Q60 / 90 · JEE Main 2019 (12 Jan Shift 2) · Mathematics

If sin4α+4cos4β+2=42sinαcosβ, α, β∈0,π, then cosα+β-cosα-β is equal to

  • A
    -1
  • Correct answer
    B
    -2
  • C
    2
  • D
    0

Solution

We have,

sin4α+4cos4β+2=42sin αcosβ

Applying AM≥GM, we get

sin4α+4cos4β+1+14≥4sin4α·cos4β·1·114

⇒sin4α+4cos4β+2≥8sinαcosβ

And, AM=GM⇒numbers are equal and positive

⇒sin4α=4cos4β=1

∴sin4α=1⇒α=π2 and

cosβ=12⇒β=π4

Hence,

cosα+β-cosα-β

=-2sinαsinβ

=-2×1×12=-2

Hence, the correct answer is option: 'B'.

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