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)
Question 2 Important
Question 3
Question 4 Important
Question 5
Question 6 Important
Question 7 Important
Question 8
Question 9 Important
Question 10
Question 11
Question 12 Important
Question 13
Question 14
Question 15
Question 16 Important
Question 17 Important
Question 18
Question 19 Important
Question 20
Question 21 Important
Question 22
Question 23
Question 24 Important
Question 25
Question 26
Question 27 Important
Question 28 Important
Question 29
Question 30
Question 31 Important
Question 32
Question 33 Important
Question 34 Important
Question 35
Question 36 Important
Question 37 Important
Question 38
Question 39 Important
Question 40
Question 41
Question 42 Important
Question 43 Important
Question 44
Question 45 Important You are here
Question 46 (Case Based Question) Important
Question 47 (Case Based Question) Important
Question 48 (Case Based Question) Important
Question 49 (Case Based Question) Important
Question 50 (Case Based Question)
CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
Last updated at Sept. 4, 2021 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