$$ \lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2}= $$

If $f(x)=\left\{\begin{array}{l}\frac{x^2-16}{x-4} \text { if } x>4 \\ 2 x \quad \text { if } x \leq 4\end{array}\right.$ then $f^{\prime}\left(4^{-}\right)+f^{\prime}\left(4^{+}\right)=$

If $f(x)=\log _e\left(e^{2 x}\left(\frac{3 x+5}{5-3 x}\right)^{2 / 3}\right), x \neq \frac{-5}{3}, \frac{5}{3}$, then the value of $\frac{d f}{d x}$ at $x=1$, is

If $x=\operatorname{cosec} \theta-\sin \theta, y=\operatorname{cosec}^{2022} \theta-\sin ^{2022} \theta$ and $\left(\frac{d y}{d x}\right)^2=\frac{k\left(y^2+4\right)}{g(x)}$ where $k \in R$, then $10+k-g(2022)=$