All Tasks - Asymptote

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).

Consider $f(x)= -3\pi +a \arccos(3x+1)$ with $a \in \mathbb R$ and domain $D_f=\left[-\dfrac{2}{3}, 0\right]$. Knowing that the value of the differential of $f(x)$ at the point of abscissa $x_0=-\dfrac{1}{3}$ with $\Delta x=0.1$, is $df\left(-\dfrac{1}{3}\right)=-9$, then the value of $a$ is:

Consider the function $f(x)= 5 \arccos\left(2x\right)$ with domain $D_f = \left[ -\dfrac{1}{2}, \dfrac{1}{2} \right]$. The differential of the function $y=f(x)$ at the point of abscissa $x=0$ is:

Consider the function $f(x)= 5 \arccos\left(2x\right)$ with domain $D_f = \left[ -\dfrac{1}{2}, \dfrac{1}{2} \right]$. Using the notion of differential, the approximate value of $5 \arccos\left(0.2\right)$ is: