Let $d \in R$, and $A=\left[\begin{array}{ccc}-2 & 4+d & (\sin \theta)-2 \\ 1 & (\sin \theta)+2 & d \\ 5 & (2 \sin \theta)-d & (-\sin \theta)+2+2 d\end{array}\right], \theta \in[0,2 \pi]$. If the minimum value of $\operatorname{det}(A)$ is 8, then a value of $d$ is:

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Determinants

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

det (A)

is 8, then a value of

d

is:

Answer: (A)

Clearing the awkward entries with the row operation

R_{3} \to R_{3} - 2 R_{2} + R_{1}

turns the bottom row into

(1, 0, 0)

, and expanding along it collapses the determinant to

(4 + d) d - sin^{2} θ + 4 = (d + 2)^{2} - sin^{2} θ

. Over

θ

this is smallest when

sin^{2} θ = 1

, giving the minimum

(d + 2)^{2} - 1

. Setting that to

8

gives

(d + 2)^{2} = 9

, so

d + 2 = \pm 3

and

d = 1

d = - 5

; the value offered is

- 5

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option