Let M = sin 4 ⁡ θ - 1 - sin 2 ⁡ θ 1 + cos 2 ⁡ θ cos 4 ⁡ θ =…

Let M=sin4⁡θ-1-sin2⁡θ1+cos2⁡θcos4⁡θ=αI+βM-1, where α=αθ and β=βθ are real number, and I is the 2×2 identity matrix. If α* is the minimum of the set { αθ:θ∈0, 2π} and β* is the minimum of the set βθ:θ∈0, 2π, then the value of α*+β* is
  1. -3716
  2. -3116
  3. -2916
  4. -1716

Solution

Method I
As given, M=sin4θ-1-sin2θ1+cos2θcos4θ=αI+βM-1
adj M=cos4θ1+sin2θ-1-cos2θsin4θ
M-1=1|M|cos4θ1+sin2θ-1-cos2θsin4θ
sin4θ-1-sin2θ1+cos2θcos4θ=α1001+βMcos4θ1+sin2θ-1-cos4θsin4θ
Compare element a12 in L.H.S. and R.H.S.
-1-sin2θ=0+βM1+sin2θ
β= |M| ...(iii)
Compare element a11 in LHS. and R.H.S.
sin4θ=α+βMcos4θ
From equation (iii)
sin4θ=α-cos4θ
α=sin4θ+cos4θ ...(iv)
Now from equation (iv)
α=sin4θ+cos4θ
α=sin2θ+cos2θ2-2sin2θcos2θ
α=1-sin22θ2
0≤sin22θ≤1
-12≤-sin22θ2≤0
12≤1-sin22θ2≤1
α∈12,1 , αmin=12=α*
For β , consider
M=sin4θcos4θ+sin2θcos2θ+2
Msin2θcos2θ+122+74
Msin22θ4+122+74
0≤sin22θ≤1
Hence, M ∈2,3716
β=-M
β∈-3716,-2
βmin=-3716=β*
α*+β*=12-3716=-2916
Method II
M=αI+βM-1
Multiply with M both side
M2=αM+βI
M2-αM-BI=0 ...(i)
Equation (i) represents characteristic equation of matrix ‘M’
Hence, Trace (M)=α
M =-β
Sum of roots of characteristic equation = Trace of matrix
Product of roots of characteristic equation = determined of matrix
∴ α=sin4θ+cos4θ
β=-sin4θcos4θ+sin2θcos2θ+2
Now proceed as Method 1

Asked in: JEE Advanced 2019 (Paper 1)

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