Daily Math · 2026-07-31

Question 1

The point of inflection of $f(x) = x^3 - 3x$ is at:

Accepted Answer

$f'(x) = 3x^2 - 3$, $f''(x) = 6x$. $f''$ changes sign at $x = 0$, so the inflection is at $x = 0$.

Answer

$x = -1$

Answer

$x = 0$

Answer

$x = 1$

Answer

$f$ has no inflection

Question 2

Dacă $\log_{10} 2 = a$ și $\log_{10} 3 = b$, exprimați $\log_{10} 18$ în funcție de $a$ și $b$.

Accepted Answer

Avem $18 = 2 \cdot 3^2$, deci $\log_{10} 18 = \log_{10} 2 + \log_{10} 3^2 = \log_{10} 2 + 2\log_{10} 3 = a + 2b$.

Answer

$a + b$

Answer

$a + 2b$

Answer

$2a + b$

Answer

$ab$

Question 3

The numbers $4, 12, 36, 108, \ldots$ form a geometric progression. The common ratio $q$ equals:

Accepted Answer

$q = \dfrac{b_2}{b_1} = \dfrac{12}{4} = 3$.

Answer

$2$

Answer

$\dfrac{1}{3}$

Answer

$3$

Answer

$4$

Question 4

The area of the triangle with vertices $(0,0), (3,0), (1,4)$ is:

Accepted Answer

Base along the $x$-axis has length $3$; the third vertex sits at height $4$. Area $= 	frac{1}{2} \cdot 3 \cdot 4 = 6$.

Answer

$6$

Answer

$12$

Answer

$5$

Answer

$7$

Question 5

For $A = \begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix}$, the matrix $A^2$ equals:

Accepted Answer

$A^2 = \begin{pmatrix} 0 \cdot 0 + 1 \cdot 0 & 0 \cdot 1 + 1 \cdot 0 \ 0 \cdot 0 + 0 \cdot 0 & 0 \cdot 1 + 0 \cdot 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}$. ($A$ is nilpotent.)

Answer

$A$

Answer

$I_2$

Answer

$\begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}$

Answer

$\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}$

Question 6

Compute $f'(1)$ where $f(x) = \dfrac{\ln x}{x}$.

Accepted Answer

$f'(x) = \dfrac{(1/x) \cdot x - \ln x \cdot 1}{x^2} = \dfrac{1 - \ln x}{x^2}$. At $x = 1$: $\dfrac{1 - 0}{1} = 1$.

Answer

$0$

Answer

$1$

Answer

$-1$

Answer

$	frac{1}{2}$

Question 7

Triangle $ABC$ has sides $(BC):3x+2y+1=0$, $(AB):x-2y+3=0$, $(AC):2x-y-3=0$. Find its area.

Accepted Answer

$\bullet$ Find vertex $A=AB\cap AC$ by solving $x-2y=-3$ and $2x-y=3$:

$$A=(3,3)$$

$\bullet$ Find vertex $B=AB\cap BC$ by solving $x-2y=-3$ and $3x+2y=-1$:

$$B=(-1,1)$$

$\bullet$ Find vertex $C=AC\cap BC$ by solving $2x-y=3$ and $3x+2y=-1$:

$$C=\bigl(	frac{5}{7},-	frac{11}{7}\bigr)$$

$\bullet$ Apply the shoelace-style area formula:

$$\mathcal{A}=	frac{1}{2}\left|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)ight|$$

$\bullet$ Plugging the coordinates gives $	frac{1}{2}\cdot	frac{96}{7}$:

$$\mathcal{A}=	frac{48}{7}$$

Answer

$5$

Answer

$6$

Answer

$\dfrac{48}{7}$

Answer

$10$

Answer

$7$

Answer

$\dfrac{24}{7}$

Question 8

$\displaystyle\int_0^{\infty} e^{-x^2} \,dx$ equals:

Accepted Answer

$\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$. By even symmetry, $\int_0^{\infty} e^{-x^2}\,dx = \dfrac{\sqrt{\pi}}{2}$.

Answer

$	frac{\sqrt{\pi}}{2}$

Answer

$\sqrt{\pi}$

Answer

$	frac{\pi}{2}$

Answer

$1$

Question 9

A linear regression of test score $y$ on hours studied $x$ gives $\hat{y} = 50 + 8x$. By how much does the predicted score increase when $x$ increases by $0.5$?

Accepted Answer

The slope is $8$ score points per additional hour. For a $0.5$-hour increase, the predicted change is $0.5 	imes 8 = 4$.

Answer

$0.5$

Answer

$4$

Answer

$8$

Answer

$54$

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 — Applications of Derivatives

Problem 2 — Logarithms

Problem 3 — Geometric Sequences

Problem 4 — Geometry

Problem 5 — Matrix Equations

Problem 6 — Differentiation

Problem 7 — Analytic Geometry

Problem 8 — Calculus

Problem 9 — Linear Regression Interpretation

Problem 10 — Local Extrema