Let $r$ and $s$ be two strictly positive natural integers such that $r > s$, and
$$J _ { r , s } = \sum _ { k = 0 } ^ { + \infty } \frac { 1 } { ( r + k + 1 ) ( s + k + 1 ) }$$
Deduce that
$$J _ { r , s } = \frac { 1 } { r - s } \sum _ { k = 0 } ^ { + \infty } \left( \frac { 1 } { s + k + 1 } - \frac { 1 } { r + k + 1 } \right)$$

In this subsection, we assume that $J_n = J_n^{(\mathrm{C})}$, the matrix introduced in subsection A-II.
We set $G_h : x \longmapsto \frac{(x-h)^2}{2\beta} - \ln(2\operatorname{ch}(x))$.
Show that $$Z_n(h) = \sqrt{\frac{n}{2\mathrm{e}^{\beta}\pi\beta}} \int_{-\infty}^{+\infty} \mathrm{e}^{-nG_h(x)} \mathrm{d}x$$

In this subsection, we assume that $J_n = J_n^{(\mathrm{C})}$, the matrix introduced in subsection A-II.
We set $G_h : x \longmapsto \frac{(x-h)^2}{2\beta} - \ln(2\operatorname{ch}(x))$. We now assume that $h > 0$.
We denote $\widehat{G}_h : x \longmapsto G_h(x + u_h) - \min G_h$. Show that $$\psi_n(h) = -G_h(u_h) - \frac{1}{2n}\ln\left(2\mathrm{e}^{\beta}\pi\beta\right) + \frac{1}{n}\ln\left(\int_{-\infty}^{+\infty} \mathrm{e}^{-n\widehat{G}_h\left(\frac{t}{\sqrt{n}}\right)} \mathrm{d}t\right)$$

We define on $[ 0,1 ]$ the function $P _ { n }$ by:
$$\forall x \in [ 0,1 ] , \quad P _ { n } ( x ) = \frac { 1 } { n ! } \frac { \mathrm { d } ^ { n } \left( x ^ { n } ( 1 - x ) ^ { n } \right) } { \mathrm { d } x ^ { n } } .$$
Let $n \in \mathbb { N } ^ { * }$. Show that for all $y \in ] 0,1 [$,
$$\int _ { 0 } ^ { 1 } \frac { P _ { n } ( x ) } { 1 - x y } \mathrm {~d} x = ( - y ) ^ { n } \int _ { 0 } ^ { 1 } \frac { x ^ { n } ( 1 - x ) ^ { n } } { ( 1 - x y ) ^ { n + 1 } } \mathrm {~d} x$$

We define on $[ 0,1 ]$ the function $P _ { n }$ by:
$$\forall x \in [ 0,1 ] , \quad P _ { n } ( x ) = \frac { 1 } { n ! } \frac { \mathrm { d } ^ { n } \left( x ^ { n } ( 1 - x ) ^ { n } \right) } { \mathrm { d } x ^ { n } } .$$
We set $I _ { n } = \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 1 } \frac { ( 1 - y ) ^ { n } P _ { n } ( x ) } { 1 - x y } \mathrm {~d} x \mathrm {~d} y$.
Deduce that
$$I _ { n } = ( - 1 ) ^ { n } \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 1 } \frac { x ^ { n } ( 1 - x ) ^ { n } y ^ { n } ( 1 - y ) ^ { n } } { ( 1 - x y ) ^ { n + 1 } } \mathrm {~d} x \mathrm {~d} y$$

In this subsection, we assume that $J_n = J_n^{(\mathrm{C})}$, the matrix introduced in subsection A-II.
We set $G_h : x \longmapsto \frac{(x-h)^2}{2\beta} - \ln(2\operatorname{ch}(x))$. We now assume that $h > 0$.
We set $\gamma_h = G_h''(u_h)$ and we denote $f_h : x \longmapsto \frac{\widehat{G}_h(x)}{x^2}$.
Show then that $$\int_{-\infty}^{+\infty} \mathrm{e}^{-n\widehat{G}_h\left(\frac{t}{\sqrt{n}}\right)} \mathrm{d}t \xrightarrow[n \rightarrow +\infty]{} \sqrt{\frac{2\pi}{\gamma_h}}$$ then conclude that $\psi(h) = -G_h(u_h)$.

Let $a$ and $\beta$ be two positive real numbers. For every integer $n > 0$, define $a_n = \int_{\beta}^{n} \frac{a}{u(u^{a}+2+u^{-a})} du$. Then $\lim_{n \to \infty} a_n$ is equal to
(a) $1/(1+\beta^{a})$
(b) $\beta^{a}/(1+\beta^{-a})$
(c) $\beta^{a}/(1+\beta^{a})$
(d) $\beta^{-a}/(1+\beta^{a})$

If $b = \int _ { 0 } ^ { 1 } \frac { e ^ { t } } { t + 1 } d t$ then $\int _ { a - 1 } ^ { a } \frac { e ^ { - t } } { t - a - 1 } d t$ is
(A) $b e ^ { a }$
(B) $b e ^ { - a }$
(C) $- b e ^ { - a }$
(D) $- b e ^ { a }$

