B

9000062910

Level: 
B
Consider the square of the side \(4\, \mathrm{cm}\). The second square is inscribed into this first square by joining the centers of all sides. In a similar way, the third square is inscribed into the second square by joining the centers of the sides of the second square and this process continues up to infinity. Find the sum of the squares of all squares.
\(32\)
\(40\)
\(\frac{32} {3} \)
\(\infty \)

9000062907

Level: 
B
A quarter circle is an arc formed by one quarter of the full circle. An infinite spiral is built from quarter circles with an increasing radius. The radius of the first quarter circle is \(1\, \mathrm{cm}\). The radius of each quarter circle in the spiral is bigger by one half than the radius of the previous quarter circle. Find the total length of the spiral.
\(\infty \)
\(4\pi \)
\(\frac{2} {5}\pi \)
\(\frac{1} {3}\pi \)

9000063105

Level: 
B
Differentiate the following function. \[ f(x) = \frac{\sqrt{x} - 1} {\sqrt{x} + 1} \]
\(f'(x) = \frac{1} {\sqrt{x}(\sqrt{x}+1)^{2}} ,\ x > 0\)
\(f'(x) = \frac{\sqrt{x}} {(\sqrt{x}+1)^{2}} ,\ x > 0\)
\(f'(x) = \frac{2} {x(\sqrt{x}+1)^{2}} ,\ x > 0\)
\(f'(x) = \frac{1} {(\sqrt{x}+1)^{2}} ,\ x > 0\)

9000045704

Level: 
B
Given a right triangle \(ABC\) with the right angle at $C$ and an altitude $v$ (see the picture). Find the valid relation between the angle \(\beta \) and the lengths in the triangle.
\(\sin \beta = \frac{v} {a}\)
\(\mathop{\mathrm{tg}}\nolimits \beta = \frac{a} {v}\)
\(\cos \beta = \frac{v} {a}\)
\(\mathop{\mathrm{tg}}\nolimits \beta = \frac{v} {a}\)

9000046403

Level: 
B
Consider an isosceles triangle, i.e. the triangle with two sides of equal length. The length of the third side is \(4\, \mathrm{cm}\). One of the interior angles is \(120^{\circ }\). Find the area of this triangle.
\(\frac{4\sqrt{3}} {3} \, \mathrm{cm}^{2}\)
\(4\sqrt{3}\, \mathrm{cm}^{2}\)
\(\frac{8\sqrt{3}} {3} \, \mathrm{cm}^{2}\)

9000046506

Level: 
B
Identify the optimal first step convenient to solve the following trigonometric equation. Do not consider the step which is possible but does not help to solve the equation. \[ \sin 2x =\mathop{\mathrm{tg}}\nolimits x \]
\(2\sin x\cdot \cos x = \frac{\sin x} {\cos x}\)
substitution \( 2x = z\)
\(\sin x = \frac{\mathop{\mathrm{tg}}\nolimits x} {2} \)
\(\cos ^{2}x -\sin ^{2}x =\mathop{\mathrm{tg}}\nolimits x\)