Problem 1: calculation of an integral
For $x \geqslant 0$ we define $$f ( x ) = \int _ { 0 } ^ { \infty } \frac { e ^ { - t x } } { 1 + t ^ { 2 } } \mathrm {~d} t \quad \text { and } \quad g ( x ) = \int _ { 0 } ^ { \infty } \frac { \sin t } { t + x } \mathrm {~d} t$$
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 $+ \infty$.

Let $\Gamma$ denote a curve $y = y ( x )$ which is in the first quadrant and let the point $( 1,0 )$ lie on it. Let the tangent to $\Gamma$ at a point $P$ intersect the $y$-axis at $Y _ { P }$. If $P Y _ { P }$ has length 1 for each point $P$ on $\Gamma$, then which of the following options is/are correct?
(A) $y = \log _ { e } \left( \frac { 1 + \sqrt { 1 - x ^ { 2 } } } { x } \right) - \sqrt { 1 - x ^ { 2 } }$
(B) $\quad x y ^ { \prime } + \sqrt { 1 - x ^ { 2 } } = 0$
(C) $y = - \log _ { e } \left( \frac { 1 + \sqrt { 1 - x ^ { 2 } } } { x } \right) + \sqrt { 1 - x ^ { 2 } }$
(D) $x y ^ { \prime } - \sqrt { 1 - x ^ { 2 } } = 0$

Let $f ( x )$ be a continuously differentiable function on the interval $( 0 , \infty )$ such that $f ( 1 ) = 2$ and
$$\lim _ { t \rightarrow x } \frac { t ^ { 10 } f ( x ) - x ^ { 10 } f ( t ) } { t ^ { 9 } - x ^ { 9 } } = 1$$
for each $x > 0$. Then, for all $x > 0 , f ( x )$ is equal to
(A) $\frac { 31 } { 11 x } - \frac { 9 } { 11 } x ^ { 10 }$
(B) $\frac { 9 } { 11 x } + \frac { 13 } { 11 } x ^ { 10 }$
(C) $\frac { - 9 } { 11 x } + \frac { 31 } { 11 } x ^ { 10 }$
(D) $\frac { 13 } { 11 x } + \frac { 9 } { 11 } x ^ { 10 }$

The normal to a curve at $P ( x , y )$ meets the $x$-axis at $G$. If the distance of $G$ from the origin is twice the abscissa of $P$, then the curve is a
(1) ellipse
(2) parabola
(3) circle
(4) pair of straight lines

Let $f : R \rightarrow R$ be a continuous function such that $f ( 3 x ) - f ( x ) = x$. If $f ( 8 ) = 7$, then $f ( 14 )$ is equal to:
(1) 4
(2) 10
(3) 11
(4) 16

Let a curve $y = y ( x )$ pass through the point $( 3,3 )$ and the area of the region under this curve, above the $x$-axis and between the abscissae 3 and $x ( > 3 )$ be $\left( \frac { y } { x } \right) ^ { 3 }$. If this curve also passes through the point $( \alpha , 6 \sqrt { 10 } )$ in the first quadrant, then $\alpha$ is equal to $\_\_\_\_$.

Consider a function $\mathrm { f } : \mathbb { N } \rightarrow \mathbb { R }$, satisfying $f ( 1 ) + 2 f ( 2 ) + 3 f ( 3 ) + \ldots + x f ( x ) = x ( x + 1 ) f ( x ) ; x \geq 2$ with $f ( 1 ) = 1$. Then $\frac { 1 } { f ( 2022 ) } + \frac { 1 } { f ( 2028 ) }$ is equal to (1) 8200 (2) 8000 (3) 8400 (4) 8100

Let $Y = Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the line $Y - y = Y'(x)(X - x)$ and the coordinate axes, where $(x, y)$ is any point on the curve, is always $\frac{-y^2}{2Y'(x)} + 1$, $Y'(x) \neq 0$. If $Y(1) = 1$, then $12\,Y(2)$ equals $\underline{\hspace{1cm}}$.

Let $f ( x )$ be a real differentiable function such that $f ( 0 ) = 1$ and $f ( x + y ) = f ( x ) f ^ { \prime } ( y ) + f ^ { \prime } ( x ) f ( y )$ for all $x , y \in \mathbf { R }$. Then $\sum _ { \mathrm { n } = 1 } ^ { 100 } \log _ { \mathrm { e } } f ( \mathrm { n } )$ is equal to:
(1) 2525
(2) 5220
(3) 2384
(4) 2406