Daily math problems for June 15, 2026 — Combinatorics, Calculus, Analytic Geometry

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

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

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

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

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

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

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

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

On $[0,1]$: $1\le 1+x\le 2$, hence ${\displaystyle \frac{1}{2}}\le {\displaystyle \frac{1}{1+x}}\le 1$. So ${\displaystyle \frac{x^n}{1+x}}\le x^n$, giving $I_n\le {\int }_0^1{x}^{n}dx={\displaystyle \frac{1}{n+1}}$.

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