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If f ( x ) = c o s - 1 s i n x 4 c o s 2 x - 1 , then 1 π f ' π 3 ⋅ f π 10 is

Correct answer

0.60

Step-by-step solution

f x = c o s - 1 sin ⁡ x 4 1 - s i n 2 x - 1 f x = c o s - 1 sin ⁡ x 3 - 4 s i n 2 x = c o s - 1 3 sin ⁡ x - 4 s i n 3 x f x = c o s - 1 sin ⁡ 3 x = π 2 - s i n - 1 sin ⁡ 3 x In the neighbourhood of x = π 3 ⇒ f x = π 2 - π - 3 x f x = 3 x - π 2 ∴ f ' π 3 = 3 In the neighbourhood of x = π 10 ⇒ f x = π 2 - 3 x f π 10 = π 2 - 3 π 10 = π 5

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