Matura 2026, August

Solution: \(-2^2=-4\) is true (the power is applied before the minus sign), and \(-3\gt -2\) is false. An implication from a true premise to a false conclusion is false, so statement B is false. Implications with a false premise (C and D) are always true.

Solution: \(|z|=|1+i|^3=\left(\sqrt{1^2+1^2}\right)^3=(\sqrt{2})^3=2\sqrt{2}\).

Solution: \(9x^{10}:3x^{2}=3x^{8}\), so \(\left(3x^{8}\right)^3=27x^{24}\).

Solution: Subtract 1 from every part: \(-4\le x\lt 4\), so \(x\in[-4,4)\).

Solution: \(1+\frac{1}{-2}=\frac{1}{2}\), \(2:\frac{1}{2}=4\), so \(1-4=-3\).

Solution: \(a^2+b^2=(a+b)^2-2ab=16-8=8\).

Solution: If the first number is \(x\), the second is \(x+4\) and the third is \(3x\). \(x+(x+4)+3x=-6\Rightarrow 5x=-10\Rightarrow x=-2\). The numbers are \(-2,\ 2,\ -6\), so the smallest is \(-6\).

Solution: \(f(x)=(2+m)x-3m\). The graph of \(g(x)=2\) is a horizontal line, so the coefficient of \(x\) must be 0: \(2+m=0\Rightarrow m=-2\).

Solution: The zeros are \(-5\) and \(3\), so the vertex is halfway between them at \(x=-1\). \(f(-1)=-2\cdot4\cdot(-4)=32\).

Solution: \(2(5+b)=34\Rightarrow b=12\), so \(A=5\cdot12=60\).

Solution: \(\overrightarrow{FA}+\overrightarrow{AD}=\overrightarrow{FD}\). In a regular hexagon \(FD\) and \(AC\) are parallel, equal in length and point the same way, so \(\overrightarrow{FD}=\overrightarrow{AC}\).

Solution: The cylinder has \(r=H=a\), so \(V=a^2\pi\cdot a=a^3\pi=64\pi\Rightarrow a=4\) cm.

Solution: The central angle is twice the inscribed angle: \(2\beta+\beta=105^\circ\Rightarrow\beta=35^\circ\), so the central angle is \(70^\circ\).

Solution: \(V=0.7\cdot0.3\cdot1=0.21\ \text{m}^3=210\ \text{l}\). Half of that is 105 l.

Solution: \(\cos20^\circ=\sin70^\circ\gt \sin20^\circ\), so D is true. A is false because sine increases on \([0^\circ,90^\circ]\); B is false because \(\tan\alpha=\cot\alpha\) only at \(45^\circ\); C is false because cotangent decreases, so \(\cot11^\circ\gt \cot13^\circ\).

Solution: \(0.04\cdot x=840\ \text{kg}\Rightarrow x=21\,000\ \text{kg}=21\ \text{t}\).

Solution: A number is divisible by 6 if it is divisible by 2 and by 3. The digit sum \(3+0+1+1+a=5+a\) is divisible by 3 for \(a\in\{1,4,7\}\), and the number is even only for \(a=4\).

Solution: There is exactly one solution when the discriminant is 0: \(D=25-16a=0\Rightarrow a=\dfrac{25}{16}\).

Solution: By Vieta's formulas \(x_1+x_2=\dfrac{5}{2}\) and \(x_1x_2=\dfrac{2a}{2}=a\). \(2\cdot\dfrac{5}{2}=a\Rightarrow a=5\).

Solution: \(r=3\) cm, slant height \(s=\sqrt{3^2+4^2}=5\) cm, so \(M=r\pi s=3\cdot5\cdot\pi=15\pi\ \text{cm}^2\).