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

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