Consider the function $\mathrm{f}:\left[\frac{1}{2}, 1\right] \rightarrow \mathrm{R}$ defined by $f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1$. Consider the statements

$(I)$ The curve $y=f(x)$ intersects the $x$-axis exactly at one point

$(II)$ The curve $y=f(x)$ intersects the $x$-axis at $\mathrm{x}=\cos \frac{\pi}{12}$

Then

Let for $a \ne {a_1} \ne 0,$ $f\left( x \right) = a{x^2} + bx + c\;,g\left( x \right) = {a_1}{x^2} + {b_1}x + {c_1},p\left( x \right) = f\left( x \right) – g\left( x \right),$ If $p\left( x \right) = 0$ only for  $ x=-1 $ and $p\left( { – 2} \right) = 2$ then value of $p\left( 2 \right)$ is

The value of $b$ and $c$ for which the identity $f(x + 1) – f(x) = 8x + 3$ is satisfied, where $f(x) = b{x^2} + cx + d$, are

Let $f\,:\,R \to R$ be a function such that $f\left( x \right) = {x^3} + {x^2}f'\left( 1 \right) + xf''\left( 2 \right) + f'''\left( 3 \right)$, $x \in R$. Then $f(2)$ equals

Let, $f(x)=\left\{\begin{array}{l} x \sin \left(\frac{1}{x}\right) \text { when } x \neq 0 \\ 1 \text { when } x=0 \end{array}\right\}$ and $A=\{x \in R: f(x)=1\} .$ Then, $A$ has