The function $h$ is differentiable, and for all values of $x$, $h ( x ) = h ( 2 - x )$. Which of the following statements must be true?
I. $\int _ { 0 } ^ { 2 } h ( x ) d x > 0$
II. $h ^ { \prime } ( 1 ) = 0$
III. $h ^ { \prime } ( 0 ) = h ^ { \prime } ( 2 ) = 1$
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Let $f : \mathbb { R } _ { \geq 0 } \longrightarrow \mathbb { R }$ be the function
$$f ( x ) = \begin{cases} 1 , & x = 0 \\ x ^ { - x } , & x > 0 \end{cases}$$
Determine whether the following statement is true:
$$\int _ { 0 } ^ { 1 } f ( x ) \mathrm { d } x = \sum _ { i = 0 } ^ { \infty } n ^ { - n }$$

The following is the graph of a continuous function $y = f ( x )$.
When the inverse function $g ( x )$ of function $f ( x )$ exists and is continuous on the interval $[ 0,1 ]$, the limit value $$\lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } \left\{ g \left( \frac { k } { n } \right) - g \left( \frac { k - 1 } { n } \right) \right\} \frac { k } { n }$$ has the same value as which of the following? [4 points]
(1) $\int _ { 0 } ^ { 1 } g ( x ) d x$
(2) $\int _ { 0 } ^ { 1 } x g ( x ) d x$
(3) $\int _ { 0 } ^ { 1 } f ( x ) d x$
(4) $\int _ { 0 } ^ { 1 } x f ( x ) d x$
(5) $\int _ { 0 } ^ { 1 } \{ f ( x ) - g ( x ) \} d x$

Show that $x \mapsto \int _ { 0 } ^ { \pi } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) \mathrm { d } \theta$ is an even function of the variable $x \in \mathbb { R }$.

Show that $\int _ { 0 } ^ { \pi } \ln ( \sin \theta ) \mathrm { d } \theta = 2 \int _ { 0 } ^ { \pi / 2 } \ln ( \sin \theta ) \mathrm { d } \theta = 2 \int _ { 0 } ^ { \pi / 2 } \ln ( \cos \theta ) \mathrm { d } \theta$.

Let $f$ be the function from $\mathbb { R }$ to $\mathbb { R }$ defined by $f ( x ) = \int _ { 0 } ^ { \pi } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) \mathrm { d } \theta$.
Deduce that $\forall x \in \mathbb { R } \backslash \{ - 1,1 \}$ $$f ^ { \prime } ( x ) = 4 \int _ { 0 } ^ { + \infty } \frac { ( x + 1 ) t ^ { 2 } + ( x - 1 ) } { \left( ( x + 1 ) ^ { 2 } t ^ { 2 } + ( x - 1 ) ^ { 2 } \right) \left( t ^ { 2 } + 1 \right) } \mathrm { d } t$$

Deduce $\int _ { 0 } ^ { \pi } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) \mathrm { d } \theta$ for $x \in \mathbb { R } \backslash \{ - 1,1 \}$.

Deduce that $$\int _ { 0 } ^ { \pi / 2 } \ln ( \sin \theta ) d \theta = - \pi \frac { \ln 2 } { 2 }$$
Then recover the result from question III.B.6.

For $(x,y) \in D(0,1)$ fixed, we define the complex number $z = x + iy$ and we set for $t$ real: $$\mathrm{N}(x,y,t) = \frac{1 - |z|^2}{|z - e^{it}|^2} = \frac{1 - (x^2 + y^2)}{(x - \cos t)^2 + (y - \sin t)^2}$$
Deduce that $\dfrac{1}{2\pi} \int_0^{2\pi} \mathrm{N}(x,y,t)\, \mathrm{d}t = 1$.
One may write $\dfrac{1}{1 - ze^{-it}}$ in the form of the sum of a series of functions.

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$.
Show that for all real $x > 0$ and $y > 0 , \beta ( x , y ) = \beta ( y , x )$.

