Let f(x)=3sin (2π x+(π)/(4)). Find f' ((π)/(3))+f'' ((π)/(6)).

bullet Differentiate by the chain rule: f'(x)=6πcos(2π x+(π)/(4)) bullet Differentiate again: f''(x)=-12π^2sin(2π x+(π)/(4)) bullet Evaluate at x=π/3 and x=π/6: f' ((π)/(3))=6πcos ((2π^2)/(3)+(π)/(4)) f'' ((π)/(6))=-12π^2sin ((π^2)/(3)+(π)/(4)) bullet Add the two: 6πcos ((2π^2)/(3)+(π)/(4))-12π^2sin ((π^2)/(3)+(π)/(4))

6πcos ((2π^2)/(3)+(π)/(4))-12π^2sin ((π^2)/(3)+(π)/(4))

Daily Math Problems — May 22, 2026: Calculus, Analytic Geometry, Combinatorics

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Problems & worked solutions

RO M1

Problem 1 — Calculus

Let $f (x) = 3 \sin ⁣ (2 π x + \frac{π}{4})$ . Find $f^{'} ⁣ (\frac{π}{3}) + f^{''} ⁣ (\frac{π}{6})$ .

Show answer & worked solution

A. $6 π \cos ⁣ (\frac{2 π^{2}}{3} + \frac{π}{4}) - 12 π^{2} \sin ⁣ (\frac{π^{2}}{3} + \frac{π}{4})$ ✓ correct
B. $0$
C. $6 π - 12 π^{2}$
D. $3 π$
E. $- 6 π$
F. $6 π + 12 π^{2}$

$∙$ Differentiate by the chain rule:

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

$∙$ Differentiate again:

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

$∙$ Evaluate at $x = π / 3$ and $x = π / 6$ :

$f^{'} ⁣ (\frac{π}{3}) = 6 π \cos ⁣ (\frac{2 π^{2}}{3} + \frac{π}{4})$

$f^{''} ⁣ (\frac{π}{6}) = - 12 π^{2} \sin ⁣ (\frac{π^{2}}{3} + \frac{π}{4})$

$∙$ Add the two:

$6 π \cos ⁣ (\frac{2 π^{2}}{3} + \frac{π}{4}) - 12 π^{2} \sin ⁣ (\frac{π^{2}}{3} + \frac{π}{4})$

RO M1

Problem 2 — Distances & Areas

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

Show answer & worked solution

A. $y = 2$
B. $y = 0$
C. $x = 2$ ✓ correct
D. $x = 0$

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

RO M1

Problem 3 — Permutations & Combinations

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

Show answer & worked solution

A. $35$
B. $56$
C. $70$ ✓ correct
D. $128$

By Pascal: $(\binom{7}{3}) + (\binom{7}{4}) = (\binom{8}{4}) = \frac{8!}{4! 4!} = 70$ .

Problems & worked solutions

RO M1

Problem 1 — Calculus

Let $f (x) = 3 \sin ⁣ (2 π x + \frac{π}{4})$ . Find $f^{'} ⁣ (\frac{π}{3}) + f^{''} ⁣ (\frac{π}{6})$ .

Show answer & worked solution

A. $6 π \cos ⁣ (\frac{2 π^{2}}{3} + \frac{π}{4}) - 12 π^{2} \sin ⁣ (\frac{π^{2}}{3} + \frac{π}{4})$ ✓ correct
B. $0$
C. $6 π - 12 π^{2}$
D. $3 π$
E. $- 6 π$
F. $6 π + 12 π^{2}$

$∙$ Differentiate by the chain rule:

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

$∙$ Differentiate again:

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

$∙$ Evaluate at $x = π / 3$ and $x = π / 6$ :

$f^{'} ⁣ (\frac{π}{3}) = 6 π \cos ⁣ (\frac{2 π^{2}}{3} + \frac{π}{4})$

$f^{''} ⁣ (\frac{π}{6}) = - 12 π^{2} \sin ⁣ (\frac{π^{2}}{3} + \frac{π}{4})$

$∙$ Add the two:

$6 π \cos ⁣ (\frac{2 π^{2}}{3} + \frac{π}{4}) - 12 π^{2} \sin ⁣ (\frac{π^{2}}{3} + \frac{π}{4})$

RO M1

Problem 2 — Distances & Areas

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

Show answer & worked solution

A. $y = 2$
B. $y = 0$
C. $x = 2$ ✓ correct
D. $x = 0$

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

RO M1

Problem 3 — Permutations & Combinations

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

Show answer & worked solution

A. $35$
B. $56$
C. $70$ ✓ correct
D. $128$

By Pascal: $(\binom{7}{3}) + (\binom{7}{4}) = (\binom{8}{4}) = \frac{8!}{4! 4!} = 70$ .