All Tasks - Asymptote

Let $f(x)= -3\pi +2 \arccos\left( e^{3x+1}\right) $ with domain $D_f=\left]-\infty, -\dfrac{1}{3}\right] $. $f^{-1}(x)= -\dfrac{1}{3} + \dfrac{1}{3} \ln\left( \cos\left( \dfrac{x+3\pi}{2}\right)\right)$ is the analytic expression of the inverse function with domain $D_{f^{-1}}=\left[ -3\pi, -2\pi\right[$. The solution of the equation $f\left(\dfrac{x-1}{3}\right) -f\left(-\dfrac{1}{3}\right) -\arccos(0)=\sec\left( \arccos\left(\dfrac{1}{3}\right) \right) +\dfrac{1}{f^{-1}(-3\pi)}$ is:

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

Consider $g(x)= \arccos\left(\dfrac{x}{2}\right)-3\pi$ with domain $D_g=[-2, 2]$. The first derivative of the function $g(x)$ is given by $g'\left(x\right)=\dfrac{b}{\sqrt{a-x^2}}$. Then, the values of $a$ and $b$ are, respectively:

Let $y= (a+2) \arccos(x)$ with $a \in \mathbb{R}$. $y'$ and $y''$ are the first and second derivatives of $y$. Knowing that $ x \in \left]-1, 1\right[$ and $\left( 1-x^2\right) y''+y'=0$, then, the value of $a$ is:

Let $f(x)= -3\pi +a \arccos(3x+1)$ with $a \in \mathbb R$ and domain $D_f=\left[-\dfrac{2}{3}, 0\right]$. Knowing that the tangent line to the graph of $f$ at the point of abscissa $-\dfrac{1}{3}$ is paralel to the line $15x+y=10$, we conclude that the value of $a$ is:

In the figure are represented: - a function $f(x)=\arccos(x)$ - a line $r$ tangent to the graph of $f$ at the point $P$ of abscissa $x_0=-\dfrac{1}{2}$ State whether the following statements are true (T) or false (F).