Maxima and MinimaHard
Question
If θ = sin2θ + cos4θ , then for all real values of θ
Options
A.
≤ p ≤ 1
≤ p ≤ 1B.1 < p ≤ 2
C.
≤ p ≤ 
≤ p ≤ D.
≤ p ≤ 2
≤ p ≤ 2Solution
θ = sin2θ + cos4θ
θ = sin2θ + (1 - sin2θ)2
θ = sin2θ + 1 + sin4θ - 2sin2θ
θ = sin4θ - sin2θ + 1
sin4θ - sin2θ + 1 - p = 0
for all real value of q
b2 - 4ac ≥ 0
1 - 4 × 1(1 - p) ≥ 0
1 - 4 + 4p ≥ 0
4p ≥ 3
p ≥
sin θ max value = 1
so
θ = sin2θ + (1 - sin2θ)2
θ = sin2θ + 1 + sin4θ - 2sin2θ
θ = sin4θ - sin2θ + 1
sin4θ - sin2θ + 1 - p = 0
for all real value of q
b2 - 4ac ≥ 0
1 - 4 × 1(1 - p) ≥ 0
1 - 4 + 4p ≥ 0
4p ≥ 3
p ≥
sin θ max value = 1
so
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