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Daily · 2026-08-16

Daily math problems for August 16, 2026 — Calculus, Logic, Quadratic Function & more

One bite-sized math problem set for the day. Solve the ten multiple-choice problems and reveal the worked solutions.

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Beginnercalculus
equals:

Problems & worked solutions

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Problem 1 — Antiderivatives

∫ex dx\displaystyle\int e^x \, dx∫exdx equals:

Show answer & worked solution
  1. A. ex+1+Ce^{x+1} + Cex+1+C
  2. B. xex+Cx e^x + Cxex+C
  3. C. ex+Ce^x + Cex+C✓ correct
  4. D. ln⁡x+C\ln x + Clnx+C

The exponential is its own antiderivative: ∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C.

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Problem 2 — Logic & Induction

Which statement is logically equivalent to "if n2n^2n2 is even, then nnn is even"?

Show answer & worked solution
  1. A. If nnn is even, then n2n^2n2 is even
  2. B. If nnn is odd, then n2n^2n2 is odd✓ correct
  3. C. If n2n^2n2 is odd, then nnn is even
  4. D. nnn is even if and only if n2n^2n2 is odd

The contrapositive of "n2n^2n2 even ⇒\Rightarrow⇒ nnn even" is "nnn odd ⇒\Rightarrow⇒ n2n^2n2 odd". A statement and its contrapositive always have the same truth value.

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Problem 3 — Quadratic Function

The vertex (extremum point) of f(x)=x2−4x+3f(x) = x^2 - 4x + 3f(x)=x2−4x+3 has xxx-coordinate:

Show answer & worked solution
  1. A. −4-4−4
  2. B. −2-2−2
  3. C. 222✓ correct
  4. D. 444

xV=−−42⋅1=2x_V = -\dfrac{-4}{2 \cdot 1} = 2xV​=−2⋅1−4​=2. (And f(2)=4−8+3=−1f(2) = 4 - 8 + 3 = -1f(2)=4−8+3=−1 is the minimum.)

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Problem 4 — Polynomial Rings

The factorization of P(X)=X3+X2−4X−4P(X) = X^3 + X^2 - 4X - 4P(X)=X3+X2−4X−4 over R\mathbb{R}R is:

Show answer & worked solution
  1. A. (X−1)(X2+4)(X - 1)(X^2 + 4)(X−1)(X2+4)
  2. B. (X+1)(X−2)(X+2)(X + 1)(X - 2)(X + 2)(X+1)(X−2)(X+2)✓ correct
  3. C. (X−1)(X2−X−4)(X - 1)(X^2 - X - 4)(X−1)(X2−X−4)
  4. D. (X3+1)(X−4)(X^3 + 1)(X - 4)(X3+1)(X−4)

P(−1)=0P(-1) = 0P(−1)=0, so (X+1)(X + 1)(X+1) is a factor. Polynomial division: P(X)=(X+1)(X2−4)=(X+1)(X−2)(X+2)P(X) = (X + 1)(X^2 - 4) = (X + 1)(X - 2)(X + 2)P(X)=(X+1)(X2−4)=(X+1)(X−2)(X+2).

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Problem 5 — Matrices

For general matrices AAA and BBB in Mn(R)M_n(\mathbb{R})Mn​(R) (with n≥2n \ge 2n≥2), which is true?

Show answer & worked solution
  1. A. AB=BAAB = BAAB=BA always
  2. B. AB≠BAAB \ne BAAB=BA in general✓ correct
  3. C. ABABAB does not exist
  4. D. A+B≠B+AA + B \ne B + AA+B=B+A

Matrix multiplication is not commutative in general. (Concrete example: A=(0100)A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}A=(00​10​), B=(0010)B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}B=(01​00​) — check AB≠BAAB \ne BAAB=BA.)

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Problem 6 — Asymptotes

The function f(x)=tan⁡xf(x) = \tan xf(x)=tanx has vertical asymptotes at:

Show answer & worked solution
  1. A. x=nπx = n\pix=nπ, n∈Zn \in \mathbb{Z}n∈Z
  2. B. x=π2x = \dfrac{\pi}{2}x=2π​ only
  3. C. x=π2+nπx = \dfrac{\pi}{2} + n\pix=2π​+nπ, n∈Zn \in \mathbb{Z}n∈Z✓ correct
  4. D. nowhere

tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{\sin x}{\cos x}tanx=cosxsinx​ has vertical asymptotes where cos⁡x=0\cos x = 0cosx=0, i.e. x=π2+nπx = \dfrac{\pi}{2} + n\pix=2π​+nπ.

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Problem 7 — Distances & Areas

The distance from the origin to the point P(−3,4)P(-3, 4)P(−3,4) is:

Show answer & worked solution
  1. A. 111
  2. B. 555✓ correct
  3. C. 777
  4. D. 252525

d=(−3)2+42=25=5d = \sqrt{(-3)^2 + 4^2} = \sqrt{25} = 5d=(−3)2+42​=25​=5.

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Problem 8 — Polynomials in ℂ

The number of real solutions of X4−5X2+4=0X^4 - 5X^2 + 4 = 0X4−5X2+4=0 is:

Show answer & worked solution
  1. A. 111
  2. B. 222
  3. C. 333
  4. D. 444✓ correct

With Y=X2Y = X^2Y=X2: Y2−5Y+4=0⇒Y∈{1,4}Y^2 - 5Y + 4 = 0 \Rightarrow Y \in \{1, 4\}Y2−5Y+4=0⇒Y∈{1,4}. Then X2=1⇒X=±1X^2 = 1 \Rightarrow X = \pm 1X2=1⇒X=±1 and X2=4⇒X=±2X^2 = 4 \Rightarrow X = \pm 2X2=4⇒X=±2. Four real solutions.

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Problem 9 — Definite Integrals

For fff continuous and non-negative on [a,b][a, b][a,b] with b>ab > ab>a:

Show answer & worked solution
  1. A. ∫abf<0\int_a^b f < 0∫ab​f<0 always
  2. B. ∫abf=0\int_a^b f = 0∫ab​f=0 always
  3. C. ∫abf≥0\int_a^b f \ge 0∫ab​f≥0✓ correct
  4. D. ∫abf\int_a^b f∫ab​f depends only on bbb

The integral of a non-negative continuous function over [a,b][a, b][a,b] is non-negative (representing the area under the curve).

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Problem 10 — Trigonometric Equations

The maximum value of f(x)=sin⁡x+3cos⁡xf(x) = \sin x + \sqrt{3}\cos xf(x)=sinx+3​cosx is:

Show answer & worked solution
  1. A. 1+31 + \sqrt{3}1+3​
  2. B. 222✓ correct
  3. C. 3\sqrt{3}3​
  4. D. 444

R=1+3=2R = \sqrt{1 + 3} = 2R=1+3​=2, so f(x)=2sin⁡(x+π/3)f(x) = 2\sin(x + \pi/3)f(x)=2sin(x+π/3). Max value: 222.

2026-08-15
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