Let d ∈ R, and A = [-2&4+d&( sinθ)-2 1&( sinθ)+2&d 5&(2 sinθ)-d&(- sinθ)+2 +2d] , θ∈ [0,2π] . If the minimum value of det(A) is 8, then a value of d is :

Question

Let d ∈ R, and A = [-2&4+d&( sinθ)-2 1&( sinθ)+2&d 5&(2 sinθ)-d&(- sinθ)+2 +2d] , θ∈ [0,2π] . If the minimum value of det(A) is 8, then a value of d is : (A) -7 (B) 2 ( √2 + 2) (C) -5 (D) 2 ( √2 + 1)

Tardigrade Admin · Accepted Answer

operatornamedet A=| beginarrayccc -2 4+d sin θ-2 1 sin θ+2 d 5 2 sin θ-d - sin θ+2+2 d endarray| ( R 1 arrow R 1+ R 3-2 R 2) =| beginarrayccc 1 0 0 1 sin θ+2 d 5 2 sin θ-d 2+2 d- sin θ endarray| =(2+ sin θ)(2+2 d - sin θ)- d (2 sin θ- d ) =4+4 d -2 sin θ+2 sin θ+2 dsin θ- sin 2 θ-2 d sin θ+ d 2 = d 2+4 d +4- sin 2 θ =(d+2)2- sin 2 θ For a given d, minimum value of operatornamedet( A )=( d +2)2-1=8 ⇒ d =1 or -5

Q. Let $d \in R$ , and $A = ⎣ ⎡ - 2 15 4 + d (sin θ) + 2 (2 sin θ) - d (sin θ) - 2 d (- sin θ) + 2 + 2 d ⎦ ⎤, θ \in [0, 2 π]$ . If the minimum value of $d e t (A)$ is $8$ , then a value of $d$ is :

$- 7$

$2 (2 + 2)$

$- 5$

$2 (2 + 1)$

Q. Let d∈R, and A=⎣⎡​−215​4+d(sinθ)+2(2sinθ)−d​(sinθ)−2d(−sinθ)+2+2d​⎦⎤​,θ∈[0,2π]. If the minimum value of det(A) is 8, then a value of d is :

−7

2(2​+2)

−5

2(2​+1)

Q. Let $d \in R$ , and $A = ⎣ ⎡ - 2 15 4 + d (sin θ) + 2 (2 sin θ) - d (sin θ) - 2 d (- sin θ) + 2 + 2 d ⎦ ⎤, θ \in [0, 2 π]$ . If the minimum value of $d e t (A)$ is $8$ , then a value of $d$ is :

$- 7$

$2 (2 + 2)$

$- 5$

$2 (2 + 1)$