Daily Math · 2026-06-20

Question 1

By the Intermediate Value Theorem, the equation $f(x) = x^3 + x - 1 = 0$ has at least one root in:

Accepted Answer

$f(0) = -1 < 0$ and $f(1) = 1 > 0$. Since $f$ is continuous, by the IVT there exists $c \in (0, 1)$ with $f(c) = 0$.

Answer

$(-1, 0)$

Answer

$(0, 1)$

Answer

$(1, 2)$

Answer

nowhere on $\mathbb{R}$

Question 2

For $f: \mathbb{R} 	o \mathbb{R}$, $f(x) = 3x^2 - 7x + 2$, the value of $f(2025) \cdot f(2)$ is:

Accepted Answer

$f(2) = 12 - 14 + 2 = 0$. So $f(2025) \cdot f(2) = f(2025) \cdot 0 = 0$, regardless of $f(2025)$.

Answer

$0$

Answer

$1$

Answer

$f(2025)$

Answer

A non-zero value depending on $2025$

Question 3

Find $\cos 20°\cdot\cos 40°\cdot\cos 80°$.

Accepted Answer

$\bullet$ Multiply numerator and denominator by $2\sin 20°$ and use $2\sin	heta\cos	heta=\sin 2	heta$:

$$\cos 20°\cos 40°\cos 80° = \dfrac{\sin 40°\cos 40°\cos 80°}{2\sin 20°}$$

$\bullet$ Apply the identity again:

$$= \dfrac{\sin 80°\cos 80°}{4\sin 20°}$$

$\bullet$ And once more:

$$= \dfrac{\sin 160°}{8\sin 20°}$$

$\bullet$ Use $\sin 160° = \sin(180°-20°) = \sin 20°$:

$$= \dfrac{\sin 20°}{8\sin 20°} = \dfrac{1}{8}$$

Answer

$\dfrac{1}{2}$

Answer

$\dfrac{1}{4}$

Answer

$\dfrac{1}{8}$

Answer

$\dfrac{\sqrt{3}}{8}$

Answer

$\dfrac{1}{16}$

Answer

$\dfrac{\sqrt{2}}{8}$

Problems & worked solutions