All Tasks - Asymptote

Let $ \displaystyle f(x)= \pi -\dfrac{b}{3}arccot(2x)$ with $b \in \mathbb{R}$, and let $f^{\prime}(x)$ and $f^{\prime\prime}(x)$ be the first and the second derivative of the function $f$. Knowing that $\displaystyle f\left(\dfrac{\sqrt{3}}{2}\right) \cdot \dfrac{\sqrt{3} f^{\prime}\left(\dfrac{\sqrt{3}}{2}\right)}{2f^{\prime\prime}\left(\frac{\sqrt{3}}{2}\right)} = \dfrac{\pi}{4}$, then the value of $b$ is:

Let $f$ and $g$ be two differentiable functions in their domain, such that: $f(x)= \dfrac{\pi}{2} -\dfrac{1}{3}arccot(2x-1)$ and $g(x)=\left( x^2+x+2\right) f(x)$. Considering that $g^{\prime}(x)$ is the first derivative of function $g(x)$, then $g^{\prime}\left(0\right)$ is:

Solve the following trigonometric equation: $3arccot(x − \sqrt{3})− \pi=0$

Consider $a=tg(arccot(\sqrt(2))$. So the value of $a$ is:

Consider the function $f(x)=\dfrac{2\pi}{5}\text{arccot}\left(\ln(x+1)\right)$. The inverse functions of $f$, its domain ($D_{f^{-1}}$) and its range ($D'_{f^{-1}}$) are, respectively:

Let $f$ be the function defined by $f(x) =\pi-arccot(2x)$. So the inverse of $f$, $f^{-1}$, is the function defined by