A function $f ( x )$ that is continuous on the entire set of real numbers satisfies the following conditions. (가) For $x \leq b$, $f ( x ) = a ( x - b ) ^ { 2 } + c$. (Here, $a$, $b$, and $c$ are constants.) (나) For all real numbers $x$, $f ( x ) = \int _ { 0 } ^ { x } \sqrt { 4 - 2 f ( t ) } \, dt$. When $\int _ { 0 } ^ { 6 } f ( x ) \, dx = \frac { q } { p }$, find the value of $p + q$. (Here, $p$ and $q$ are coprime natural numbers.) [4 points]

We denote by $K$ the function defined from $[0,1]^2$ to $\mathbb{R}$ by the following relation: $K(s,t) = (1-s)t$ if $0 \leq t \leq s \leq 1$ and $K(s,t) = (1-t)s$ otherwise. We denote by $T$ the application defined on $E = C([0,1], \mathbb{R})$, equipped with the norm $\|.\|_2 = \sqrt{\int_0^1 |f(x)|^2 \, dx}$, by the relation: $$\forall f \in E, \quad \forall s \in [0,1], \quad T(f)(s) = \int_0^1 K(s,t) f(t) \, dt$$ Let $f \in E$. By decomposing $T(f)$ into two integrals, show that $T(f)$ is a $C^2$ function and express $(T(f))'$ then $(T(f))''$.

Let $F$ be the vector subspace of $E$ formed of polynomial functions. For $k \in \mathbb{N}$, we denote by $p_k$ the function defined by $p_k(x) = x^k$. Deduce $(T(p))''$ for all $p \in F$.

For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t)\,\mathrm{d}t$$ where $k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$ For all $f \in E$, show that $T(f)$ is of class $\mathcal{C}^2$ then that $T(f)'' = -f$.

For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t)\,\mathrm{d}t$$ Show that $T$ is injective.

For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t)\,\mathrm{d}t$$ Determine the image of $T$.

We denote by $E$ the vector space of functions with real values continuous on $\mathbb { R } _ { + }$. For every element $f$ of $E$ and all $x \in \mathbb { R } _ { + }$ we set $F ( x ) = \int _ { 0 } ^ { x } f ( u ) \mathrm { d } u$.

