If  $y=f\left(x\right)$ be a polynomial of least degree passing through  $A(-1,1)$ and whose graph has two points of inflection  $B(1,1)$ and C with abscissa 0 at which curve is inclined at an angle $\frac{\pi }{4}$  to positive direction of x-axis then which of the following option(s) is/are CORRECT?

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=a(x^2-x) \\ f\text{'}\text{'}\left(x\right)=a(\frac{x^3}{3}-\frac{x^2}{2})+b \\ f\text{'}\left(0\right)=1\Rightarrow b=1 \\ f\text{'}\left(x\right)=a(\frac{x^3}{3}-\frac{x^2}{2})+1 \\ f\left(x\right)=a(\frac{x^4}{12}-\frac{x^3}{6})+x+c \\ f\left(1\right)=1 And f(-1)=1 \end{gathered}$$

$$\begin{gathered} 1=a(\frac{1}{12}-\frac{1}{6})+1+c\Rightarrow c-\frac{a}{12}=0\Rightarrow a=12c \\ 1=a(\frac{1}{12}-\frac{1}{6})+1+c\Rightarrow c-\frac{a}{4}=2 \\ 4c=2\Rightarrow c=\frac{1}{2},a=6 \\ f\left(x\right)=\frac{x^4}{2}-x^3+x+\frac{1}{2} \\ f\text{'}\left(x\right)=2x^3-3x^2+1 \\ f\left(x\right)=(x-1)(2x+1)(x-1) \end{gathered}$$

Local minima at and neither maxima nor minima at  $x=1$

$f\left(x\right)=\frac{5}{2}$