Daily math problems for August 3, 2026 — Quadratic equations, Limits, Calculus & more

By Viète's formulas, $x_1+x_2=-(-7)=7$. (The product is $10$, so the roots are $2$ and $5$.)

$\bullet$ Factor the numerator and cancel:

$$\frac{x^2+x}{x}=\frac{x(x+1)}{x}=x+1(x\mathrm{\ne }0)$$

$\bullet$ Take the limit of the simplified expression:

$$\underset{x\to 0}{\lim }(x+1)=1$$

$\bullet$ Factor the numerator. Roots of $x^2+5x-14=0$ are $x=2$ and $x=-7$:

$$x^2+5x-14=(x-2)(x+7)$$

$\bullet$ Cancel and evaluate:

$$\underset{x\to 2}{\lim }\frac{(x-2)(x+7)}{x-2}=\underset{x\to 2}{\lim }(x+7)=9$$

$P(-1)>0$ and $P(1)<0$: $2+m>0$ and $-2+m<0$, giving $-2<m<2$.

The positive divisors of $12$ are $\{1,2,3,4,6,12\}$, so $\mathrm{\mid }A\mathrm{\mid }=6$.

Subtract: $5y=15\Rightarrow y=3$. Then $3x=y=3\Rightarrow x=1$. So $A=1+9=10$.

$\mathrm{\mid }x-2\mathrm{\mid }<3\iff -3<x-2<3\iff -1<x<5$. The solution set is $(-1,5)$.

Substitute $x=\tan \theta$. Then $dx={\sec }^2\theta d\theta$ and $1+x^2={\sec }^2\theta$, so the ${\sec }^2\theta$ on top and bottom cancel cleanly. The bounds $0\to 1$ map to $\theta :0\to \pi \mathrm{/}4$:

$$I={\int }_0^{\pi \mathrm{/}4}\ln (1+\tan \theta )d\theta \mathrm{.}$$

Reflect by $\varphi =\pi \mathrm{/}4-\theta$. The angle-difference formula gives $\tan (\pi \mathrm{/}4-\varphi )={\displaystyle \frac{1-\tan \varphi }{1+\tan \varphi }}$, so

$$1+\tan \theta =1+\frac{1-\tan \varphi }{1+\tan \varphi }=\frac{2}{1+\tan \varphi }\mathrm{.}$$

Taking logs: $\ln (1+\tan \theta )=\ln 2-\ln (1+\tan \varphi )$.

Pair the integral with its reflected twin. Renaming the dummy variable $\varphi \to \theta$ (the bounds are unchanged because the substitution is a reflection of $[0,\pi \mathrm{/}4]$ onto itself):

$$I={\int }_0^{\pi \mathrm{/}4}[\ln 2-\ln (1+\tan \theta )]d\theta =\frac{\pi }{4}\ln 2-I\mathrm{.}$$

Solve. $2I={\displaystyle \frac{\pi }{4}}\ln 2$, so

$$\boxed{{\displaystyle I=\frac{\pi }{8}\ln 2\mathrm{.}}}$$

The reflection trick — pairing $f(\theta )$ with $f(\pi \mathrm{/}4-\theta )$ — works whenever the integrand simplifies under that reflection. Worth keeping in your toolbox alongside the half-angle and Weierstrass substitutions.

$n\ln \text{\hspace{0.17em}⁣}(1+\frac{1}{n^2})\sim n\cdot \frac{1}{n^2}=\frac{1}{n}\to 0$, so the limit equals $e^0=1$.

Euler's classical result: ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}={\displaystyle \frac{{\pi }^2}{6}}}$.