The number of solutions of the equation $x^{3} + x^{2} + 4 x + 2 sin x = 0$ in $0 \leq x \leq 2 π$ is

z e r o

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o n e

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t w o

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f o u r

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Solution

The correct option is B $o n e$
$\begin{matrix} f (x) = x^{3} + x^{2} + 4 x + 2 sin x = 0 0 ⩽ x ⩽ 2 x f^{^{'}} (x) = 3 x^{2} + 2 x + 4 + 2 cos x > 0 f (x) i s i n c r e a s i n g f; x f (0) = 0 f (x) ⩾ 0 0 ⩽ x ⩽ 2 x o p t i o n (b) \end{matrix}$

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The number of solutions of the equation $x^{3}$ + $x^{2}$ + 4x + 2sinx = 0 in 0 ≤ x ≤ 2 $π$

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