Q: The number of solutions of the equation x3 + 2x2 + 5x + 2 cosx = 0 in [0, 2π] is
(A) 0
(B) 1
(C) 2
(D) 3
Sol: Let f(x) = x3 + 2x2 + 5x + 2 cosx
f'(x) = 3x2 + 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
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