Deduce that, for all $t \in \mathbb{R}$, the application $(x,y) \mapsto \dfrac{1 - \left((x + \cos t)^2 + (y + \sin t)^2\right)}{x^2 + y^2}$ is harmonic on $\mathbb{R}^2 \backslash \{(0,0)\}$.

For $(x,y) \in D(0,1)$ fixed, we define the complex number $z = x + iy$ and we set for $t$ real (when the expression makes sense): $$\mathrm{N}(x,y,t) = \frac{1 - |z|^2}{|z - e^{it}|^2} = \frac{1 - (x^2 + y^2)}{(x - \cos t)^2 + (y - \sin t)^2}$$
Show that, for all $t \in \mathbb{R}$, the application $$\mathrm{N}_t : \left|\begin{array}{rll} D(0,1) & \rightarrow & \mathbb{R} \\ (x,y) & \mapsto & \mathrm{N}(x,y,t) \end{array}\right.$$ is harmonic.
One may use question II.B.3.

Let $f : C(0,1) \rightarrow \mathbb{R}$ be a continuous application. We define: $$\mathrm{N}_f(x,y) = \frac{1}{2\pi} \int_0^{2\pi} \mathrm{N}(x,y,t) f(\cos t, \sin t)\, \mathrm{d}t$$ on $D(0,1)$, and $$u(x,y) = \begin{cases} \mathrm{N}_f(x,y) & \text{if } (x,y) \in D(0,1) \\ f(x,y) & \text{if } (x,y) \in C(0,1) \end{cases}$$ on $\bar{D}(0,1)$.
a) Show that $\mathrm{N}_f$ admits a second-order partial derivative $\partial_{11} \mathrm{N}_f$ with respect to $x$.
Similarly, one can show that $\mathrm{N}_f$ admits second-order partial derivatives with respect to all its variables, continuous on $D(0,1)$. This result is admitted for the rest.
Express, for all $(x,y) \in D(0,1)$, for all $(i,j) \in \{1,2\}^2$, $\partial_{ij} \mathrm{N}_f(x,y)$ in terms of $\partial_{ij} \mathrm{N}(x,y,t)$.
b) Deduce that $u$ is harmonic on $D(0,1)$.

Let $f$ be defined on $I = ]-\pi/2, \pi/2[$ by $f(x) = \frac{\sin x + 1}{\cos x}$. Show $$\forall x \in I, \quad 2f^{\prime}(x) = f(x)^2 + 1.$$

Prove that the family of curves
$$\frac{x^{2}}{a^{2} + \lambda} + \frac{y^{2}}{b^{2} + \lambda} = 1$$
satisfies
$$\frac{dy}{dx}\left(a^{2} - b^{2}\right) = \left(x + y\frac{dy}{dx}\right)\left(x\frac{dy}{dx} - y\right).$$

Consider the functions defined implicitly by the equation $y ^ { 3 } - 3 y + x = 0$ on various intervals in the real line. If $x \in ( - \infty , - 2 ) \cup ( 2 , \infty )$, the equation implicitly defines a unique real valued differentiable function $y = f ( x )$. If $x \in ( - 2,2 )$, the equation implicitly defines a unique real valued differentiable function $y = g ( x )$ satisfying $g ( 0 ) = 0$.
If $f ( - 10 \sqrt { 2 } ) = 2 \sqrt { 2 }$, then $f ^ { \prime \prime } ( - 10 \sqrt { 2 } ) =$
(A) $\frac { 4 \sqrt { 2 } } { 7 ^ { 3 } 3 ^ { 2 } }$
(B) $- \frac { 4 \sqrt { 2 } } { 7 ^ { 3 } 3 ^ { 2 } }$
(C) $\frac { 4 \sqrt { 2 } } { 7 ^ { 3 } 3 }$
(D) $- \frac { 4 \sqrt { 2 } } { 7 ^ { 3 } 3 }$

The normal to the curve $x^2 + 2xy - 3y^2 = 0$ at $(1, 1)$:
(1) does not meet the curve again
(2) meets the curve again in the second quadrant
(3) meets the curve again in the third quadrant
(4) meets the curve again in the fourth quadrant

