Edexcel International A Level Maths · P4 · Pure Mathematics 4
The binomial series
28 free, high-quality past-paper-style questions on The binomial series for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 264 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P4-3-B-01P4 · A2Easy7 marks
f(x) = 1/(1 − x)3, which can be written as (1 − x)−3.
(a)
Find the binomial expansion of (1 − x)−3 in ascending powers of x, up to and including the term in x3.
(4)
(b)
State the range of values of x for which this expansion is valid.
(1)
(c)
Hence find the coefficient of x2 in the expansion of (1 + 2x)/(1 − x)3.
(2)
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Answer
Mark
Notes
Part (a)
1 + (−3)(−x) + [(−3)(−4)/2](−x)2 + [(−3)(−4)(−5)/6](−x)3, with n = −3 and −x in place of x
M1
1 + 3x
A1
+ 6x2
A1
+ 10x3
A1
Part (b)
|x| < 1
B1
Part (c)
(1 + 2x)(1 + 3x + 6x2 + …): x2 terms are 1 × 6x2 and 2x × 3x
M1
coefficient = 6 + 6 = 12
A1
Common error: Losing the minus sign inside the bracket. With (1 − x)−3 the term to substitute is (−x), so (−x)3 = −x3 and two negatives give +10x3.
Show that h(x) can be written as ¼(1 + x/2)−2, and hence find the binomial expansion of h(x) in ascending powers of x, up to and including the term in x3. Give each coefficient as a simplified fraction.
In the binomial expansion of (1 − ax)−4, where a is a positive constant, the coefficient of x2 is 90.
(a)
Find the value of a.
(3)
(b)
Find the coefficient of x3.
(2)
(c)
State the range of values of x for which the expansion is valid.
(1)
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Answer
Mark
Notes
Part (a)
x2 term: [(−4)(−5)/2](−ax)2 = 10a2x2
M1
10a2 = 90, so a2 = 9
M1
a = 3 (as a > 0)
A1
Part (b)
[(−4)(−5)(−6)/6](−3x)3 = (−20)(−27x3)
M1
coefficient = 540
A1
Part (c)
|3x| < 1, so |x| < 1/3
B1
Exam tip: To find one coefficient you only need that one term: write it out in full with the unknown inside the bracket, then set it equal to the given value.
Figure 1 shows the curve with equation y = 1/√(1 + kx), where k is a positive constant. The curve exists for x > −0.25, and the line x = −0.25 is a vertical asymptote to the curve.
(a)
Use Figure 1 to explain why k = 4.
(2)
(b)
Find the binomial expansion of y in ascending powers of x, up to and including the term in x3. Simplify each coefficient.
(4)
(c)
State the range of values of x for which this expansion is valid.
(1)
(d)
The curve in Figure 1 is defined at x = 0.5. Explain why the expansion in part (b) should not be used to estimate y when x = 0.5.
(1)
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Answer
Mark
Notes
Part (a)
the asymptote is where 1 + kx = 0, so 1 + k(−0.25) = 0
g(x) = (a − 3x)−½, where a is a positive constant. Figure 1 shows, on a number line, the set of values of x for which the binomial expansion of g(x) in ascending powers of x is valid. The point P marks x = −1.
(a)
Find the value of a.
(2)
(b)
Find the binomial expansion of g(x) in ascending powers of x, up to and including the term in x2. Give each coefficient as a simplified fraction.
(4)
(c)
Explain, with reference to Figure 1, why the expansion may be used when x = −1.
(1)
(d)
g(−1) = 1/√7. Use your expansion with x = −1 to estimate 1/√7, and find the percentage error in this estimate.
(the estimate is poor because x = −1 is close to the end of the interval, so the terms shrink slowly)
—
Exam tip: Valid is not the same as accurate. Near the ends of the interval the series still converges, but you need many more terms for a good estimate.
Figure 1 shows the coefficients of xr in the binomial expansion of (1 + x)n, for n = 1, 0, −1, −2, −3 and −4, and r = 0, 1, 2, 3 and 4. Three entries have been replaced by the letters p, q and s.
(a)
Use the formula n(n − 1)(n − 2)/3! to show that p = −10.
(2)
(b)
In the table, every entry with r ≥ 1, below the top row, equals the entry directly above it minus the entry to its left. Use this rule to find q and s.
