Let $f\left(t\right)=\underset{-1}{\overset{t}{\int }}[\sin (x^3-9x^2+18x)+x^2-6x+\frac{11}{3}]dx$. Given four statements based on the given f(t).
STATEMENT-I:  For $t\in [-3,12],f\left(t\right)=0$ has at least two real roots.
STATEMENT-II:  For $t\in [\sqrt{11},11],f\left(t\right)=0$ has at least one real root.
STATEMENT-III:  The maximum value of t where  $f\left(t\right)=0$  is less than 7.
STATEMENT-IV:  The maximum value of t where  $f\left(t\right)=0$ is greater than 7.
The number of statement(s) among above given statements is/are CORRECT?

only STATEMENT-I and II are true.
Let $x=y+3,$ the integral at hand can be rewritten as  $I\left(t\right)def={\int }_{-1}^t(\sin (x^3-9x^2+18)+x^2-6x+\frac{11}{3})dx$
$={\int }_{-4}^{t-3}(\sin (y^3-9y)+y^2-\frac{16}{3})dy$

Notice $\sin (y^3-9y)$ is an odd function in y, we have 
 $I\left(7\right)={\int }_{-4}^4(\sin (y^3-9y)+y^2-\frac{16}{3})dy={\int }_{-4}^4(y^2-\frac{16}{3})dy={\left[\frac{y^3-16y}{3}\right]}_{-4}^4=0$
For $t>7,$  we have 
$\frac{dI\left(t\right)}{dt}=\sin (y^3-9y)+y^2-\frac{16}{3}{|}_{y=t-3}\ge -1+{(t-3)}^2-\frac{16}{3}>4^2-\frac{19}{3}>0$ 
This means $I\left(t\right)$ is strictly increasing on $[7,\infty )$ and $I\left(t\right)>I\left(7\right)=0$  whenever t >7.
As a result, 7 is the largest t where  $I\left(t\right)=0$