Daily Math · 2026-06-15

Question 1

Find the term $T_7$ in the expansion of $\left(\sqrt[3]{x}+\dfrac{1}{\sqrt[4]{x}}\right)^{12}$.

Accepted Answer

$\bullet$ Write the general term with $a = x^{1/3}$, $b = x^{-1/4}$, $n = 12$:

$$T_{k+1} = C_{12}^k\,(x^{1/3})^{12-k}\,(x^{-1/4})^k$$

$\bullet$ Simplify the exponent of $x$:

$$T_{k+1} = C_{12}^k\,x^{(48-7k)/12}$$

$\bullet$ For $T_7$ take $k = 6$. Exponent:

$$\frac{48 - 42}{12} = \frac{1}{2}$$

$\bullet$ Compute $C_{12}^6 = 924$:

$$T_7 = 924\,x^{1/2}$$

Answer

$C_{12}^6\,x$

Answer

$84\,x^{1/2}$

Answer

$924\,x^{1/2}$

Answer

$C_{12}^6\,x^2$

Answer

$792\,x^{1/2}$

Answer

$924\,x$

Question 2

For $I_n = \displaystyle\int_0^1 \dfrac{x^n}{1 + x} \, dx$, the inequality $0 \le I_n \le \dfrac{1}{n + 1}$ holds because:

Accepted Answer

On $[0, 1]$: $1 \le 1 + x \le 2$, hence $\dfrac{1}{2} \le \dfrac{1}{1 + x} \le 1$. So $\dfrac{x^n}{1 + x} \le x^n$, giving $I_n \le \int_0^1 x^n \, dx = \dfrac{1}{n + 1}$.

Answer

$\dfrac{1}{1 + x} \ge 0$ on $[0, 1]$

Answer

$\dfrac{x^n}{1 + x} \le 1$ on $[0, 1]$

Answer

$\dfrac{1}{1 + x} \le 1$ on $[0, 1]$

Answer

$\dfrac{x^n}{1 + x} \le x^n$

Question 3

The perpendicular bisector of the segment from $A(0, 0)$ to $B(4, 0)$ has equation:

Accepted Answer

Midpoint: $(2, 0)$. Segment $AB$ is horizontal, so the bisector is vertical through $(2, 0)$: $x = 2$.

Answer

$y = 2$

Answer

$y = 0$

Answer

$x = 2$

Answer

$x = 0$

Problems & worked solutions

Problem 1 — Combinatorics

Problem 2 — Recursive Integrals

Problem 3 — Lines in the Plane