1. Justify that $F$ is of class $C ^ { 1 }$ on $\mathbb { R } _ { + }$ and give for all $x \in \mathbb { R } _ { + }$ the expression of $F ^ { \prime } ( x )$.
   Let $\Psi : f \in E \mapsto \Psi ( f )$ defined by: $\forall x \in \mathbb { R } _ { + } , \Psi ( f ) ( x ) = \int _ { 0 } ^ { 1 } f ( x t ) \mathrm { d } t$.
2. Express, for all strictly positive real $x$, $\Psi ( f ) ( x )$ using $F ( x )$.
3. Justify that the function $\Psi ( f )$ is continuous on $\mathbb { R } _ { + }$ and give the value of $\Psi ( f ) ( 0 )$.
4. Show that $\Psi$ is an endomorphism of $E$.
5. Surjectivity of $\Psi$
   Let $h : x \in \mathbb { R } _ { + } \longmapsto h ( x ) = \left\{ \begin{array} { l l } x \sin \left( \frac { 1 } { x } \right) & \text { for } x > 0 \\ 0 & \text { for } x = 0 \end{array} \right.$.
   1. [5.1.] Show that the function $h$ is continuous on $\mathbb { R } _ { + }$.
   2. [5.2.] Is the function $h$ of class $C ^ { 1 }$ on $\mathbb { R } _ { + }$?
   3. [5.3.] Let $g \in \operatorname { Im } ( \Psi )$. Show that the function $x \mapsto x g ( x )$ is of class $C ^ { 1 }$ on $\mathbb { R } _ { + }$.
   4. [5.4.] Do we have $h \in \operatorname { Im } ( \Psi )$?
   5. [5.5.] Conclude.
6. Show that $\Psi$ is injective.
7. Search for the eigenvectors of $\Psi$
   1. [7.1.] Justify that 0 is not an eigenvalue of $\Psi$.
      Let $\mu \in \mathbb { R }$. We consider the differential equation $(L)$ on $\mathbb { R } _ { + } ^ { * }$: $$y ^ { \prime } + \frac { \mu } { x } y = 0$$
   2. [7.2.] Solve $(L)$ on $\mathbb { R } _ { + } ^ { * }$.
   3. [7.3.] Determine the solutions of $(L)$ that can be extended by continuity on $\mathbb { R } _ { + }$.
   4. [7.4.] Then determine the eigenvalues of $\Psi$ and the associated eigenspaces.
8. Let $n \in \mathbb { N } , n > 1$. For $i \in \llbracket 1 , n \rrbracket$, we set: $$f _ { i } : x \in \mathbb { R } _ { + } \longmapsto f _ { i } ( x ) = x ^ { i } \text { and } g _ { i } : x \in \mathbb { R } _ { + } \longmapsto g _ { i } ( x ) = \begin{cases} x ^ { i } \ln ( x ) & \text { for } x > 0 \\ 0 & \text { for } x = 0 \end{cases}$$ We denote $\mathscr { B } = \left( f _ { 1 } , \ldots , f _ { n } , g _ { 1 } , \ldots , g _ { n } \right)$ and $F _ { n }$ the vector subspace of $E$ generated by $\mathscr { B }$.
   1. [8.1.] We want to show that the family $\mathcal { B } = \left( f _ { 1 } , \ldots , f _ { n } , g _ { 1 } , \ldots , g _ { n } \right)$ is a basis of $F _ { n }$.
      Let $\left( \alpha _ { i } \right) _ { i \in \llbracket 1 , n \rrbracket }$ and $\left( \beta _ { j } \right) _ { j \in \llbracket 1 , n \rrbracket }$ be scalars such that $\sum _ { i = 1 } ^ { n } \alpha _ { i } f _ { i } + \sum _ { j = 1 } ^ { n } \beta _ { j } g _ { j } = 0$.
      1. [8.1.1.] Show that $\alpha _ { 1 } = \beta _ { 1 } = 0$. One may simplify the expression (*) by $x$ when $x$ is non-zero.
      2. [8.1.2.] Let $p \in \llbracket 1 , n - 1 \rrbracket$. Suppose that $\alpha _ { 1 } = \cdots = \alpha _ { p } = \beta _ { 1 } = \cdots = \beta _ { p } = 0$. Prove that $\alpha _ { p + 1 } = \beta _ { p + 1 } = 0$.
      3. [8.1.3.] Conclude and determine the dimension of the vector space $F _ { n }$.
   2. [8.2.] Where we prove that $\Psi$ induces an endomorphism on $F _ { n }$
      1. [8.2.1.] Let $x > 0$ and $p \in \mathbb { N } ^ { * }$. Show that the integral $\int _ { 0 } ^ { x } t ^ { p } \ln ( t ) \mathrm { d } t$ is convergent and calculate it.
      2. [8.2.2.] Deduce that $\Psi$ induces an endomorphism $\Psi _ { n }$ on $F _ { n }$.
   3. [8.3.] Give the matrix of the application $\Psi _ { n }$ in the basis $\mathcal { B }$.
   4. [8.4.] Prove that $\Psi _ { n }$ is an automorphism of $F _ { n }$.
   5. [8.5.] Let $z : x \in \mathbb { R } _ { + } \longmapsto z ( x ) = \left\{ \begin{array} { l l } \left( x + x ^ { 2 } \right) \ln ( x ) & \text { for } x > 0 \\ 0 & \text { for } x = 0 \end{array} \right.$. After verifying that $z \in F _ { n }$, determine $\Psi _ { n } ^ { - 1 } ( z )$.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that $$\frac{1}{2y} \int_{x-y}^{x+y} f(t)\, dt = f(x), \quad \text{for all } x \in \mathbb{R},\ y > 0$$ Show that there exist $a, b \in \mathbb{R}$ such that $f(x) = ax + b$ for all $x \in \mathbb{R}$.

Consider a differentiable function $u : [ 0,1 ] \rightarrow \mathbb { R }$. Assume the function $u$ satisfies $$u ( a ) = \frac { 1 } { 2 r } \int _ { a - r } ^ { a + r } u ( x ) d x , \quad \text{for all } a \in ( 0,1 ) \text{ and all } r < \min ( a , 1 - a ).$$ Which of the following four statements must be true?
(A) $u$ attains its maximum but not its minimum on the set $\{ 0,1 \}$.
(B) $u$ attains its minimum but not maximum on the set $\{ 0,1 \}$.
(C) If $u$ attains either its maximum or its minimum on the set $\{ 0,1 \}$, then it must be constant.
(D) $u$ attains both its maximum and its minimum on the set $\{ 0,1 \}$.

