Question

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

• A. one
• B. two
• C. three
• D. zero

Answer 1

The correct option is D zero

Given : $x^3+2x^2+5x+2\cos x=0in[0,\pi ]$

$\Rightarrow x^3+2x^2+5x=-2\cos x=f\left(x\right)$

$\Rightarrow f\left(x\right)=x^3+2x^2+5x$ $g\left(x\right)=-2\cos x$

So, we are taking for $x\in [0,\pi ]$ such that $f\left(x\right)=g\left(x\right)$

Claim 1 : $f\left(x\right)\ge 0in[0,2\pi ]$

All coefficients are positive and the interval is positive as well

Observation 1 : $g\left(x\right)>0$ only in $[\frac{\pi }{2},\frac{3\pi }{2}]$

If $f\left(x\right)=g\left(x\right)$ then that x must lie in the first interval.

Claim 2 : $f\left(x\right)$ is strictly increasing

Proof : $f^1\left(x\right)=3x^2+4x+5>0in[0,2\pi ]$

Observation 2 : $f\left(\frac{3}{2}\right)\approx 15.375$

Note that $\frac{3}{2}<\frac{\pi }{2}$

Therefore, $f\left(\frac{\pi }{2}\right)>15.375$

Therefore, $f\left(x\right)>15.375$ in $[\frac{\pi }{2},\frac{3\pi }{2}]$

However, the maximum of $g\left(x\right)$ is $2$.

Therefore, $f\left(x\right)\ne g\left(x\right)in[\frac{\pi }{2},\frac{3\pi }{2}]$

From observations $1$ and $2$, we can see that $f\left(x\right)\ne g\left(x\right)in[0,2\pi ]$

Hence, the total number of solutions to the equations is $0$.