We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, and $f$ is $\alpha$-convex with $\alpha > 0$, that is $g(x) := f(x) - \frac{1}{2}\alpha x^2$ is a convex function on $\mathbb{R}$. Show that $f(x) \geq f(0) + f'(0)x + \alpha x^2/2$ for all $x \in \mathbb{R}$. Deduce that $f$ admits a minimizer on $\mathbb{R}$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, $f$ is $\alpha$-convex with $\alpha > 0$, and $x_*$ denotes a minimizer of $f$. Show that for all $x, y \in \mathbb{R}$ $$\alpha|x-y|^2 \leq \left(f'(x) - f'(y)\right)(x-y)$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. In this question only, we suppose that $f \in \mathcal{C}^1(\mathbb{R})$. Show that $x_1 := p_f(x_0)$ satisfies $$x_1 = x_0 - \tau f'(x_1)$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$, where $p_f(x_0)$ is the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. In this question only, we set $f(x) := |x|$ for all $x \in \mathbb{R}$. a) Show that $$p_f(x) = \begin{cases} x - \tau & \text{if } x \geq \tau \\ x + \tau & \text{if } x \leq -\tau \\ 0 & \text{otherwise} \end{cases}$$ b) Deduce that $x_n \rightarrow 0$ as $n \rightarrow \infty$, for all $x_0 \in \mathbb{R}$ and $\tau > 0$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$. Show that $\frac{1}{2}|x_1 - x_0|^2 + \tau f(x_1) \leq \tau f(x_0)$. Deduce that for all integers $N > M \geq 0$ $$\frac{1}{2}\sum_{M < n \leq N}|x_n - x_{n-1}|^2 \leq \tau\left(f(x_M) - f(x_N)\right)$$ Deduce that $|x_{n+1} - x_n| \rightarrow 0$ as $n \rightarrow \infty$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Let $x \in \mathbb{R}$ and $\tilde{x} := p_f(x)$. Show that for all $v \in \mathbb{R}$ and $t \in \mathbb{R}$ $$\tau f(\tilde{x}) + \frac{1}{2}|\tilde{x} - x|^2 \leq \tau f(\tilde{x} + tv) + \frac{1}{2}|\tilde{x} + tv - x|^2$$ Let also $y \in \mathbb{R}$, and $\tilde{y} := p_f(y)$. Deduce that $$2\tau(f(\tilde{x}) + f(\tilde{y}) - f(\tilde{x} + tv) - f(\tilde{y} - tv)) \leq |\tilde{x} + tv - x|^2 + |\tilde{y} - tv - y|^2 - |\tilde{x} - x|^2 - |\tilde{y} - y|^2$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Let $x, y \in \mathbb{R}$, $\tilde{x} := p_f(x)$, $\tilde{y} := p_f(y)$. Show that the right-hand side in inequality $$2\tau(f(\tilde{x}) + f(\tilde{y}) - f(\tilde{x} + tv) - f(\tilde{y} - tv)) \leq |\tilde{x} + tv - x|^2 + |\tilde{y} - tv - y|^2 - |\tilde{x} - x|^2 - |\tilde{y} - y|^2$$ admits the asymptotic expansion $2tv(\tilde{x} - x + y - \tilde{y}) + o(t)$ as $t \rightarrow 0$.

3. Show that on $] 0 , \infty [$ we have $f = g$. For this you may use a differential equation satisfied by $( f - g )$ and use the behavior of $f$ and $g$ at $+ \infty$.

32. The differential equation representing the family of curves $y 2 = 2 c ( x + \sqrt { } c )$, where $c$ is a positive parameter, is of :

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(A) order 1
(B) order 2
(C) degree 3
(D) 10

Consider the family of all circles whose centers lie on the straight line $y = x$. If this family of circles is represented by the differential equation $P y ^ { \prime \prime } + Q y ^ { \prime } + 1 = 0$, where $P , Q$ are functions of $x , y$ and $y ^ { \prime }$ (here $y ^ { \prime } = \frac { d y } { d x } , y ^ { \prime \prime } = \frac { d ^ { 2 } y } { d x ^ { 2 } }$), then which of the following statements is (are) true?
(A) $P = y + x$
(B) $P = y - x$
(C) $P + Q = 1 - x + y + y ^ { \prime } + \left( y ^ { \prime } \right) ^ { 2 }$
(D) $P - Q = x + y - y ^ { \prime } - \left( y ^ { \prime } \right) ^ { 2 }$

