Daily Math · 2026-05-22

Question 1

Let $f(x)=3\sin\!\left(2\pi x+\dfrac{\pi}{4}\right)$. Find $f'\!\left(\dfrac{\pi}{3}\right)+f''\!\left(\dfrac{\pi}{6}\right)$.

Accepted Answer

$\bullet$ Differentiate by the chain rule:

$$f'(x)=6\pi\cos(2\pi x+	frac{\pi}{4})$$

$\bullet$ Differentiate again:

$$f''(x)=-12\pi^2\sin(2\pi x+	frac{\pi}{4})$$

$\bullet$ Evaluate at $x=\pi/3$ and $x=\pi/6$:

$$f'\!\left(	frac{\pi}{3}ight)=6\pi\cos\!\left(	frac{2\pi^2}{3}+	frac{\pi}{4}ight)$$

$$f''\!\left(	frac{\pi}{6}ight)=-12\pi^2\sin\!\left(	frac{\pi^2}{3}+	frac{\pi}{4}ight)$$

$\bullet$ Add the two:

$$6\pi\cos\!\left(	frac{2\pi^2}{3}+	frac{\pi}{4}ight)-12\pi^2\sin\!\left(	frac{\pi^2}{3}+	frac{\pi}{4}ight)$$

Answer

$6\pi\cos\!\left(\frac{2\pi^2}{3}+\frac{\pi}{4}\right)-12\pi^2\sin\!\left(\frac{\pi^2}{3}+\frac{\pi}{4}\right)$

Answer

$0$

Answer

$6\pi-12\pi^2$

Answer

$3\pi$

Answer

$-6\pi$

Answer

$6\pi+12\pi^2$

Question 2

The set of points $(x, y)$ equidistant from $A(0, 0)$ and $B(4, 0)$ is the line:

Accepted Answer

Set $\sqrt{x^2 + y^2} = \sqrt{(x - 4)^2 + y^2}$. Squaring: $x^2 = (x - 4)^2 = x^2 - 8x + 16$, so $8x = 16$, $x = 2$.

Answer

$y = 2$

Answer

$y = 0$

Answer

$x = 2$

Answer

$x = 0$

Question 3

Using Pascal's identity $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$, compute $\binom{7}{3} + \binom{7}{4}$.

Accepted Answer

By Pascal: $\binom{7}{3} + \binom{7}{4} = \binom{8}{4} = \dfrac{8!}{4!\,4!} = 70$.

Answer

$35$

Answer

$56$

Answer

$70$

Answer

$128$

Problems & worked solutions