B

1003134603

Level: 
B
The sum of the first five terms of a geometric sequence is less than $1$ and the common ratio is $10$. Determine all possible values of the first term.
$ a_1 < \frac1{11111}$, $a_1\in\mathbb{R}$
$ -\frac1{10^5} < a_1 < \frac1{10^5}$, $a_1\in\mathbb{R}$
$a_1 < \frac1{99999}$, $a_1\in\mathbb{R}$
$a_1 < 10^{-4}$, $a_1\in\mathbb{R}$
$a_1 < 10^{-5}$, $a_1\in\mathbb{R}$

1003107809

Level: 
B
Solve an indefinite integral $\int\frac{8x^7-30x^5}{x^8-5x^6+2}\,\mathrm{d}x$ on $(3;\infty)$.
$\ln\left|x^8-5x^6+2\right|+c$, $c\in\mathbb{R}$
$\ln\left|8x^7-30x^5+2x\right|+c$, $c\in\mathbb{R}$
$\log\left|x^8-5x^6+2\right|+c$, $c\in\mathbb{R}$
$\log\left|8x^7-30x^5+2x\right|+c$, $c\in\mathbb{R}$

1003107804

Level: 
B
Four girls evaluated the integral $I=\int\sin ⁡x\cdot\cos x\,\mathrm{d}x$ on $\mathbb{R}$. Ann started to integrate by parts like this: $I=\int\sin ⁡x\cdot\cos x\,\mathrm{d}x=\sin^2⁡x-\int\cos x\cdot\sin x\,\mathrm{d}x$. Beth started to integrate by parts like this: $I=\int\sin ⁡x\cdot\cos x\,\mathrm{d}x=-\cos^2 x-\int\sin x\cdot\cos x\,\mathrm{d}x$. Claire used substitution $a=\sin ⁡x$ like this: $I=\int\sin ⁡x\cdot\cos x\,\mathrm{d}x=\int a\,\mathrm{d}a$. Diana integrated $\int\sin ⁡x\cdot\cos x\,\mathrm{d}x=-\cos x\cdot\sin⁡ x+c$, $c\in\mathbb{R}$. Which of the girls made a mistake?
Diana
Ann
Beth
Claire