The number of solutions of the equation $x^{3} + 2 x^{2} + 5 x + 2 c o s x = 0$ in $[0, 2 π]$ is

2

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1

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3

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0

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Solution

The correct option is D $0$

Given: $x^{3} + 2 x^{2} + 5 x + 2 c o s x = 0$
$x^{3} + 2 x^{2} + 5 x = - 2 c o s x$
Let $y = x^{3} + 2 x^{2} + 5 x = - 2 c o s x$
$y = x (x^{2} + 2 x + 5) = - 2 c o s x$
$y = x (x^{2} + 2 x + 5)$

From the graph, we can see that if $x \in [0, π]$ , there is no intersection point of the two graphs $y = - 2 c o s x$ and $y = x (x^{2} + 2 x + 5)$

If $x \in [0, 2 π]$ , then the number of solution(s) of the equation $x^{3} + 2 x^{2} + 5 x + 2 c o s x = 0$ is $0$ .

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