If $f : [ 0 , \infty ) \rightarrow \mathbb { R }$ is a continuous function such that $$f ( x ) + \ln 2 \int _ { 0 } ^ { x } f ( t ) d t = 1 , x \geq 0$$ then for all $x \geq 0$,
(A) $f ( x ) = e ^ { x } \ln 2$.
(B) $f ( x ) = e ^ { - x } \ln 2$.
(C) $f ( x ) = 2 ^ { x }$.
(D) $f ( x ) = \left( \frac { 1 } { 2 } \right) ^ { x }$.

Let $f$ be a non-negative function defined on the interval $[ 0,1 ]$. If
$$\int _ { 0 } ^ { x } \sqrt { 1 - \left( f ^ { \prime } ( t ) \right) ^ { 2 } } d t = \int _ { 0 } ^ { x } f ( t ) d t , \quad 0 \leq x \leq 1 ,$$
and $f ( 0 ) = 0$, then
(A) $f \left( \frac { 1 } { 2 } \right) < \frac { 1 } { 2 }$ and $f \left( \frac { 1 } { 3 } \right) > \frac { 1 } { 3 }$
(B) $f \left( \frac { 1 } { 2 } \right) > \frac { 1 } { 2 }$ and $f \left( \frac { 1 } { 3 } \right) > \frac { 1 } { 3 }$
(C) $f \left( \frac { 1 } { 2 } \right) < \frac { 1 } { 2 }$ and $f \left( \frac { 1 } { 3 } \right) < \frac { 1 } { 3 }$
(D) $f \left( \frac { 1 } { 2 } \right) > \frac { 1 } { 2 }$ and $f \left( \frac { 1 } { 3 } \right) < \frac { 1 } { 3 }$

Let $f:\mathbf{R}\rightarrow\mathbf{R}$ be a continuous function which satisfies $$f(x)=\int_{0}^{x}f(t)\,dt.$$ Then the value of $f(\ln5)$ is

Let $f : [ 0 , \infty ) \rightarrow \mathbb { R }$ be a continuous function such that $$f ( x ) = 1 - 2 x + \int _ { 0 } ^ { x } e ^ { x - t } f ( t ) d t$$ for all $x \in [ 0 , \infty )$. Then, which of the following statement(s) is (are) TRUE?
(A) The curve $y = f ( x )$ passes through the point $( 1,2 )$
(B) The curve $y = f ( x )$ passes through the point $( 2 , - 1 )$
(C) The area of the region $\left\{ ( x , y ) \in [ 0,1 ] \times \mathbb { R } : f ( x ) \leq y \leq \sqrt { 1 - x ^ { 2 } } \right\}$ is $\frac { \pi - 2 } { 4 }$
(D) The area of the region $\left\{ ( x , y ) \in [ 0,1 ] \times \mathbb { R } : f ( x ) \leq y \leq \sqrt { 1 - x ^ { 2 } } \right\}$ is $\frac { \pi - 1 } { 4 }$

Let $f : [ 1 , \infty ) \rightarrow \mathbb { R }$ be a differentiable function such that $f ( 1 ) = \frac { 1 } { 3 }$ and $3 \int _ { 1 } ^ { x } f ( t ) d t = x f ( x ) - \frac { x ^ { 3 } } { 3 } , x \in [ 1 , \infty )$. Let $e$ denote the base of the natural logarithm. Then the value of $f ( e )$ is
(A) $\frac { e ^ { 2 } + 4 } { 3 }$
(B) $\frac { \log _ { e } 4 + e } { 3 }$
(C) $\frac { 4 e ^ { 2 } } { 3 }$
(D) $\frac { e ^ { 2 } - 4 } { 3 }$

For $x \in R , x \neq 0$, if $y ( x )$ is a differentiable function such that $x \int _ { 1 } ^ { x } y ( t ) d t = ( x + 1 ) \int _ { 1 } ^ { x } t y ( t ) d t$, then $y ( x )$ equals (where $C$ is a constant)
(1) $C x ^ { 3 } e ^ { \frac { 1 } { x } }$
(2) $\frac { C } { x ^ { 2 } } e ^ { - \frac { 1 } { x } }$
(3) $\frac { C } { x } e ^ { - \frac { 1 } { x } }$
(4) $\frac { C } { x ^ { 3 } } e ^ { - \frac { 1 } { x } }$