The normal to the curve $x ^ { 2 } + 2 x y - 3 y ^ { 2 } = 0$, at $( 1,1 )$
(1) Meets the curve again in the fourth quadrant
(2) Does not meet the curve again
(3) Meets the curve again in the second quadrant
(4) Meets the curve again in the third quadrant

If $x ^ { 2 } + y ^ { 2 } + \sin y = 4$, then the value of $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ at the point $( - 2,0 )$ is :
(1) $-34$
(2) 4
(3) $-2$
(4) $-32$

If $x ^ { 2 } + y ^ { 2 } + \sin y = 4$, then the value of $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ at the point $( - 2,0 )$ is
(1) - 34
(2) - 32
(3) - 2
(4) 4

Let $y = y ( x )$ be a function of $x$ satisfying $y \sqrt { 1 - x ^ { 2 } } = k - x \sqrt { 1 - y ^ { 2 } }$ where $k$ is a constant and $y \left( \frac { 1 } { 2 } \right) = - \frac { 1 } { 4 }$. Then $\frac { d y } { d x }$ at $x = \frac { 1 } { 2 }$, is equal to
(1) $- \frac { \sqrt { 5 } } { 4 }$
(2) $- \frac { \sqrt { 5 } } { 2 }$
(3) $\frac { 2 } { \sqrt { 5 } }$
(4) $\frac { \sqrt { 5 } } { 2 }$

Let $x ^ { k } + y ^ { k } = a ^ { k }$, $(a, k > 0)$ and $\frac { d y } { d x } + \left( \frac { y } { x } \right) ^ { \frac { 1 } { 3 } } = 0$, then $k$ is
(1) $\frac { 3 } { 2 }$
(2) $\frac { 4 } { 3 }$
(3) $\frac { 2 } { 3 }$
(4) $\frac { 1 } { 3 }$

If $y ^ { 2 } + \log _ { e } \left( \cos ^ { 2 } x \right) = y , \quad x \in \left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$ then:
(1) $y \prime \prime ( 0 ) = 0$
(2) $| y \prime ( 0 ) | + | y \prime \prime ( 0 ) | = 1$
(3) $| y \prime \prime ( 0 ) | = 2$
(4) $| y \prime ( 0 ) | + | y \prime \prime ( 0 ) | = 3$

Which of the following points lies on the tangent to the curve $x^4 e^y + 2\sqrt{y+1} = 3$ at the point $(1, 0)$?
(1) $(2, 2)$
(2) $(2, 6)$
(3) $(-2, 6)$
(4) $(-2, 4)$

For the curve $C : \left( x ^ { 2 } + y ^ { 2 } - 3 \right) + \left( x ^ { 2 } - y ^ { 2 } - 1 \right) ^ { 5 } = 0$, the value of $3 y ^ { \prime } - y ^ { 3 } y ^ { \prime \prime }$, at the point $( \alpha , \alpha ) , \alpha > 0$, on $C$, is equal to $\_\_\_\_$ .

Let the function $f ( x ) = \left( x ^ { 2 } + 1 \right) \left| x ^ { 2 } - a x + 2 \right| + \cos | x |$ be not differentiable at the two points $x = \alpha = 2$ and $x = \beta$. Then the distance of the point $( \alpha , \beta )$ from the line $12 x + 5 y + 10 = 0$ is equal to :
(1) 5
(2) 4
(3) 3
(4) 2

In the analytic plane
$$x y ^ { 2 } - x ^ { 3 } y - 6 = 0$$
Given that the tangent line passing through the point $\mathbf { P } \left( \mathbf { x } _ { 0 } , \mathbf { y } _ { 0 } \right)$ on the curve given by the equation is parallel to the x-axis, what is $\mathrm { x } _ { 0 }$?
A) $- 3$
B) $- 2$
C) $\frac { - 3 } { 2 }$

In the rectangular coordinate plane
$$y ^ { 2 } + \sin \left( x ^ { 2 } - 1 \right) = 4$$
What is the slope of the tangent line to the curve given by this equation at the point $\mathbf { P } ( - \mathbf { 1 } , - \mathbf { 2 } )$?
A) - 1
B) $\frac { 1 } { 2 }$
C) 2
D) $\frac { - 1 } { 2 }$