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The equation x 3 + 3 x 2 + 6 x + 3 - 2 c o s x = 0 has n solution(s) in 0,1 , then the value of n + 2 is equal to

Correct answer

2

Step-by-step solution

Let f x = x 3 + 3 x 2 + 6 x + 3 - 2 c o s x f ' x = 3 x 2 + 6 x + 6 + 2 sinx f ′ x = 3 x 2 + 2 x + 2 + 2 sinx f ′ x is always positive as the minimum value of 3 x 2 + 2 x + 2 is 3 and that of 2 sin ⁡ x is - 2 , so f x is increasing in 0,1 f 0 = 1 ,   f 1 = 13 - 2 cos ⁡ x > 0 f x = 0 has no solution in 0,1 n = 0 ⇒ n + 2 = 2

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