Daily Math · 2026-07-24

Question 1

The negation of "for every real number $x$, $x^2 + 1 > 0$" is:

Accepted Answer

Negation of $\forall$ is $\exists$; negation of $>$ is $\le$. So the negation is $\exists x \in \mathbb{R}:\ x^2 + 1 \le 0$. (This statement is itself false, but the question asked for the negation, not its truth value.)

Answer

For every real $x$, $x^2 + 1 \le 0$

Answer

There exists a real $x$ such that $x^2 + 1 \le 0$

Answer

There exists a real $x$ such that $x^2 + 1 > 0$

Answer

For every real $x$, $x^2 + 1 < 0$

Question 2

The numbers $3$, $x$, $27$ are in geometric progression (with positive ratio). Determine $x$.

Accepted Answer

$x^2 = 3 \cdot 27 = 81 \Rightarrow x = 9$ (positive ratio).

Answer

$5$

Answer

$9$

Answer

$15$

Answer

$\sqrt{30}$

Question 3

Solve $25^x - 6\cdot5^x + 10\sin\dfrac{\pi}{6}=0$ and find $x_1^2+x_2^2$.

Accepted Answer

$\bullet$ Simplify the constant term: $\sin	frac{\pi}{6} = 	frac{1}{2}$, so $10 \cdot 	frac{1}{2} = 5$:

$$25^x - 6 \cdot 5^x + 5 = 0$$

$\bullet$ Substitute $t = 5^x > 0$:

$$t^2 - 6t + 5 = 0$$

$\bullet$ Factor:

$$(t - 1)(t - 5) = 0 \;\Rightarrow\; t \in \{1, 5\}$$

$\bullet$ Back-substitute: $5^x = 1 \Rightarrow x = 0$ and $5^x = 5 \Rightarrow x = 1$.

$\bullet$ Compute the sum of squares:

$$x_1^2 + x_2^2 = 0^2 + 1^2 = 1$$

Answer

$1$

Answer

$0$

Answer

$2$

Answer

$5$

Answer

$-1$

Answer

$\dfrac{1}{2}$

Question 4

The smallest non-abelian group has order:

Accepted Answer

$S_3$ (or equivalently $D_3$) has $6$ elements and is non-abelian. All groups of order $1, 2, 3, 4, 5$ are abelian.

Answer

$3$

Answer

$4$

Answer

$6$

Answer

$8$

Question 5

The area of an equilateral triangle with side $a = 6$ equals:

Accepted Answer

$A = \dfrac{36\sqrt{3}}{4} = 9\sqrt{3}$.

Answer

$9$

Answer

$6\sqrt{3}$

Answer

$9\sqrt{3}$

Answer

$18$

Question 6

The function $f(x) = x^3 - 3x^2$ is decreasing on:

Accepted Answer

$f'(x) < 0 \Leftrightarrow 0 < x < 2$. So $f$ is decreasing on $(0, 2)$. (Increasing on $(-\infty, 0)$ and $(2, +\infty)$.)

Answer

$\mathbb{R}$

Answer

$(-\infty, 0)$

Answer

$(0, 2)$

Answer

$(2, +\infty)$

Question 7

$\displaystyle\int_0^{\pi/2} \sin^2 x \,dx$ equals:

Accepted Answer

$\int_0^{\pi/2} \sin^2 x\,dx = \int_0^{\pi/2} \dfrac{1-\cos 2x}{2}\,dx = \dfrac{1}{2}\left[x - 	frac{\sin 2x}{2}ight]_0^{\pi/2} = \dfrac{\pi}{4}$.

Answer

$	frac{\pi}{4}$

Answer

$	frac{\pi}{2}$

Answer

$1$

Answer

$	frac{1}{2}$

Question 8

$\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2}$ equals:

Accepted Answer

Euler's classical result: $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.

Answer

$	frac{\pi^2}{6}$

Answer

$	frac{\pi^2}{4}$

Answer

$	frac{\pi}{4}$

Answer

$1$

Question 9

By the rational-root theorem, possible rational roots of $P(X) = 2X^3 + 3X^2 - 1$ are of the form $\dfrac{p}{q}$ with $p \mid 1$ and $q \mid 2$. They are:

Accepted Answer

$\dfrac{p}{q}$ runs over $\!\left\{\pm 1, \pm\dfrac{1}{2}\right\}$.

Answer

$\pm 1, \pm 2$

Answer

$\pm 1, \pm \dfrac{1}{2}$

Answer

$\pm 1, \pm 3, \pm 2$

Answer

$\pm 1$ only

Question 10

Solve the equation $C_x^5 = C_x^{\,i^{20}+1}$ for $x \in \mathbb{N}$, $x \geq 5$.

Accepted Answer

$\bullet$ Simplify the exponent on the right: $i^{20} = (i^4)^5 = 1$, so $i^{20}+1 = 2$:

$$C_x^5 = C_x^2$$

$\bullet$ Recall the combinatorial symmetry:

$$C_x^k = C_x^{x-k}$$

$\bullet$ Since $5 
eq 2$, the symmetry forces $5 + 2 = x$:

$$x = 7$$

Answer

$x = 5$

Answer

$x = 7$

Answer

$x = 8$

Answer

$x = 9$

Answer

$x = 6$

Answer

$x = 10$

Problems & worked solutions

Problem 1 — Logic & Induction

Problem 2 — Geometric Sequences

Problem 3 — Algebra

Problem 4 — Groups

Problem 5 — Solving Triangles

Problem 6 — Applications of Derivatives

Problem 7 — Calculus

Problem 8 — Calculus

Problem 9 — Polynomial Rings

Problem 10 — Combinatorics