Let d ∈ R , and A = - 2 4 + d sin θ - 2 1 sin θ + 2 d 5 2 sin θ - d - sin θ + 2 + 2 d , θ ∈ 0 , 2 π . If the minimum value of d e t A is 8 , then a value of d is:

Step-by-step solution

A = - 2 4 + d s i n θ - 2 1 sin θ + 2 d 5 2 sin θ - d - sin θ + 2 + 2 d R 3 → R 3 - 2 R 2 + R 1 = - 2 4 + d s i n θ - 2 1 sin θ + 2 d 1 0 0 = 1 4 + d d - sin θ + 2 sin θ - 2 = 4 d + d 2 - sin 2 θ + 4 = d + 2 2 - sin 2 θ We know that 0 ≤ sin 2 θ ≤ 1 Given, the minimum value of A = 8 , which is possible, when sin 2 θ = 1 ⇒ d + 2 2 = 9 ⇒ d + 2 = ± 3 ⇒ d = 1 or d = - 5 .