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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 9 Choose the correct answer. Let A = [■8(1&sin⁡θ&1@−sin⁡θ&1&sin⁡θ@−1&〖−sin〗⁡θ&1)] , where 0 ≤ θ≤ 2π, then A. Det (A) = 0 B. Det (A) ∈ (2, ∞) C. Det (A) ∈ (2, 4) D. Det (A)∈ [2, 4] A = [■8(1&sin⁡θ&1@−sin⁡θ&1&sin⁡θ@−1&〖−sin〗⁡θ&1)] |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) Thus 2 ≤ |A| ≤ 4 |A|∈ [2 , 4] Det (A) ∈ [2 , 4] Thus, D is the correct answer Putting sin2 θ = 0 in |A| |A| = 2(1 + 0) |A| = 2 Thus minimum value of |A| is 2 Putting sin2 θ = 1 in |A| |A| = 2 (1 + 1) = 2 (2) = 4 Thus maximum value of |A| is 4

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.