Let $f \in \mathcal{S}$. We assume that $\mathcal{F}(f)$ is integrable on $\mathbb{R}$. For every positive natural number $n$, we set
$$I_{n} = \int_{-\infty}^{+\infty} \mathcal{F}(f)(\xi) \theta\left(\frac{\xi}{n}\right) \mathrm{d}\xi \quad J_{n} = \int_{-\infty}^{+\infty} f\left(\frac{t}{n}\right) \mathcal{F}(\theta)(t) \mathrm{d}t$$
Prove that $\forall n \in \mathbb{N}^{*}, I_{n} = J_{n}$.
We will admit the Fubini formula:
$$\int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} f(t) \theta\left(\frac{\xi}{n}\right) e^{-2\pi\mathrm{i} t\xi} \mathrm{d}\xi\right) \mathrm{d}t = \int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} f(t) \theta\left(\frac{\xi}{n}\right) e^{-2\pi\mathrm{i} t\xi} \mathrm{d}t\right) \mathrm{d}\xi$$

For every natural number $n$, we denote by $S_{n}$ the function defined on $\mathbb{R}$ by
$$\forall x \in \mathbb{R}, \quad S_{n}(x) = \sum_{k=-n}^{n} e^{2\pi\mathrm{i} kx}$$
Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be a function of class $C^{\infty}$ on $\mathbb{R}$ and 1-periodic. We consider the function $g$ defined on $[-1,1]$ by
$$\forall x \in ]-1,1[\backslash\{0\}, \quad g(x) = \frac{f(x)-f(0)}{\sin(\pi x)} \quad g(0) = \frac{f'(0)}{\pi} \quad g(1) = g(-1) = -g(0)$$
and the sequence of complex numbers $(c_{n}(f))_{n \in \mathbb{Z}}$ defined by
$$\forall n \in \mathbb{Z}, \quad c_{n}(f) = \int_{-1/2}^{1/2} f(x) e^{-2\pi\mathrm{i} nx} \mathrm{d}x$$
Justify that
$$\forall n \in \mathbb{N}^{*}, \quad \sum_{k=-n}^{n} c_{k}(f) = f(0) + \int_{-1/2}^{1/2} g(x) \sin((2n+1)\pi x) \mathrm{d}x$$

By observing that the function $|\cos|$ is $2\pi$-periodic, calculate, for $\omega \in \mathbf{R}$, the integral $$\int_{-\pi}^{\pi} |\cos(u - \omega)| \mathrm{d}u$$ Deduce that, if $(a,b) \in \mathbf{R}^{2}$, $$\int_{-\pi}^{\pi} |a\cos(u) + b\sin(u)| \mathrm{d}u = 4\sqrt{a^{2} + b^{2}}$$

The value of the integral $$\int _ { \pi / 2 } ^ { 5 \pi / 2 } \frac { e ^ { \tan ^ { - 1 } ( \sin x ) } } { e ^ { \tan ^ { - 1 } ( \sin x ) } + e ^ { \tan ^ { - 1 } ( \cos x ) } } d x$$ equals (A) 1 (B) $\pi$ (C) $e$ (D) none of these

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a twice differentiable one-to-one function. If $f ( 2 ) = 2 , f ( 3 ) = - 8$ and $$\int _ { 2 } ^ { 3 } f ( x ) d x = - 3$$ then $$\int _ { - 8 } ^ { 2 } f ^ { - 1 } ( x ) d x$$ equals
(A) $- 25$.
(B) $25$.
(C) $- 31$.
(D) $31$.

The value of the integral $\int _ { \pi / 2 } ^ { 5 \pi / 2 } \frac { e ^ { \tan ^ { - 1 } ( \sin x ) } } { e ^ { \tan ^ { - 1 } ( \sin x ) } + e ^ { \tan ^ { - 1 } ( \cos x ) } } d x$ equals
(a) 1 .
(B) $\pi$.
(C) $e$.
(D) none of these.

Match the statements in Column I with the values in Column II.
Column I
(A) $\int_{-\pi}^{\pi} \cos^2 x\,\frac{1}{1+a^x}\,dx$, $a > 0$
(B) $\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$
(C) $\int_{-2}^{2} \frac{x^2}{1+5^x}\,dx$
(D) $\int_1^2 \frac{\sqrt{\ln(3-x)}}{\sqrt{\ln(3-x)}+\sqrt{\ln(x+1)}}\,dx$
Column II
(p) $\frac{1}{2}$
(q) $0$
(r) $\frac{\pi}{4}$
(s) $\frac{\pi}{2}$

