If $f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x$, then $f' \left( \frac{\pi}{4} \right)$ is equal to

Let $f: R \rightarrow R$ be a differentiable function such that $f(0)=0$,$f(\frac{\pi}{2})=3$ and $f^{\prime}(0)=1$. If $g(x)=\int_x^{\pi / 2} [f^{\prime}(t) \operatorname{cosec} t - f(t) \operatorname{cosec} t \cot t] dt$ for $x \in (0, \frac{\pi}{2}]$,then $\lim _{x \rightarrow 0} g(x)=$

Let $f:[0, \infty) \rightarrow [0, 3]$ be a function defined by
$f(x) = \begin{cases} \max \{\sin t : 0 \leq t \leq x\}, & 0 \leq x \leq \pi \\ 2 + \cos x, & x > \pi \end{cases}$
Then which of the following is true?

If $f(x) = \begin{cases} \frac{8}{x^3} - 6x, & x \le 1 \\ \sqrt{x} + 1, & x > 1 \end{cases}$, then at $x = 1$, $f$ is:

Let $f(x) = x^{13} + x^{11} + x^{9} + x^{7} + x^{5} + x^{3} + x + 12$. Then

Let $f$ be a differentiable function and the equation of the normal to the graph of $y = f(x)$ at $x = 3$ is $3y = x + 18$. If $L = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( {3 + {{\left( {4{{\tan }^{ - 1}}x - \pi } \right)}^2}} \right) - f\left( {3 + {{\left( {f\left( 3 \right) - x - 6} \right)}^2}} \right)}}{{{{\sin }^2}\left( {x - 1} \right)}}$,then: