Verified Solution • 2025–26 Session moderate

QJEEM19Maeefb: JEE Main 2019 Mathematics - Class 12 JEEM19Maeefb Solution: Let  d R, and  A = [ − 2 4 + d ( sin ⁡ θ ) − 2 1 …

Let  d ∈\in R, and 

A=[−24+d(sin⁡θ)−21(sin⁡θ)+2d5(2sin⁡θ)−d(−sin⁡θ)+2+2d],

θ∈[0,2π]\theta \in \left[ {0,2\pi } \right] If the minimum value of det(A) is 8, then a value of d is -

Multiple Choice Options

  • A −- 7
  • B 2(2+2)2\left( {\sqrt 2 + 2} \right)
  • C −- 5
  • D 2(2+1)2\left( {\sqrt 2 + 1} \right)

Step-by-Step Verified Solution

detA=|−24+dsin⁡θ−21sin⁡θ+2d52sin⁡θ−d−sin⁡θ+2+2d|

(R1 →\to R1 + R3 −- 2R2)

=|1001sin⁡θ+2d52sin⁡θ−d2+2d−sin⁡θ|

= (2 + sin θ\theta ) ( 2 + 2d −- sin θ\theta ) −- d(2sin θ\theta −- d)

= 4 + 4d −- 2sin θ\theta + 2sin θ\theta + 2dsin θ\theta −- sin2 θ\theta −- 2dsin θ\theta + d2

d2 + 4d + 4 −- sin2 θ\theta

= (d + 2)2 −- sin2 θ\theta

For a given d, minimum value of

det(A) = (d + 2)2 −- 1 = 8

⇒\Rightarrow   d = 1 or −- 5

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