Let $f ( x )$ be a non-constant twice differentiable function defined on $( - \infty , \infty )$ such that $f ( x ) = f ( 1 - x )$ and $f ^ { \prime } \left( \frac { 1 } { 4 } \right) = 0$. Then,
(A) $f ^ { \prime \prime } ( x )$ vanishes at least twice on $[ 0,1 ]$
(B) $f ^ { \prime } \left( \frac { 1 } { 2 } \right) = 0$
(C) $\quad \int _ { - 1 / 2 } ^ { 1 / 2 } f \left( x + \frac { 1 } { 2 } \right) \sin x d x = 0$
(D) $\int _ { 0 } ^ { 1 / 2 } f ( t ) e ^ { \sin \pi t } d t = \int _ { 1 / 2 } ^ { 1 } f ( 1 - t ) e ^ { \sin \pi t } d t$

If $$I_{n}=\int_{-\pi}^{\pi}\frac{\sin nx}{\left(1+\pi^{x}\right)\sin x}dx,\quad n=0,1,2,\ldots,$$ then
(A) $I_{n}=I_{n+2}$
(B) $\sum_{m=1}^{10}I_{2m+1}=10\pi$
(C) $\sum_{m=1}^{10}I_{2m}=0$
(D) $I_{n}=I_{n+1}$

List I	List II
P. The number of polynomials $f(x)$ with non-negative integer coefficients of degree $\leq 2$, satisfying $f(0) = 0$ and $\int_{0}^{1} f(x)\,dx = 1$, is	1. 8
Q. The number of points in the interval $[-\sqrt{13}, \sqrt{13}]$ at which $f(x) = \sin(x^2) + \cos(x^2)$ attains its maximum value, is	2. 2
R. $\int_{-2}^{2} \frac{3x^2}{1+e^x}\,dx$ equals	3. 4
S. $\dfrac{\displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos 2x \log\left(\frac{1+x}{1-x}\right)dx}{\displaystyle\int_{0}^{\frac{1}{2}} \cos 2x \log\left(\frac{1+x}{1-x}\right)dx}$ equals	4. 0

P Q R S
(A) 3241
(B) 2341
(C) 3214
(D) 2314

The option(s) with the values of $a$ and $L$ that satisfy the following equation is(are)
$$\frac { \int _ { 0 } ^ { 4 \pi } e ^ { t } \left( \sin ^ { 6 } a t + \cos ^ { 4 } a t \right) d t } { \int _ { 0 } ^ { \pi } e ^ { t } \left( \sin ^ { 6 } a t + \cos ^ { 4 } a t \right) d t } = L ?$$
(A) $\quad a = 2 , L = \frac { e ^ { 4 \pi } - 1 } { e ^ { \pi } - 1 }$
(B) $\quad a = 2 , L = \frac { e ^ { 4 \pi } + 1 } { e ^ { \pi } + 1 }$
(C) $\quad a = 4 , L = \frac { e ^ { 4 \pi } - 1 } { e ^ { \pi } - 1 }$
(D) $\quad a = 4 , L = \frac { e ^ { 4 \pi } + 1 } { e ^ { \pi } + 1 }$

The value of $\int _ { - \frac { \pi } { 2 } } ^ { \frac { \pi } { 2 } } \frac { x ^ { 2 } \cos x } { 1 + e ^ { x } } d x$ is equal to
(A) $\frac { \pi ^ { 2 } } { 4 } - 2$
(B) $\frac { \pi ^ { 2 } } { 4 } + 2$
(C) $\pi ^ { 2 } - e ^ { \frac { \pi } { 2 } }$
(D) $\pi ^ { 2 } + e ^ { \frac { \pi } { 2 } }$

If $$I = \frac { 2 } { \pi } \int _ { - \pi / 4 } ^ { \pi / 4 } \frac { d x } { \left( 1 + e ^ { \sin x } \right) ( 2 - \cos 2 x ) }$$ then $27 I ^ { 2 }$ equals

The value of the integral $$\int_0^{\pi/2} \frac{3\sqrt{\cos\theta}}{(\sqrt{\cos\theta} + \sqrt{\sin\theta})^5}\,d\theta$$ equals

Let the function $f: [0,1] \rightarrow \mathbb{R}$ be defined by $$f(x) = \frac{4^{x}}{4^{x} + 2}$$ Then the value of $$f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + f\left(\frac{3}{40}\right) + \cdots + f\left(\frac{39}{40}\right) - f\left(\frac{1}{2}\right)$$ is $\_\_\_\_$