The differential equation of all circles passing through the origin and having their centres on the $x$-axis is
(1) $x ^ { 2 } = y ^ { 2 } + x y \frac { d y } { d x }$
(2) $x ^ { 2 } = y ^ { 2 } + 3 x y \frac { d y } { d x }$
(3) $y ^ { 2 } = x ^ { 2 } + 2 x y \frac { d y } { d x }$
(4) $y ^ { 2 } = x ^ { 2 } - 2 x y \frac { d y } { d x }$

Statement-1: The slope of the tangent at any point P on a parabola, whose axis is the axis of $x$ and vertex is at the origin, is inversely proportional to the ordinate of the point P. Statement-2: The system of parabolas $y ^ { 2 } = 4 a x$ satisfies a differential equation of degree 1 and order 1.
(1) Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.
(2) Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.
(3) Statement-1 is true; Statement-2 is false.
(4) Statement-1 is false; Statement-2 is true.

If the differential equation representing the family of all circles touching $x$-axis at the origin is $\left( x ^ { 2 } - y ^ { 2 } \right) \frac { d y } { d x } = g ( x ) y$, then $g ( x )$ equals
(1) $\frac { 1 } { 2 } x ^ { 2 }$
(2) $2 x$
(3) $\frac { 1 } { 2 } x$
(4) $2 x ^ { 2 }$

A differential equation representing the family of parabolas with axis parallel to $y$-axis and whose length of latus rectum is the distance of the point $( 2 , - 3 )$ from the line $3 x + 4 y = 5$, is given by: (1) $11 \frac { d ^ { 2 } x } { d y ^ { 2 } } = 10$ (2) $11 \frac { d ^ { 2 } y } { d x ^ { 2 } } = 10$ (3) $10 \frac { d ^ { 2 } y } { d x ^ { 2 } } = 11$ (4) $10 \frac { d ^ { 2 } x } { d y ^ { 2 } } = 11$

The differential equation of the family of circles passing through the points $( 0,2 )$ and $( 0 , - 2 )$ is
(1) $2 x y \frac { d y } { d x } + \left( x ^ { 2 } - y ^ { 2 } + 4 \right) = 0$
(2) $2 x y \frac { d y } { d x } + \left( x ^ { 2 } + y ^ { 2 } - 4 \right) = 0$
(3) $2 x y \frac { d y } { d x } + \left( y ^ { 2 } - x ^ { 2 } + 4 \right) = 0$
(4) $2 x y \frac { d y } { d x } - \left( x ^ { 2 } - y ^ { 2 } + 4 \right) = 0$

The differential equation of the family of circles passing through the origin and having centre at the line $y = x$ is :
(1) $\left( x ^ { 2 } - y ^ { 2 } + 2 x y \right) \mathrm { d } x = \left( x ^ { 2 } - y ^ { 2 } - 2 x y \right) \mathrm { d } y$
(2) $\left( x ^ { 2 } + y ^ { 2 } + 2 x y \right) \mathrm { d } x = \left( x ^ { 2 } + y ^ { 2 } - 2 x y \right) \mathrm { d } y$
(3) $\left( x ^ { 2 } + y ^ { 2 } - 2 x y \right) \mathrm { d } x = \left( x ^ { 2 } + y ^ { 2 } + 2 x y \right) \mathrm { d } y$
(4) $\left( x ^ { 2 } - y ^ { 2 } + 2 x y \right) \mathrm { d } x = \left( x ^ { 2 } - y ^ { 2 } + 2 x y \right) \mathrm { d } y$

Let $f ( x )$ be a positive function such that the area bounded by $y = f ( x ) , y = 0$ from $x = 0$ to $x = a > 0$ is $e ^ { - a } + 4 a ^ { 2 } + a - 1$. Then the differential equation, whose general solution is $y = c _ { 1 } f ( x ) + c _ { 2 }$, where $c _ { 1 }$ and $c _ { 2 }$ are arbitrary constants, is
(1) $\left( 8 e ^ { x } - 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } = 0$
(2) $\left( 8 e ^ { x } - 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - \frac { d y } { d x } = 0$
(3) $\left( 8 e ^ { x } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - \frac { d y } { d x } = 0$
(4) $\left( 8 e ^ { x } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } = 0$