JEE Advanced | 2013Trigonometric EquationHard
Question
Given A = sin2 θ + cos4 θ then for all real values of θ
Options
A.1 ≤ A ≤ 2
B.
≤ A ≤ 1
≤ A ≤ 1C.
≤ A ≤ 1
≤ A ≤ 1D. ≤ A ≤
≤ A ≤ Solution
Given, Asin2 θ + (1 - sin2 θ)2
⇒ Asin4 θ - sin2 θ + 1
⇒
∵ 0 ≤ sin2 θ ≤ 1
⇒
∴
≤ A ≤ 1
⇒ Asin4 θ - sin2 θ + 1
⇒
∵ 0 ≤ sin2 θ ≤ 1
⇒
∴
≤ A ≤ 1Create a free account to view solution
View Solution FreeMore Trigonometric Equation Questions
The values of θ satisfying sin 7θ = sin 4θ - sin θ and 0 < θ < are -...The number of solutions of $\tan^{- 1}4x + \tan^{- 1}6x = \frac{\pi}{6}$, where $- \frac{1}{2\sqrt{6}} < x < \frac...If 0 2θ - 5 sin θ + 2 > 0, is...If the equation of the locus of a point equidistant from the points (a1, b1) and (a2, b2) is (a1 - a2) x + (b1 - b2) y +...The general solution of equation 4 cos2 x + 6 sin2 x = 5 is -...