Daily Math · 2026-07-29

Question 1

The radius of the sphere $(x - 1)^2 + (y + 2)^2 + (z - 3)^2 = 49$ is:

Accepted Answer

$r^2 = 49 \Rightarrow r = 7$.

Answer

$6$

Answer

$7$

Answer

$14$

Answer

$49$

Question 2

Rolle's Theorem fails for $f(x) = |x|$ on $[-1, 1]$ because:

Accepted Answer

$f$ is continuous on $[-1, 1]$ and $f(-1) = f(1) = 1$, but $f$ is not differentiable at $0$. Rolle's hypotheses fail.

Answer

$f$ is not continuous on $[-1, 1]$

Answer

$f(-1) 
e f(1)$

Answer

$f$ is not differentiable at $0$

Answer

Rolle does apply

Question 3

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 4

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 5

A square matrix $A$ is invertible iff:

Accepted Answer

Invertibility ⇔ non-zero determinant.

Answer

$\det(A) = 0$

Answer

$A$ is symmetric

Answer

$\det(A) 
e 0$

Answer

all entries are non-zero

Question 6

Find the $3	imes3$ determinant whose entries are all limit values (see problem image).

Accepted Answer

$\bullet$ Evaluate each of the nine limits to fill the matrix:

$$\begin{pmatrix}\sqrt{3}/2 & 2 & 1\ 3 & 4 & 4\ 0 & 3 & 1\end{pmatrix}$$

$\bullet$ Expand along row 1. The cofactor of the $(1,1)$ entry is $4\cdot 1-4\cdot 3=-8$:

$$	frac{\sqrt{3}}{2}\cdot(-8)=-4\sqrt{3}$$

$\bullet$ Cofactors of the $(1,2)$ and $(1,3)$ entries are $3\cdot 1-4\cdot 0=3$ and $3\cdot 3-4\cdot 0=9$:

$$-2\cdot 3+1\cdot 9=3$$

$\bullet$ Add the three contributions:

$$\det=3-4\sqrt{3}$$

Answer

$3-4\sqrt{3}$

Answer

$0$

Answer

$3+4\sqrt{3}$

Answer

$12$

Answer

$-4\sqrt3$

Answer

$3$

Question 7

The slant asymptote of $f(x) = \dfrac{2x^2 + 3x + 1}{x - 1}$ as $x 	o \pm\infty$ is:

Accepted Answer

$\dfrac{2x^2 + 3x + 1}{x - 1} = 2x + 5 + \dfrac{6}{x - 1}$. As $x 	o \pm\infty$, the remainder vanishes, leaving the slant asymptote $y = 2x + 5$.

Answer

$y = 2x$

Answer

$y = 2x + 3$

Answer

$y = 2x + 5$

Answer

$y = x + 1$

Question 8

Evaluate $\displaystyle\lim_{x	o 0} \dfrac{e^x - 1 - x}{x^2}$.

Accepted Answer

Direct substitution gives $	frac{0}{0}$. Differentiate top and bottom: $\displaystyle\lim_{x	o 0}\dfrac{e^x - 1}{2x}$ — still $	frac{0}{0}$. Once more: $\displaystyle\lim_{x	o 0}\dfrac{e^x}{2} = \dfrac{1}{2}$.

Answer

$0$

Answer

$\dfrac{1}{2}$

Answer

$1$

Answer

$2$

Question 9

The last two digits of $7^{2024}$ are:

Accepted Answer

$7^2 = 49$, $7^4 = 49^2 = 2401 \equiv 1 \pmod{100}$. Since $2024 = 4 \cdot 506$, we have $7^{2024} = (7^4)^{506} \equiv 1^{506} = 1 \pmod{100}$. Last two digits: $01$.

Answer

$01$

Answer

$49$

Answer

$07$

Answer

$43$

Question 10

For which value of $a$ does $f(x) = x^3 - 3ax + 1$ have a local minimum at $x = 2$?

Accepted Answer

$f'(x) = 3x^2 - 3a$, so $f'(2) = 12 - 3a = 0 \Rightarrow a = 4$. Check: $f''(x) = 6x$, so $f''(2) = 12 > 0$, confirming a local minimum.

Answer

$a = 1$

Answer

$a = 2$

Answer

$a = 3$

Answer

$a = 4$

Problems & worked solutions

Problem 1 — 3D Coordinate Geometry

Problem 2 — Rolle's Sign Method

Problem 3 — Powers, Radicals, Logarithms

Problem 4 — Polynomial Rings

Problem 5 — Determinants

Problem 6 — Matrices

Problem 7 — Asymptotes

Problem 8 — L'Hôpital's Rule

Problem 9 — Number theory

Problem 10 — Local Extrema