Daily math problems for July 10, 2026 — Calculus, Statistics, Analytic Geometry & more

Differentiating term by term: $(\sin x{)}^{\mathrm{\prime }}+(\cos x{)}^{\mathrm{\prime }}=\cos x-\sin x$.

Mean: $(2+4+4+6+9)\mathrm{/}5=25\mathrm{/}5=5$. Median (middle of the sorted set): $4$. So $m+M=5+4=9$.

If $f$ has $n$ distinct roots, between each consecutive pair Rolle gives a root of $f^{\mathrm{\prime }}$ — that's $n-1$ roots of $f^{\mathrm{\prime }}$. So if $f^{\mathrm{\prime }}$ has at most $n-1$ roots, $f$ has at most $n$.

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{\sin 2x}{x}}=2\cdot \underset{x\to 0}{\lim }{\displaystyle \frac{\sin 2x}{2x}}=2\cdot 1=2}$. So $a=2$.

$\bullet$ 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 $3x+2y=-1$:

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

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

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

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

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

$\bullet$ Evaluate each special entry:

$$i^2=-1,\lg 100=2,\ln 1=0$$

$$\sin \frac{\pi }{2}=1,i^{24}=1,C_4^3=4$$

$\bullet$ Assemble the numerical matrix:

$$A=\left(\begin{array}{cccc}1& 0& 2& 4\\ -1& 1& 3& 2\\ 0& 1& 3& 0\\ 1& 1& 4& 7\end{array}\right)$$

$\bullet$ Row-reduce. $R_2+=R_1$, $R_4-=R_1$:

$$\left(\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& 5& 6\\ 0& 1& 3& 0\\ 0& 1& 2& 3\end{array}\right)$$

$\bullet$ $R_3-=R_2$, $R_4-=R_2$:

$$\left(\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& 5& 6\\ 0& 0& -2& -6\\ 0& 0& -3& -3\end{array}\right)$$

$\bullet$ $R_4-=\frac{3}{2}{R}_3$ gives upper-triangular with pivots $1,1,-2,6$:

$$\det A=1\cdot 1\cdot (-2)\cdot 6=-12$$

$g(0)=1>0$, $g(\pi \mathrm{/}2)=0-\pi \mathrm{/}2<0$. Since $g$ is continuous, by the IVT there is $c\in (0,\pi \mathrm{/}2)$ with $g(c)=0$, i.e. $\cos c=c$.

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.

Direct substitution gives $\frac{0}{0}$. Differentiate top and bottom: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x-1}{2x}}}$ — still $\frac{0}{0}$. Once more: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x}{2}}={\displaystyle \frac{1}{2}}}$.

$\int \text{\hspace{0.17em}⁣}({\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x+1}})dx=\ln \mathrm{\mid }x\mathrm{\mid }-\ln \mathrm{\mid }x+1\mathrm{\mid }+C=\ln \text{\hspace{0.17em}⁣}\mid {\displaystyle \frac{x}{x+1}}\mid +C$.