9000062401
Level:
A
Find the derivative of \(f(x) = 3x^{4} - 2x^{3} - 3x^{2}\)
at the point \(x_{0} = -1\).
\(- 12\)
\(0\)
\(12\)
\(24\)
A
9000062402
Level:
A
Find the second derivative of \(f(x) = x^{4} - 3x^{2}\)
at the point \(x_{0} = 1\).
\(6\)
\(- 2\)
\(- 6\)
\(1\)
9000062404
Level:
A
Evaluate the following limit.
\[
\lim _{x\to +\infty } \frac{x^{3} - x + 1}
{1 - x^{2} - x^{3}}
\]
\(- 1\)
\(0.5\)
\(- 0.5\)
\(1\)
9000062902
Level:
A
Find the sum of the following geometric series.
\[
1 + \frac{3}
{2} + \frac{9}
{4} + \frac{27}
{8} + \frac{81}
{16}+\cdots
\]
\(\infty \)
\(- 2\)
\(2\)
\(\frac{2}
{5}\)
9000062405
Level:
A
Evaluate the following one-sided limit.
\[
\lim _{x\to 6^{-}}\frac{3x + 2}
{x - 6}
\]
\(-\infty \)
\(1\)
\(+\infty \)
\(0\)
9000039107
Level:
A
Find the sum of the reciprocal values of the solutions of
\(5x^{2} - 2x + 1 = 0\).
\(2\)
\(\frac{2}
{5}\)
\(\frac{2}
{5} + \frac{4}
{5}\mathrm{i}\)
\(-\frac{2}
{5}\)
9000046401
Level:
A
Consider the rectangle \(ABCD\)
with length and height \(|AB| = 6\, \mathrm{cm}\)
and \(|BC| = 2\sqrt{3}\, \mathrm{cm}\), respectively.
Find the measure of the angle \( CAB\).
\(30^{\circ }\)
\(45^{\circ }\)
\(60^{\circ }\)
9000046402
Level:
A
Consider the rectangle \(ABCD\)
with length and height \(|AB| = 6\, \mathrm{cm}\)
and \(|BC| = 2\sqrt{3}\, \mathrm{cm}\). Let
\(S\) be the intersection of
diagonals. Find the angle \(\measuredangle ASB\).
\(120^{\circ }\)
\(60^{\circ }\)
\(90^{\circ }\)
9000046502
Level:
A
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.
\[
\cos 3x = 0.5
\]
substitution \( 3x = z\)
substitution \( \cos x = z\)
\(\cos ^{3}x -\sin ^{3}x = 0.5\)
\(\cos x = \frac{0.5}
{3} \)
9000046503
Level:
A
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.
\[
\mathop{\mathrm{tg}}\nolimits \left (-x + \frac{\pi }
{6}\right ) = \sqrt{3}
\]
substitution \( - x + \frac{\pi } {6} = z\)
\(\mathop{\mathrm{tg}}\nolimits (-x) = \sqrt{3} - \frac{\pi } {6}\)
\(\mathop{\mathrm{tg}}\nolimits ^{2}\left (-x + \frac{\pi } {6}\right ) = 3\)
\(\frac{\sin \left (-x+ \frac{\pi }{6} \right )}
{\cos \left (-x+ \frac{\pi }{6} \right )} = \sqrt{3}\)