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Practice · Oxbridge maths · STEP

STEP practice questions

20 questions with worked solutions. Longer problems at STEP 2 and STEP 3 level, each with a single final answer and a full method.

5 easy · 9 medium · 6 hard · STEP 2 · STEP 3

Practise timed: 4 questions · 60 min All Oxbridge maths topics

1 Medium STEP 2
Evaluate ∫0π/2sin⁡xsin⁡x+cos⁡x dx\displaystyle\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx.
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Answer: π4\dfrac\pi4
Call it II, and let JJ be the same integral with cos⁡x\cos x on top. The substitution x↦π2−xx\mapsto\tfrac\pi2-x swaps sin⁡\sin and cos⁡\cos, so I=JI=J. But I+J=∫0π/21 dx=π2I+J=\int_0^{\pi/2}1\,dx=\tfrac\pi2. Hence I=π4I=\tfrac\pi4.
2 Medium STEP 2
Evaluate ∑r=16r⋅r!\displaystyle\sum_{r=1}^{6}r\cdot r!.
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Answer: 50395039
r⋅r!=(r+1)!−r!r\cdot r!=(r+1)!-r!, so the sum telescopes to 7!−1!=5040−17!-1!=5040-1. In general ∑r=1nr⋅r!=(n+1)!−1\sum_{r=1}^n r\cdot r!=(n+1)!-1.
3 Easy STEP 2
A sequence satisfies un+2=un+1+2unu_{n+2}=u_{n+1}+2u_n with u0=0u_0=0 and u1=1u_1=1. Find u10u_{10}.
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Answer: 341341
The characteristic equation λ2=λ+2\lambda^2=\lambda+2 has roots 2 and −1-1, so un=A2n+B(−1)nu_n=A2^n+B(-1)^n. The initial values give A=13A=\tfrac13, B=−13B=-\tfrac13: un=2n−(−1)n3u_n=\tfrac{2^n-(-1)^n}3, and u10=10233=341u_{10}=\tfrac{1023}3=341.
4 Hard STEP 3
Evaluate ∫01ln⁡(1+x)1+x2 dx\displaystyle\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx.
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Answer: π8ln⁡2\dfrac\pi8\ln2
Put x=tan⁡θx=\tan\theta: I=∫0π/4ln⁡(1+tan⁡θ) dθI=\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta. Now θ↦π4−θ\theta\mapsto\tfrac\pi4-\theta gives 1+tan⁡(π4−θ)=21+tan⁡θ1+\tan(\tfrac\pi4-\theta)=\tfrac{2}{1+\tan\theta}, so I=∫0π/4(ln⁡2−ln⁡(1+tan⁡θ))dθ=π4ln⁡2−II=\int_0^{\pi/4}\left(\ln2-\ln(1+\tan\theta)\right)d\theta=\tfrac\pi4\ln2-I. Hence I=π8ln⁡2I=\tfrac\pi8\ln2.
5 Medium STEP 2
Evaluate ∫0π/2sin⁡4x dx\displaystyle\int_0^{\pi/2}\sin^4x\,dx.
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Answer: 3π16\dfrac{3\pi}{16}
sin⁡4x=(1−cos⁡2x2)2=38−12cos⁡2x+18cos⁡4x\sin^4x=\left(\tfrac{1-\cos2x}2\right)^2=\tfrac38-\tfrac12\cos2x+\tfrac18\cos4x. The cosine terms integrate to zero over [0,π2][0,\tfrac\pi2], leaving 38⋅π2\tfrac38\cdot\tfrac\pi2. The reduction formula In=n−1nIn−2I_n=\tfrac{n-1}nI_{n-2} gives 34⋅12⋅π2\tfrac34\cdot\tfrac12\cdot\tfrac\pi2 as a check.
6 Medium STEP 2
How many ordered pairs of positive integers (x,y)(x,y) satisfy 1x+1y=16\dfrac1x+\dfrac1y=\dfrac16?
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Answer: 99
Clear denominators: xy−6x−6y=0xy-6x-6y=0, i.e. (x−6)(y−6)=36(x-6)(y-6)=36. Both factors exceed −6-6, so a negative pair would have product below 36; both must be positive divisors. 36 has 9 divisors, each giving one ordered pair.
7 Easy STEP 2
Find the minimum value of x2+y2x^2+y^2 subject to x+2y=5x+2y=5.
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Answer: 55
x2+y2x^2+y^2 is the squared distance from the origin, minimised at the foot of the perpendicular to the line: ∣0+0−5∣212+22=255=5\tfrac{|0+0-5|^2}{1^2+2^2}=\tfrac{25}5=5, at (1,2)(1,2). Cauchy–Schwarz gives the same: 25=(x+2y)2≤5(x2+y2)25=(x+2y)^2\le5(x^2+y^2).
8 Medium STEP 2
Evaluate ∑n=1∞n22n\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{2^n}.
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Answer: 66
From ∑xn=11−x\sum x^n=\tfrac1{1-x}, apply xddxx\tfrac{d}{dx} twice: ∑n2xn=x(1+x)(1−x)3\sum n^2x^n=\tfrac{x(1+x)}{(1-x)^3}. At x=12x=\tfrac12: 12⋅3218=6\tfrac{\frac12\cdot\frac32}{\frac18}=6.
9 Hard STEP 2
In how many ways can a 3×43\times4 rectangle be tiled with 1×21\times2 dominoes?
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Answer: 1111
