Q.
Let M=sin4θ−1−sin2θ1+cos2θcos4θ=αI+βM−1 , Where α=αθ and β=βθ are real numbers, 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
see full answer
Start JEE / NEET / Foundation preparation at rupees 99/day !!
21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya
a
−2916
b
−3716
c
−3116
d
−1716
answer is A.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
M=sin4θ-1-sin2θ1+cos2θcos4θ=αI+βM-1 Where α=α(θ) and β=β(θ) are real numbers |M|=(sinθcosθ)4+1+sin2θ1+cos2θ=2+sin2θcos2θ+sin4θcos4θ since M=αI+βM-1sin4θ-1+sin2θ1+cos2θcos4θ=α00α+β|M|cos4θ1+sin2θ-1-cos2θsin4θ Comparing on both side sin4θ=α+β|M|cos4θ-1-sin2θ=β|M|1+sin2θ⇒β=-|M| =-2+sin2θcos2θ+sin4θcos4θ therefore sin4θ=α-cos4θ ⇒α=αθ=sin4θ+cos4θ=1-2sin2θcos2θ=1-12sin22θα*= minimum of α(θ)=1-121=12now β* =Minimum of βθ=-2+sin2θcos2θ+sin4θcos4θ=-sin2θcos2θ+122+74=-14sin22θ+122+74β*=-14+122+74=-3716 when θ=nπ+π4α*+β*=-3716+12=-37+816=-2916
Watch 3-min video & get full concept clarity