Daily Math · 2026-07-10

Question 1

What is $f'(x)$ if $f(x) = \sin x + \cos x$?

Accepted Answer

Differentiating term by term: $(\sin x)' + (\cos x)' = \cos x - \sin x$.

Answer

$\sin x - \cos x$

Answer

$\cos x + \sin x$

Answer

$\cos x - \sin x$

Answer

$-\cos x - \sin x$

Question 2

The dataset $\{2, 4, 4, 6, 9\}$ has mean $m$ and median $M$. Compute $m + M$.

Accepted Answer

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

Answer

$8$

Answer

$9$

Answer

$10$

Answer

$11$

Question 3

A continuous function $f: [a, b] 	o \mathbb{R}$ has at most $n$ zeros on $[a, b]$ if its derivative has at most:

Accepted Answer

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

Answer

$n - 2$ zeros on $(a, b)$

Answer

$n$ zeros on $(a, b)$

Answer

$n - 1$ zeros on $(a, b)$

Answer

no zeros on $(a, b)$

Question 4

Find $a \in \mathbb{R}$ so that $f(x) = \begin{cases} \dfrac{\sin(2x)}{x}, & x 
e 0 \ a, & x = 0 \end{cases}$ is continuous at $0$:

Accepted Answer

$\displaystyle\lim_{x 	o 0} \dfrac{\sin 2x}{x} = 2 \cdot \lim_{x 	o 0} \dfrac{\sin 2x}{2x} = 2 \cdot 1 = 2$. So $a = 2$.

Answer

$0$

Answer

$1$

Answer

$2$

Answer

$\dfrac{1}{2}$

Question 5

Triangle $ABC$ has sides $(BC):3x+2y+1=0$, $(AB):x-2y+3=0$, $(AC):2x-y-3=0$. Find its area.

Accepted Answer

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

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

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

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

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

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

Answer

$5$

Answer

$6$

Answer

$\dfrac{48}{7}$

Answer

$10$

Answer

$7$

Answer

$\dfrac{24}{7}$

Question 6

Let $A=\begin{pmatrix}1&0&2&4\ i^2&1&3&\lg 100\ 0&1&3&\ln 1\ \sin\frac{\pi}{2}&i^{24}&C_4^3&7\end{pmatrix}$. Find $\det(A)$.

Accepted Answer

$\bullet$ Evaluate each special entry:

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

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

$\bullet$ Assemble the numerical matrix:

$$A = \begin{pmatrix}1 & 0 & 2 & 4\ -1 & 1 & 3 & 2\ 0 & 1 & 3 & 0\ 1 & 1 & 4 & 7\end{pmatrix}$$

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

$$\begin{pmatrix}1 & 0 & 2 & 4\ 0 & 1 & 5 & 6\ 0 & 1 & 3 & 0\ 0 & 1 & 2 & 3\end{pmatrix}$$

$\bullet$ $R_3 \mathrel{-}= R_2$, $R_4 \mathrel{-}= R_2$:

$$\begin{pmatrix}1 & 0 & 2 & 4\ 0 & 1 & 5 & 6\ 0 & 0 & -2 & -6\ 0 & 0 & -3 & -3\end{pmatrix}$$

$\bullet$ $R_4 \mathrel{-}= 	frac{3}{2} R_3$ gives upper-triangular with pivots $1, 1, -2, 6$:

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

Answer

$-12$

Answer

$1$

Answer

$-3$

Answer

$6$

Answer

$-1$

Answer

$3$

Question 7

The equation $\cos x = x$ has at least one solution in:

Accepted Answer

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

Answer

$(-1, 0)$

Answer

$(0, \pi/2)$

Answer

$(\pi, 2\pi)$

Answer

nowhere

Question 8

Evaluate $\int_0^1 \frac{\ln(1+x)}{1+x^2}\,dx$

Accepted Answer

**Substitute $x = 	an	heta$.** Then $dx = \sec^2	heta\,d	heta$ and $1 + x^2 = \sec^2	heta$, so the $\sec^2	heta$ on top and bottom cancel cleanly. The bounds $0 	o 1$ map to $	heta : 0 	o \pi/4$:

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

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

$$1 + 	an	heta \;=\; 1 + \frac{1 - 	an\phi}{1 + 	an\phi} \;=\; \frac{2}{1 + 	an\phi}.$$

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

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

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

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

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

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

Answer

$\dfrac{\pi \ln 2}{4}$

Answer

$\dfrac{\pi \ln 2}{8}$

Answer

$\dfrac{\pi \ln 2}{16}$

Answer

$\dfrac{\pi^2 \ln 2}{8}$

Answer

$\dfrac{\ln 2}{2}$

Answer

$\dfrac{\pi}{8}$

Question 9

Evaluate $\displaystyle\lim_{x	o 0} \dfrac{e^x - 1 - x}{x^2}$.

Accepted Answer

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

Answer

$0$

Answer

$\dfrac{1}{2}$

Answer

$1$

Answer

$2$

Question 10

$\displaystyle\int \dfrac{1}{x(x + 1)} \, dx$ equals:

Accepted Answer

$\int \!\left(\dfrac{1}{x} - \dfrac{1}{x+1}\right) dx = \ln|x| - \ln|x + 1| + C = \ln\!\left|\dfrac{x}{x + 1}\right| + C$.

Answer

$\ln|x(x + 1)| + C$

Answer

$\arctan x + C$

Answer

$\ln\!\left|\dfrac{x}{x+1}\right| + C$

Answer

$\dfrac{1}{x} - \dfrac{1}{x+1} + C$

Problems & worked solutions

Problem 1 — Differentiation

Problem 2 — Mean, Median, Mode

Problem 3 — Rolle's Sign Method

Problem 4 — Continuity

Problem 5 — Analytic Geometry

Problem 6 — Matrices

Problem 7 — Continuity

Problem 8 — Integrals

Problem 9 — L'Hôpital's Rule

Problem 10 — Antiderivatives