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

Question

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, 10, 2, 40, 20

MARKS App · Accepted Answer

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 }

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