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0 votes
8.1k views
in Trigonometry by (53.1k points)

The number of solutions of the equation x3 + 2x2 + 5x + 2cosx = 0 in [0,2π] is

(A)  0 

(B)  1 

(C)  2 

(D)  3

1 Answer

+1 vote
by (53.3k points)
selected by
 
Best answer

Correct option (A) 0

Let f(x) = x3 + 2x2 + 5x +2 cos x. Differentiating w.r.t. x we get 

Now   11/3 - 2sin x >0 ∀ x ( as −1 ≤ sinx ≤ 1)

⇒f′(x) > 0 ∀x 

 ⇒f(x) is an increasing function

Now f(0) = 2. Hence, f(x) = 0 has no solution in [ 0, 2π].

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