Daily math problems for August 1, 2026 — Calculus, Trigonometry, Analytic Geometry & more

The two images $(-\infty, 1)$ and $[1, +\infty)$ are disjoint and together cover all of $\mathbb{R}$, so $f$ is surjective. Each branch is strictly increasing, and the ranges don't overlap, so no two distinct $x$-values give the same $f(x)$ — $f$ is injective. Hence $f$ is bijective.

(Note: $f(-1) = 0$ and $f(0) = 1$, so distractor A is factually false.)

$\bullet$ Substitute $(x,y)=(1,1)$:

$$2 + 1 + y' + 2y' = 0$$

$\bullet$ Solve for $y'$:

$$3y' = -3 \Rightarrow y' = -1$$

$$I = \int_0^{\pi/4} \ln(1 + \tan\theta)\,d\theta.$$

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

$$1 + \tan\theta \;=\; 1 + \frac{1 - \tan\phi}{1 + \tan\phi} \;=\; \frac{2}{1 + \tan\phi}.$$

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

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

$$I \;=\; \int_0^{\pi/4} \bigl[\ln 2 - \ln(1 + \tan\theta)\bigr]\,d\theta \;=\; \frac{\pi}{4}\ln 2 \;-\; I.$$

Solve. $\;2I = \dfrac{\pi}{4}\ln 2$, so

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

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