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

Slide117.JPG

Slide118.JPG
Slide119.JPG

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


Transcript

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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.