Daily Math · 2026-07-02

Question 1

After rationalizing the denominator, $\dfrac{1}{\sqrt{3} - 1}$ equals:

Accepted Answer

$\dfrac{1}{\sqrt{3} - 1} \cdot \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} = \dfrac{\sqrt{3} + 1}{3 - 1} = \dfrac{\sqrt{3} + 1}{2}$.

Answer

$\sqrt{3} - 1$

Answer

$\dfrac{\sqrt{3} - 1}{2}$

Answer

$\dfrac{\sqrt{3} + 1}{2}$

Answer

$\dfrac{1}{2}$

Question 2

The solution set of $x^2 > 9$ in $\mathbb{R}$ is:

Accepted Answer

$x^2 > 9 \Leftrightarrow |x| > 3 \Leftrightarrow x < -3$ or $x > 3$. Solution: $(-\infty, -3) \cup (3, +\infty)$.

Answer

$(-3, 3)$

Answer

$[-3, 3]$

Answer

$(-\infty, -3) \cup (3, +\infty)$

Answer

$(-\infty, -3] \cup [3, +\infty)$

Question 3

$\displaystyle\int x \cos x \, dx$ equals:

Accepted Answer

$u = x$, $du = dx$, $v = \sin x$. $\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C$.

Answer

$\dfrac{x^2}{2} \cos x + C$

Answer

$-x \sin x + \cos x + C$

Answer

$x \sin x + \cos x + C$

Answer

$\sin x - x \cos x + C$

Problems & worked solutions