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AP Calculus BC AP Calculus BC Practice Test 2025

AP Calculus BCModel questions
108marks
195min
10questions

This is the official AP Calculus BC AP Calculus BC question paper for 2025, as set in the Model questions examination. It carries 108 full marks and a time allowance of 195 minutes, across 10 questions. On Kekkei you can attempt this AP Calculus BC past paper online with a timer, get instant AI feedback and step-by-step solutions, and track the topics where you lose marks — completely free. Whether you are revising for your AP Calculus BC AP Calculus BC exam or solving previous years' question papers, this 2025 paper is a great way to practise under real exam conditions.

LevelAP Calculus BC
SubjectAP Calculus BC
Year2025 BS
Exam sessionModel questions
Full marks108
Time allowed195 minutes
Questions10, all with step-by-step solutions
A

Multiple Choice

Select the best answer.

10 questions·1 mark each
1Multiple choice1 mark

Which of the following is the Maclaurin series for e−x2e^{-x^2}?

  • a
    ∑n=0∞(−1)nx2nn!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}
  • b
    ∑n=0∞(−1)nxnn!\sum_{n=0}^{\infty} \frac{(-1)^n x^{n}}{n!}
  • c
    ∑n=0∞x2n(2n)!\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}
  • d
    ∑n=0∞(−1)nx2n(2n)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}
Correct answer: a
∑n=0∞(−1)nx2nn!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}

The Maclaurin series for eue^u is ∑n=0∞unn!\sum_{n=0}^{\infty} \frac{u^n}{n!}. Substituting u=−x2u = -x^2:

e−x2=∑n=0∞(−1)nx2nn!e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}

The correct answer is (a).

seriestaylor-series
2Multiple choice1 mark

A curve is defined by x(t)=t2−1x(t) = t^2 - 1 and y(t)=t3−3ty(t) = t^3 - 3t. Find dydx\frac{dy}{dx} at t=2t = 2.

  • a
    32\frac{3}{2}
  • b
    94\frac{9}{4}
  • c
    92\frac{9}{2}
  • d
    33
Correct answer: b
94\frac{9}{4}

dydx=dy/dtdx/dt=3t2−32t\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2-3}{2t}. At t=2t=2: 94\frac{9}{4}.

The correct answer is (b) 94\frac{9}{4}.

parametric-equationsderivatives
3Multiple choice1 mark

Find the area enclosed by one petal of the rose curve r=cos⁡(3θ)r = \cos(3\theta).

  • a
    π6\frac{\pi}{6}
  • b
    π4\frac{\pi}{4}
  • c
    π12\frac{\pi}{12}
  • d
    π3\frac{\pi}{3}
Correct answer: c
π12\frac{\pi}{12}

One petal: θ∈[−π/6,π/6]\theta \in [-\pi/6, \pi/6].

A=12∫−π/6π/6cos⁡2(3θ) dθ=π12A = \frac{1}{2}\int_{-\pi/6}^{\pi/6} \cos^2(3\theta)\,d\theta = \frac{\pi}{12}

The correct answer is (c) π12\frac{\pi}{12}.

polar-coordinatesintegrals
4Multiple choice1 mark

Which series converges?

I. ∑n=1∞n!nn\sum_{n=1}^{\infty} \frac{n!}{n^n}

II. ∑n=1∞(−1)nn\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}

III. ∑n=1∞1nln⁡n\sum_{n=1}^{\infty} \frac{1}{n \ln n}

  • a

    I only

  • b

    I and II only

  • c

    II and III only

  • d

    I, II, and III

Correct answer: b

I and II only

I: Ratio test gives e−1<1e^{-1}<1, converges. II: AST applies, converges. III: Integral test diverges.

The correct answer is (b) I and II only.

seriesconvergence-tests
5Multiple choice1 mark

Evaluate ∫1∞1x3 dx\displaystyle\int_1^{\infty} \frac{1}{x^3}\,dx.

  • a
    12\frac{1}{2}
  • b
    11
  • c
    13\frac{1}{3}
  • d

    The integral diverges

Correct answer: a
12\frac{1}{2}
∫1∞x−3 dx=[x−2−2]1∞=12\int_1^{\infty} x^{-3}\,dx = \left[\frac{x^{-2}}{-2}\right]_1^{\infty} = \frac{1}{2}

The correct answer is (a) 12\frac{1}{2}.

integralsimproper-integrals
6Multiple choice1 mark

Find the particular solution of dydx=2xy\frac{dy}{dx} = 2xy given y(0)=3y(0) = 3.

