Question

The number of solutions of the equation $x^3+2x^2+5x+2+\cos \left(x\right)=0$ in $\left[0,2\pi \right]$ is :

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A

$0$

Explanation for correct option

Given, $f\left(x\right)=x^3+2x^2+5x+2\cos \left(x\right)$

Now differentiating with respect to $x$ we get

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=3x^2+4x+5-2\sin \left(x\right) \\ \Rightarrow f\text{'}\left(x\right)=3{\left(x+\frac{2}{3}\right)}^2+\frac{11}{3}-2\sin \left(x\right) \end{gathered}$$

$\frac{11}{3}-2\sin \left(x\right)>0\forall x$ [as $-1\le \sin \left(\theta \right)\le 1$]

$\Rightarrow f\text{'}\left(x\right)>0\forall x$

Therefore $f\left(x\right)$ is an increasing function.

$f\left(0\right)=2$

$f\left(x\right)=0$ has no solution in $\left[0,2\pi \right]$

Hence, the correct option is $\left(\mathrm{A}\right)$