If the function $f(x)=\frac{e^x\left(e^{\tan x-x}-1\right)+\log _e(\sec x+\tan x)-x}{\tan x-x}$ is continuous at $x=0$, then the value of $f(0)$ is equal to

If $f(x)=\left\{\begin{array}{cc}\frac{a|x|+x^2-2(\sin |x|)(\cos |x|)}{x} & , x \neq 0 \\ b & , x=0\end{array}\right.$

is continuous at $x=0$, then $a+b$ is equal to :

Let $f(x)= \begin{cases}\frac{\mathrm{a} x^2+2 \mathrm{a} x+3}{4 x^2+4 x-3} & , x \neq-\frac{3}{2}, \frac{1}{2} \\ \mathrm{~b} & , x=-\frac{3}{2}, \frac{1}{2}\end{cases}$ be continuous at $x=-\frac{3}{2}$. If $f \circ f(x)=\frac{7}{5}$, then $x$ is equal to:

If $\lim\limits_{x \rightarrow 0} \frac{\mathrm{e}^{(\mathrm{a}-1) x}+2 \cos \mathrm{~b} x+(\mathrm{c}-2) \mathrm{e}^{-x}}{x \cos x-\log _{\mathrm{e}}(1+x)}=2$, then $\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2$ is equal to :