  • a
    y=3e2xy = 3e^{2x}
  • b
    y=ex2+2y = e^{x^2}+2
  • c
    y=3ex2/2y = 3e^{x^2/2}
  • d
    y=3ex2y = 3e^{x^2}
Correct answer: d
y=3ex2y = 3e^{x^2}

Separate: dyy=2x dx\frac{dy}{y}=2x\,dx. Integrate: ln⁡∣y∣=x2+C\ln|y|=x^2+C, so y=Aex2y=Ae^{x^2}. y(0)=3⇒A=3y(0)=3\Rightarrow A=3.

y=3ex2y = 3e^{x^2}

The correct answer is (d).

differential-equationsseparation-of-variables
7Multiple choice1 mark

Evaluate ∫x2ex dx\displaystyle\int x^2 e^x\,dx.

  • a
    ex(x2−2x)+Ce^x(x^2-2x)+C
  • b
    ex(x2−2x+2)+Ce^x(x^2-2x+2)+C
  • c
    ex(x2+2x+2)+Ce^x(x^2+2x+2)+C
  • d
    ex(x2−x+1)+Ce^x(x^2-x+1)+C
Correct answer: b
ex(x2−2x+2)+Ce^x(x^2-2x+2)+C

IBP twice: ∫x2exdx=ex(x2−2x+2)+C\int x^2 e^x dx = e^x(x^2-2x+2)+C.

The correct answer is (b).

integralsintegration-by-parts
8Multiple choice1 mark

Radius of convergence of ∑n=1∞(x−2)nn⋅3n\displaystyle\sum_{n=1}^{\infty} \frac{(x-2)^n}{n \cdot 3^n}?

  • a
    11
  • b
    22
  • c
    33
  • d
    ∞\infty
Correct answer: c
33

Ratio Test: L=∣x−2∣/3<1⇒∣x−2∣<3L=|x-2|/3<1 \Rightarrow |x-2|<3. R=3R=3.

The correct answer is (c) 3.

seriesradius-of-convergence
9Multiple choice1 mark

Evaluate lim⁡x→0sin⁡x−xx3\displaystyle\lim_{x \to 0} \frac{\sin x - x}{x^3}.

  • a
    00
  • b
    −16-\frac{1}{6}
  • c
    16\frac{1}{6}
  • d
    −13-\frac{1}{3}
Correct answer: b
−16-\frac{1}{6}

Apply L'Hopital three times: cos⁡x−13x2→−sin⁡x6x→−cos⁡x6=−16\frac{\cos x-1}{3x^2}\to\frac{-\sin x}{6x}\to\frac{-\cos x}{6}=-\frac{1}{6}.

The correct answer is (b) −16-\frac{1}{6}.

limitslhopital-rule
10Multiple choice1 mark

Arc length integral for y=ln⁡(cos⁡x)y=\ln(\cos x) from x=0x=0 to x=π4x=\frac{\pi}{4}.

  • a

    ∫0π/41+ln⁡2(cos⁡x) dx\int_0^{\pi/4}\sqrt{1+\ln^2(\cos x)}\,dx

  • b

    ∫0π/41+cos⁡2x dx\int_0^{\pi/4}\sqrt{1+\cos^2 x}\,dx

  • c

    ∫0π/4tan⁡x dx\int_0^{\pi/4}\tan x\,dx

  • d

    ∫0π/4sec⁡x dx\int_0^{\pi/4}\sec x\,dx

Correct answer: d

∫0π/4sec⁡x dx\int_0^{\pi/4}\sec x\,dx

y′=−tan⁡xy'=-\tan x. 1+tan⁡2x=sec⁡x\sqrt{1+\tan^2 x}=\sec x.

L=∫0π/4sec⁡x dxL=\int_0^{\pi/4}\sec x\,dx

The correct answer is (d).

integralsarc-length

Frequently asked questions

Where can I find the AP Calculus BC AP Calculus BC question paper 2025?
The full AP Calculus BC AP Calculus BC 2025 (Model questions) question paper is available free on Kekkei. You can read every question online and attempt the paper under timed exam conditions.
Does the AP Calculus BC 2025 paper come with solutions?
Yes. Every question on this AP Calculus BC past paper includes a step-by-step solution, plus instant AI feedback when you attempt it on Kekkei.
How many marks is the AP Calculus BC AP Calculus BC 2025 paper?
The AP Calculus BC AP Calculus BC 2025 paper carries 108 full marks and is meant to be completed in 195 minutes, across 10 questions.
Is practising this AP Calculus BC past paper free?
Yes — reading and attempting this AP Calculus BC past paper on Kekkei is completely free.