Let A = [8(1 sinα 1 -sinα 1 sinα -1 -sinα 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
CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
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CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
Last updated at April 16, 2024 by Teachoo
This question is inspired from Misc 19 (MCQ) - Chapter 4 Class 12 - Determinants
Question 45 Let A = [■8(1&sin𝛼&1@−sin𝛼&1&sin𝛼@−1&−sin𝛼&1)], where 0 ≤ α ≤ 2π, then: (a) |A|= 0 (b) |A| ∈ (2,∞) (c) |A| ∈ (2,4) (d) |A| ∈ [2, 4] |A| = |■8(1&sinθ&1@−sinθ&1&sinθ@−1&〖−sin〗θ&1)| = 1 |■8(1&sinθ@−sinθ&1)| – sin θ |■8(−sinθ&sinθ@−1&1)| + 1 |■8(−sinθ&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