NTA Abhyas JEE Main2020MathematicsTrigonometric EquationsPractice
Let f x = 25 x 25 x + 5 , then the number of solution(s) of the equation f s i n 2 θ + f c o s 2 θ = t a n 2 θ , θ ∈ 0,10 π is/are
Options
- A10
- B2
- C40
- D20
Correct answer
D. 20
Step-by-step solution
f s i n 2 θ + f 1 - s i n 2 θ Let, s i n 2 θ = t ∴ f t + f 1 - t = 25 t 25 t + 5 + 25 1 - t 25 1 - t + 5 = 25 t 25 t + 5 + 25 25 + 5 25 t = 25 t + 5 25 t + 5 = 1 Therefore, the given equation will become t a n 2 θ = 1 ∴ tan θ = ± 1 ⇒ θ = n π ± π 4 , n ∈ Z Hence, the number of solutions are 4 × 5 = 20 ∵ 4 solutions in 0,2 π