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 Misc 19 - Let A = [1 sin 1 -sin 1 sin -1 -sin 1], then - Miscellaneous

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Misc. 19 Choose the correct answer. Let A = 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, 唴 C. Det (A) ∈ (2, 4) D. Det (A)∈ [2, 4] A = 1﷮ sin﷮θ﷯﷮1﷮− sin﷮θ﷯﷮1﷮ sin﷮θ﷯﷮−1﷮ −sin﷮θ﷯﷮1﷯﷯ |A| = 1﷮ sin﷮θ﷯﷮1﷮− sin﷮θ﷯﷮1﷮ sin﷮θ﷯﷮−1﷮ −sin﷮θ﷯﷮1﷯﷯ = 1 1﷮ sin﷮θ﷯﷮− sin﷮θ﷯﷮1﷯﷯ – sin θ − sin﷮θ﷯﷮ sin﷮θ﷯﷮−1﷮1﷯﷯ + 1 − sin﷮θ﷯﷮1﷮1﷮ −sin﷮θ﷯﷯﷯ = 1 (1 + sin2θ) – sin θ ( –sin θ + sin θ ) + 1 (sin2 θ + 1) = 1 (1 + sin2θ) – sin θ (0) + 1(sin2θ + 1) = 2 (1 + sin2θ) Thus, |A| = 2 (1 + sin2θ) 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

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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