Question

The equation $x^3+3x^2+6x+3-2cosx=0$ has $n$ solution(s) in $\left(0,1\right)$ , then the value of $\left(n+2\right)$ is equal to

Answer 1

Answer: 2

Let $f\left(x\right)=x^3+3x^2+6x+3-2cosx$
${\left(\text{f}\right)}^{{}^{\prime }}\left(\text{x}\right)=3{\left(\text{x}\right)}^2+6\text{x}+6+2\text{sinx}$
${\left(\text{f}\right)}^{{}^{\prime }}\left(\text{x}\right)=3\left({\left(\text{x}\right)}^2+2\text{x}+2\right)+2\text{sinx}$
${\left(\text{f}\right)}^{{}^{\prime }}\left(\text{x}\right)$ is always positive as the minimum value of $3\left(x^2+2x+2\right)$ is $3$ and that of $2sinx$ is $-2,$ so $f\left(x\right)$ is increasing in $\left(0,1\right)$
$f\left(0\right)=1,f\left(1\right)=13-2cosx>0$
$f\left(x\right)=0$ has no solution in $\left(0,1\right)$
$n=0\Rightarrow n+2=2$