If $b = \int _ { 0 } ^ { 1 } \frac { e ^ { t } } { t + 1 } d t$ then $\int _ { a - 1 } ^ { a } \frac { e ^ { - t } } { t - a - 1 } d t$ is
(A) $b e ^ { a }$
(B) $b e ^ { - a }$
(C) $- b e ^ { - a }$
(D) $- b e ^ { a }$

If $b = \int _ { 0 } ^ { 1 } \frac { e ^ { t } } { t + 1 } d t$ then $\int _ { a - 1 } ^ { a } \frac { e ^ { - t } } { t - a - 1 } d t$ is
(A) $b e ^ { a }$
(B) $b e ^ { - a }$
(C) $- b e ^ { - a }$
(D) $- b e ^ { a }$

Let $$f ( x ) = e ^ { - | x | } , x \in \mathbb { R }$$ and $$g ( \theta ) = \int _ { - 1 } ^ { 1 } f \left( \frac { x } { \theta } \right) d x , \theta \neq 0$$ Then, $$\lim _ { \theta \rightarrow 0 } \frac { g ( \theta ) } { \theta }$$ (A) equals 0 .
(B) equals $+ \infty$.
(C) equals 2 .
(D) does not exist.

6. Consider the integral function $F ( x ) = \int _ { a } ^ { x } \frac { \cos \left( \frac { 1 } { t } \right) } { t ^ { 2 } } d t , \text{ with } x \geq a$, in which $a$ denotes a positive real parameter. Determine the largest value of $a$ so that $F \left( \frac { 2 } { \pi } \right) = - \frac { 1 } { 2 }$.

5. Evaluate $\int \sin - 1 ( ( 2 \mathrm { x } + 2 ) / \sqrt { } ( 4 \mathrm { x } 2 + 8 \mathrm { x } + 13 ) ) d \mathrm { x }$.

9. For any natural number m , evaluate
$$\int \left( x ^ { 3 m } + x ^ { 2 m } + x ^ { m } \right) \left( 2 x ^ { 2 m } + 3 x ^ { m } + 6 \right) ^ { \frac { 1 } { m } } d x , x > 0$$

1. Let

$$\begin{aligned} & \text { Let } f ( x ) = \left\{ \begin{array} { c l } x + a , & x < 0 \\ | x - 1 | , & x \geq 0 , \end{array} \right. \\ & \text { And } g ( x ) = \left\{ \begin{array} { c l } x + 1 & \text { if } x < 0 \\ ( x - 1 ) ^ { 2 } + b & \text { if } x \geq 0 \end{array} \right. \end{aligned}$$
where a and b aneegetive real numbers. Determine the composite function gof. (If (gof) (x) is continuous for all real x , determine the values of a and b . Further, for these values of a and b , is gof differentiable at $\mathrm { x } = 0$ ? Justify your answer.

20. If $I ( m , n ) = \int _ { 0 } { } ^ { 1 } t ^ { m } ( 1 + t ) ^ { n } \mathrm { dt }$, then the expression for $I ( m , n )$ in terms of $I ( m + 1 , n$ 1) is:
(a) $\left. \left. 2 ^ { n } / m + 1 \right) - n / m + 1 \right) * I ( m + 1 , n - 1 )$
(b) $n / m + 1 ) * \mid ( m + 1 , n - 1 )$
(c) $\quad 2 ^ { n } / ( m + 1 ) + n / ( m + 1 ) * I ( m + 1 , n - 1 )$
(b) $\quad m / ( m + 1 ) * I ( m + 1 , n - 1 )$

2. $\quad \int \frac { x ^ { 2 } - 1 } { x ^ { 3 } \sqrt { 2 x ^ { 4 } - 2 x ^ { 2 } + 1 } } d x$ is equal to
(A) $\frac { \sqrt { 2 x ^ { 4 } - 2 x ^ { 2 } + 1 } } { x ^ { 2 } } + c$
(B) $\frac { \sqrt { 2 x ^ { 4 } - 2 x ^ { 2 } + 1 } } { x ^ { 3 } } + c$
(C) $\frac { \sqrt { 2 x ^ { 4 } - 2 x ^ { 2 } + 1 } } { x } + c$
(D) $\frac { \sqrt { 2 x ^ { 4 } - 2 x ^ { 2 } + 1 } } { 2 x ^ { 2 } } + c$
Sol. (D)
$$\int \frac { \left( \frac { 1 } { x ^ { 3 } } - \frac { 1 } { x ^ { 5 } } \right) d x } { \sqrt { 2 - \frac { 2 } { x ^ { 2 } } + \frac { 1 } { x ^ { 4 } } } }$$
Let $2 - \frac { 2 } { x ^ { 2 } } + \frac { 1 } { x ^ { 4 } } = z \Rightarrow \frac { 1 } { 4 } \int \frac { d z } { \sqrt { z } }$ $\Rightarrow \quad \frac { 1 } { 2 } \times \sqrt { \mathrm { z } } + \mathrm { c } \Rightarrow \frac { 1 } { 2 } \sqrt { 2 - \frac { 2 } { \mathrm { x } ^ { 2 } } + \frac { 1 } { \mathrm { x } ^ { 4 } } } + \mathrm { c }$.

