = 2 (1 + sin2 θ) We know that –1 ≤ sin θ ≤ 1 So, value of sin θ can be from –1 to 1 Suppose, Hence, value of sin2 θ can be from 0 to 1 (negative not possible) Putting sin2 θ = 0 in |A| |A| = 2(1 + 0) = 2 ∴ Minimum value of |A| is 2 Putting sin2 θ = 1 in |A| |A| = 2 (1 + 1) = 2 (2) = 4 ∴ Maximum value of |A| is 4

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo