Daily Math · 2026-05-28

Question 1

If $x^2 + y^2 = 13$ and $xy = 6$, the value of $|x+y|$ is:

Accepted Answer

$(x+y)^2 = (x^2+y^2) + 2xy = 13 + 12 = 25$, so $|x+y| = 5$.

Answer

$5$

Answer

$7$

Answer

$\sqrt{13}$

Answer

$1$

Question 2

Let $f(x)=\lim_{k	o x}\dfrac{k^2-16}{k^2-5k+4}+\begin{vmatrix}C_7^x&3x+1\4\log_3 i&0\end{vmatrix}$. Find $f(4)+f(0)$.

Accepted Answer

$\bullet$ Simplify the limit by factoring numerator and denominator:

$$\lim_{k	o x}\frac{(k-4)(k+4)}{(k-4)(k-1)}=\frac{x+4}{x-1}$$

$\bullet$ The determinant has a row of zeros (with $\log_3 1=0$), so it contributes $0$:

$$f(x)=\frac{x+4}{x-1}$$

$\bullet$ Evaluate at $x=4$ and $x=0$:

$$f(4)=	frac{8}{3},\quad f(0)=-4$$

$\bullet$ Sum them:

$$f(4)+f(0)=	frac{8}{3}-4=-	frac{4}{3}$$

Answer

$-3$

Answer

$\dfrac{5}{3}$

Answer

$-\dfrac{4}{3}$

Answer

$5$

Answer

$\dfrac{8}{3}$

Answer

$-4$

Question 3

Calculați produsul $\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{255} 256$.

Accepted Answer

Folosind schimbarea de bază, $\log_k (k+1) = \dfrac{\ln(k+1)}{\ln k}$. Produsul telescopează: $\dfrac{\ln 3}{\ln 2} \cdot \dfrac{\ln 4}{\ln 3} \cdots \dfrac{\ln 256}{\ln 255} = \dfrac{\ln 256}{\ln 2} = \log_2 256 = 8$.

Answer

$8$

Answer

$16$

Answer

$128$

Answer

$256$

Problems & worked solutions

Problem 1 — Algebra

Problem 2 — Calculus

Problem 3 — Logarithms