Let ana_n count tilings of a 3×n3\times n board (nn even) and bnb_n those of a 3×n3\times n board with one corner square removed. Looking at the left edge: an=an−2+2bn−1a_n=a_{n-2}+2b_{n-1} and bn=an−1+bn−2b_n=a_{n-1}+b_{n-2}. Eliminating bb gives an=4an−2−an−4a_n=4a_{n-2}-a_{n-4}. With a0=1a_0=1 and a2=3a_2=3: a4=12−1=11a_4=12-1=11.
10 Easy STEP 2
Evaluate ∫01x31−x2 dx\displaystyle\int_0^1x^3\sqrt{1-x^2}\,dx.
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Answer: 215\dfrac2{15}
Put u=1−x2u=1-x^2, du=−2x dxdu=-2x\,dx, x2=1−ux^2=1-u: the integral becomes 12∫01(1−u)u du=12(23−25)=215\tfrac12\int_0^1(1-u)\sqrt u\,du=\tfrac12\left(\tfrac23-\tfrac25\right)=\tfrac2{15}.
11 Medium STEP 2
How many real roots does x5−5x+1=0x^5-5x+1=0 have?
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Answer: 33
f′(x)=5(x4−1)f'(x)=5(x^4-1) is zero only at x=±1x=\pm1. f(−1)=5>0f(-1)=5>0 and f(1)=−3<0f(1)=-3<0, with f→∓∞f\to\mp\infty as x→∓∞x\to\mp\infty. So there is exactly one root in each of (−∞,−1)(-\infty,-1), (−1,1)(-1,1) and (1,∞)(1,\infty), and ff is monotonic on each.
12 Hard STEP 3
Find lim⁡n→∞∑k=1nkn2+k2\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{n^2+k^2}.
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Answer: 12ln⁡2\tfrac12\ln2
Write each term as 1n⋅k/n1+(k/n)2\tfrac1n\cdot\tfrac{k/n}{1+(k/n)^2}: a Riemann sum for ∫01x1+x2 dx=12ln⁡2\int_0^1\tfrac{x}{1+x^2}\,dx=\tfrac12\ln2.
13 Medium STEP 2
Evaluate ∫01ln⁡x dx\displaystyle\int_0^1\ln x\,dx.
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Answer: −1-1
The integral is improper at 0. By parts, ∫ln⁡x dx=xln⁡x−x\int\ln x\,dx=x\ln x-x, and xln⁡x→0x\ln x\to0 as x→0+x\to0^+, so the value is (0−1)−0=−1(0-1)-0=-1.
14 Medium STEP 2
Evaluate ∑n=1∞1n(n+2)\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+2)}.
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Answer: 34\dfrac34
1n(n+2)=12(1n−1n+2)\tfrac1{n(n+2)}=\tfrac12\left(\tfrac1n-\tfrac1{n+2}\right). The sum telescopes in two interleaved chains to 12(1+12)=34\tfrac12\left(1+\tfrac12\right)=\tfrac34.
15 Hard STEP 3
Evaluate ∫0π/2ln⁡(sin⁡x) dx\displaystyle\int_0^{\pi/2}\ln(\sin x)\,dx.
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Answer: −π2ln⁡2-\dfrac\pi2\ln 2
Call it II. The substitution x↦π2−xx\mapsto\tfrac\pi2-x shows ∫0π/2ln⁡(cos⁡x) dx=I\int_0^{\pi/2}\ln(\cos x)\,dx=I too. Adding, 2I=∫0π/2ln⁡(12sin⁡2x) dx=−π2ln⁡2+12∫0πln⁡(sin⁡u) du=−π2ln⁡2+I2I=\int_0^{\pi/2}\ln(\tfrac12\sin 2x)\,dx=-\tfrac\pi2\ln2+\tfrac12\int_0^{\pi}\ln(\sin u)\,du=-\tfrac\pi2\ln2+I. Hence I=−π2ln⁡2I=-\tfrac\pi2\ln2.
16 Easy STEP 2
How many solutions in positive integers does x+y+z=12x+y+z=12 have?
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Answer: 5555
Write x=1+ax=1+a and so on: a+b+c=9a+b+c=9 with a,b,c≥0a,b,c\ge0, which has (9+22)=55\binom{9+2}{2}=55 solutions.
17 Easy STEP 2
Evaluate ∑k=110k3\displaystyle\sum_{k=1}^{10}k^3.
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Answer: 30253025
∑k=1nk3=(n(n+1)2)2=552=3025\sum_{k=1}^n k^3=\left(\tfrac{n(n+1)}2\right)^2=55^2=3025: the square of the sum of the first ten integers.
18 Hard STEP 3
What is the smallest positive solution of tan⁡x=x\tan x=x? Give a decimal.
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Answer: x≈4.4934x\approx4.4934
On (0,π2)(0,\tfrac\pi2) the graph of tan⁡x\tan x stays above y=xy=x, so the first crossing is in (π,3π2)(\pi,\tfrac{3\pi}2), just below the asymptote. Newton's method from x0=4.5x_0=4.5 converges to 4.49344.4934.
19 Medium STEP 2
Evaluate ∫0∞e−xsin⁡x dx\displaystyle\int_0^{\infty}e^{-x}\sin x\,dx.
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Answer: 12\dfrac12
Integrate by parts twice to get I=1−II=1-I, so I=12I=\tfrac12. (Equivalently, take the imaginary part of ∫0∞e−(1−i)xdx=11−i\int_0^\infty e^{-(1-i)x}dx=\tfrac1{1-i}.)
20 Hard STEP 2
How many subsets of {1,2,…,10}\{1,2,\dots,10\} contain no two consecutive integers?
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Answer: 144144
Let ana_n be the count for {1,…,n}\{1,\dots,n\}. Splitting on whether nn is used gives an=an−1+an−2a_n=a_{n-1}+a_{n-2} with a1=2a_1=2, a2=3a_2=3: the Fibonacci numbers, and a10=144a_{10}=144.

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