(2)
(c)
Hence, or otherwise, find the binomial expansion of (1 − 2x)−4 in ascending powers of x, up to and including the term in x3.
(3)
(d)
State the range of values of x for which this expansion is valid.
(1)
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Answer
Mark
Notes
Part (a)
p is the coefficient of x3 for n = −3: (−3)(−4)(−5)/3!
M1
= −60/6 = −10
A1
Part (b)
q = 6 − (−4) = 10
B1
s = 15 − (−20) = 35
B1
Part (c)
row n = −4 with −2x in place of x: 1 + (−4)(−2x) + q(−2x)2 + (−20)(−2x)3
M1
= 1 + 8x + 40x2
A1
+ 160x3
A1
Part (d)
|−2x| < 1, so |x| < ½
B1
Common error: Forgetting that the whole term −2x is raised to the power: (−2x)3 = −8x3, which turns −20 into +160.
Find the first three terms, in ascending powers of x, of the binomial expansion of p(x), giving each coefficient as a simplified fraction.
(4)
(b)
State the range of validity of the expansion.
(1)
(c)
By substituting x = 1 into your answer to part (a), find an approximation for √2. Give your answer as a fraction in its simplest form and as a decimal to 5 decimal places.
(3)
(d)
Calculate the percentage error in this approximation.
Exam tip: Choose the substitution so that the left-hand side becomes a simple multiple of the root you want: √(9 − 1) = √8 = 2√2, so divide the series value by 2.
The function q is defined by q(x) = (2 − x)/∛(1 + 3x).
(a)
Find the binomial expansion of (1 + 3x)−⅓ in ascending powers of x, up to and including the term in x2.
(3)
(b)
Hence show that, for small values of x, q(x) ≈ 2 − 3x + 5x2.
(3)
(c)
State the values of x for which this approximation is valid.
(1)
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Answer
Mark
Notes
Part (a)
1 + (−⅓)(3x) + [(−⅓)(−4/3)/2](3x)2
M1
= 1 − x
A1
+ 2x2
A1
Part (b)
q(x) = (2 − x)(1 − x + 2x2 + …)
M1
= 2 − 2x + 4x2 − x + x2 + … (terms in x3 not needed)
M1
= 2 − 3x + 5x2 as required
A1
Part (c)
|3x| < 1, so |x| < 1/3
B1
Exam tip: When multiplying by a polynomial, collect only the powers you are asked for. Writing out x3 terms wastes time and they are incomplete anyway.
Hence, or otherwise, find the series expansion of r(x) in ascending powers of x, up to and including the term in x2. Give each coefficient as a simplified fraction.
(5)
(c)
State the range of values of x for which your expansion is valid.
(1)
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Answer
Mark
Notes
Part (a)
5 + x ≡ A(3 − x) + B(1 + x)
M1
x = −1: 4 = 4A, so A = 1
A1
x = 3: 8 = 4B, so B = 2; r(x) = 1/(1 + x) + 2/(3 − x)
A1
Part (b)
(1 + x)−1 = 1 − x + x2 + …
B1
2(3 − x)−1 = (2/3)(1 − x/3)−1
M1
= 2/3 + 2x/9 + 2x2/27 + …
A1
adds the two expansions term by term
M1
r(x) = 5/3 − 7x/9 + 29x2/27
A1
Part (c)
|x| < 1 (from 1/(1 + x); the other part needs only |x| < 3)
B1
Exam tip: Check the constant term: r(0) = 5/(1 × 3) = 5/3, which matches the first term of the series.
In the theory of relativity, a particle moving at speed v has a factor γ = (1 − u2)−½, where u = v/c and c is the speed of light. Throughout, 0 ≤ u < 1.
(a)
Find the binomial expansion of γ in ascending powers of u, up to and including the term in u4.
(3)
(b)
Explain why the expansion is valid for every value of u in the given range.
(1)
(c)
For a particle with u = 0.3, use your expansion to estimate γ, and find the percentage error in this estimate.
(3)
(d)
The kinetic energy of the particle is (γ − 1)mc2, where m is a constant. Show that, when u is small, this is approximately ½mv2.
