Q: The number of solutions of the equation x^{3} + 2x^{2} + 5x + 2 cosx = 0 in [0, 2π] is

(A) 0

(B) 1

(C) 2

(D) 3

Sol: Let f(x) = x^{3} + 2x^{2} + 5x + 2 cosx

f'(x) = 3x^{2} + 4x + 5 – 2 sinx

$\large = 3 (x+\frac{2}{3})^2 + \frac{11}{3} – 2 sinx $

Now $\large \frac{11}{3}-2sinx > 0 \forall x $ ( as -1 ≤ sinx ≤ 1)

⇒ f'(x) > 0 ∀ x

⇒ f(x) is an increasing function.

Now f(0) = 2

⇒ f(x) = 0 has no solution in [ 0, 2π]

Hence (A) is correct