Let $f:[0,1] \rightarrow R$ be such that $f(xy) = f(x) \cdot f(y)$, for all $x,y \in [0,1]$, and $f(0) \neq 0$. If $y = y(x)$ satisfies the differential equation, $\frac{dy}{dx} = f(x)$ with $y(0) = 1$ then $y\left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right)$ is equal to:
(1) 5
(2) 2
(3) 3
(4) 4

Let $f ( x ) = \int _ { 0 } ^ { x } e ^ { t } f ( t ) d t + e ^ { x }$ be a differentiable function for all $x \in R$. Then $f ( x )$ equals:
(1) $e ^ { \left( e ^ { x } - 1 \right) }$
(2) $e ^ { e ^ { x } } - 1$
(3) $2 e ^ { e ^ { x } } - 1$

Let $f$ be a differentiable function satisfying $f ( x ) = \frac { 2 } { \sqrt { 3 } } \int _ { 0 } ^ { \sqrt { 3 } } f \left( \frac { \lambda ^ { 2 } x } { 3 } \right) d \lambda , x > 0$ and $f ( 1 ) = \sqrt { 3 }$. If $y = f ( x )$ passes through the point $( \alpha , 6 )$, then $\alpha$ is equal to $\_\_\_\_$ .

Let $f$ be a differentiable function such that $x^2 f(x) - x = 4\int_0^x t\, f(t)\, dt$, $f(1) = \frac{2}{3}$. Then $18\, f(3)$ is equal to
(1) 210
(2) 160
(3) 150
(4) 180

Let $\int _ { 0 } ^ { x } \sqrt { 1 - \left( y ^ { \prime } ( t ) \right) ^ { 2 } } d t = \int _ { 0 } ^ { x } y ( t ) d t , 0 \leq x \leq 3 , y \geq 0 , y ( 0 ) = 0$. Then at $x = 2 , y ^ { \prime \prime } + y + 1$ is equal to
(1) 1
(2) 2
(3) $\sqrt { 2 }$
(4) $1 / 2$

Let $f$ be a differentiable function such that $2(x+2)^2 f(x) - 3(x+2)^2 = 10\int_0^x (t+2)f(t)\,dt$, $x \geq 0$. Then $f(2)$ is equal to $\underline{\hspace{2cm}}$.

For each of Q , S , V in the following sentences, choose the appropriate expression from among (0) $\sim$ (7) at the bottom of this page. For the other $\square$, enter the correct number.
Suppose we have a differentiable function $f ( x )$ which satisfies the equation
$$\int _ { 0 } ^ { x } f ( t ) d t = \left( 1 + e ^ { - x } \right) f ( x ) + 2 x - 4 \log 2 \tag{1}$$
We are to find $f ( x )$ and the value of $\lim _ { x \rightarrow \infty } f ( x )$.
When we differentiate each side of (1) with respect to $x$ and transform the equation, we have
$$\left( 1 + e ^ { - x } \right) ( \mathbf { Q } ) = \mathbf { R } . \tag{2}$$
Next we set $f ( x ) = e ^ { x } g ( x )$, and using (2), we obtain
$$g ^ { \prime } ( x ) = \frac { \mathbf { S } } { 1 + e ^ { - x } }$$
and hence
$$g ( x ) = \mathbf { T } \log \left( 1 + e ^ { - x } \right) + C ,$$
where $C$ is an integral constant. Furthermore, since $g ( 0 ) = f ( 0 )$, we see that $C = \mathbf { U }$. Thus we obtain $g ( x )$ and from that,
$$f ( x ) = \mathbf { V } \log \left( 1 + e ^ { - x } \right) .$$
Finally, we set $e ^ { - x } = t$ and obtain
$$f ( x ) = \mathbf { W } \log ( 1 + t ) ^ { \frac { 1 } { t } }$$
and hence
$$\lim _ { x \rightarrow \infty } f ( x ) = \lim _ { t \rightarrow \mathbf { X } } \mathbf { W } \log ( 1 + t ) ^ { \frac { 1 } { t } } = \mathbf { Y }$$
Choices: (0) $f ^ { \prime } ( x ) - f ( x )$
(1) $f ( x ) - f ^ { \prime } ( x )$
(2) $f ^ { \prime } ( x ) - 2 f ( x )$
(3) $f ( x ) - 2 f ^ { \prime } ( x )$
(4) $2 e ^ { x }$
(5) $- 2 e ^ { x }$ (6) $2 e ^ { - x }$ (7) $- 2 e ^ { - x }$