Let
$$I = \int \frac { e ^ { x } } { e ^ { 4 x } + e ^ { 2 x } + 1 } d x , \quad J = \int \frac { e ^ { - x } } { e ^ { - 4 x } + e ^ { - 2 x } + 1 } d x$$
Then, for an arbitrary constant $C$, the value of $J - I$ equals
(A) $\frac { 1 } { 2 } \log \left( \frac { e ^ { 4 x } - e ^ { 2 x } + 1 } { e ^ { 4 x } + e ^ { 2 x } + 1 } \right) + C$
(B) $\frac { 1 } { 2 } \log \left( \frac { e ^ { 2 x } + e ^ { x } + 1 } { e ^ { 2 x } - e ^ { x } + 1 } \right) + C$
(C) $\frac { 1 } { 2 } \log \left( \frac { e ^ { 2 x } - e ^ { x } + 1 } { e ^ { 2 x } + e ^ { x } + 1 } \right) + C$
(D) $\frac { 1 } { 2 } \log \left( \frac { e ^ { 4 x } + e ^ { 2 x } + 1 } { e ^ { 4 x } - e ^ { 2 x } + 1 } \right) + C$

The value(s) of $\int _ { 0 } ^ { 1 } \frac { x ^ { 4 } ( 1 - x ) ^ { 4 } } { 1 + x ^ { 2 } } d x$ is (are)
A) $\frac { 22 } { 7 } - \pi$
B) $\frac { 2 } { 105 }$
C) 0
D) $\frac { 71 } { 15 } - \frac { 3 \pi } { 2 }$

The following integral
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2\operatorname{cosec} x)^{17}\, dx$$
is equal to
(A) $\int_{0}^{\log(1+\sqrt{2})} 2\left(e^{u} + e^{-u}\right)^{16} du$
(B) $\int_{0}^{\log(1+\sqrt{2})} \left(e^{u} + e^{-u}\right)^{17} du$
(C) $\int_{0}^{\log(1+\sqrt{2})} \left(e^{u} - e^{-u}\right)^{17} du$
(D) $\int_{0}^{\log(1+\sqrt{2})} 2\left(e^{u} - e^{-u}\right)^{16} du$

Let $f : (0, \infty) \rightarrow \mathbb{R}$ be given by $$f(x) = \int_{\frac{1}{x}}^{x} e^{-\left(t + \frac{1}{t}\right)} \frac{dt}{t}$$ Then
(A) $f(x)$ is monotonically increasing on $[1, \infty)$
(B) $f(x)$ is monotonically decreasing on $(0,1)$
(C) $f(x) + f\left(\frac{1}{x}\right) = 0$, for all $x \in (0, \infty)$
(D) $f\left(2^x\right)$ is an odd function of $x$ on $\mathbb{R}$

Let $f : [0,2] \rightarrow \mathbb{R}$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0) = 1$. Let
$$F(x) = \int_{0}^{x^2} f(\sqrt{t})\, dt$$
for $x \in [0,2]$. If $F'(x) = f'(x)$ for all $x \in (0,2)$, then $F(2)$ equals
(A) $e^2 - 1$
(B) $e^4 - 1$
(C) $e - 1$
(D) $e^4$

If
$$\alpha = \int _ { 0 } ^ { 1 } \left( e ^ { 9 x + 3 \tan ^ { - 1 } x } \right) \left( \frac { 12 + 9 x ^ { 2 } } { 1 + x ^ { 2 } } \right) d x$$
where $\tan ^ { - 1 } x$ takes only principal values, then the value of $\left( \log _ { e } | 1 + \alpha | - \frac { 3 \pi } { 4 } \right)$ is

If $I = \sum _ { k = 1 } ^ { 98 } \int _ { k } ^ { k + 1 } \frac { k + 1 } { x ( x + 1 ) } d x$, then
[A] $I > \log _ { e } 99$
[B] $I < \log _ { e } 99$
[C] $I < \frac { 49 } { 50 }$
[D] $I > \frac { 49 } { 50 }$

The value of the integral
$$\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 + \sqrt { 3 } } { \left( ( x + 1 ) ^ { 2 } ( 1 - x ) ^ { 6 } \right) ^ { \frac { 1 } { 4 } } } d x$$
is $\_\_\_\_$ .

Consider the equation
$$\int _ { 1 } ^ { e } \frac { \left( \log _ { \mathrm { e } } x \right) ^ { 1 / 2 } } { x \left( a - \left( \log _ { \mathrm { e } } x \right) ^ { 3 / 2 } \right) ^ { 2 } } d x = 1 , \quad a \in ( - \infty , 0 ) \cup ( 1 , \infty )$$
Which of the following statements is/are TRUE ?
(A) No $a$ satisfies the above equation
(B) An integer $a$ satisfies the above equation
(C) An irrational number $a$ satisfies the above equation
(D) More than one $a$ satisfy the above equation