(2)
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Answer
Mark
Notes
Part (a)
1 + (−½)(−u2) + [(−½)(−3/2)/2](−u2)2, using −u2 in place of x
M1
= 1 + ½u2
A1
+ (3/8)u4
A1
Part (b)
it needs |−u2| < 1, i.e. |u| < 1, which holds for 0 ≤ u < 1
Figure 1 shows a cross-section through the centre O of a planet, modelled as a sphere of radius R. An observer at A is a height h above the point S on the surface. The observer's line of sight AT touches the surface at T, and AT = d.
(a)
Show that d = √(2Rh) (1 + h/(2R))½.
(3)
(b)
Hence show that, when h is small compared with R, d ≈ √(2Rh) [1 + h/(4R) − h2/(32R2)].
(3)
(c)
For a planet with R = 6400 km and an observer at height h = 8 km, find √(2Rh) and use the first two terms of the bracket to estimate d.
(2)
(d)
Find the size of the third term, 320 × h2/(32R2), in centimetres, and hence state whether it matters when d is given to the nearest metre.
(2)
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Answer
Mark
Notes
Part (a)
AT is a tangent, so angle OTA = 90° and OA2 = OT2 + AT2
B1
d2 = (R + h)2 − R2 = 2Rh + h2
M1
= 2Rh(1 + h/(2R)), so d = √(2Rh) (1 + h/(2R))½
A1
Part (b)
(1 + u)½ = 1 + ½u − ⅛u2 + …
M1
with u = h/(2R): ½u = h/(4R)
A1
⅛u2 = h2/(32R2), giving the result
A1
Part (c)
√(2 × 6400 × 8) = √102400 = 320
B1
d ≈ 320(1 + 8/25600) = 320.1 km
A1
Part (d)
320 × 64/(32 × 64002) = 1.5625 × 10−5 km
M1
= 1.56 cm, far less than 1 m, so it makes no difference to d = 320 100 m
A1
Exam tip: Before expanding, take out the biggest part of the expression so that the bracket is 1 + (something small). Here 2Rh is much bigger than h2, so it comes outside the root.
f(x) = (px + q)/[(1 + 2x)(3 − x)], where p and q are constants. Figure 1 shows the curve y = f(x), which has vertical asymptotes x = −0.5 and x = 3. The curve crosses the x-axis at A(−1.2, 0) and the y-axis at B(0, 2).
(a)
Use the points A and B to find the values of p and q.
(3)
(b)
Express f(x) in partial fractions.
(3)
(c)
Hence find the series expansion of f(x) in ascending powers of x, up to and including the term in x2. Give each coefficient as a simplified fraction.
(3)
(d)
State the range of values of x for which this expansion is valid, and say which asymptote in Figure 1 fixes it.
(1)
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Answer
Mark
Notes
Part (a)
f(0) = q/(1 × 3) = 2, so q = 6
B1
f(−1.2) = 0 gives −1.2p + q = 0
M1
p = 5
A1
Part (b)
5x + 6 ≡ A(3 − x) + B(1 + 2x)
M1
x = 3: 21 = 7B, so B = 3
A1
x = −½: 3.5 = 3.5A, so A = 1; f(x) = 1/(1 + 2x) + 3/(3 − x)
A1
Part (c)
(1 + 2x)−1 = 1 − 2x + 4x2
B1
3(3 − x)−1 = (1 − x/3)−1 = 1 + x/3 + x2/9
M1
f(x) = 2 − 5x/3 + 37x2/9
A1
Part (d)
|x| < ½, set by the asymptote x = −0.5, the one nearer to the origin
B1
Exam tip: The constant term of the series must equal f(0). Here f(0) = 2 from Figure 1, which checks the 2 at the start of your answer.
Sn denotes the sum of the terms, up to and including the term in xn, of the binomial expansion of (1 + x)½ in ascending powers of x. Figure 1 shows the values of Sn, for n = 0, 1, 2, …, 10, when x = 0.5 and when x = 1.5, together with the lines y = √1.5 and y = √2.5.
(a)
Find the binomial expansion of (1 + x)½, up to and including the term in x3.
(3)
(b)
Show that, when x = 0.5, S3 = 157/128, and compare this value with √1.5.
(2)
(c)
Describe how Sn behaves when x = 1.5, as shown in Figure 1, and explain this behaviour.
(2)
(d)
By writing 2.5 = (9/4)(1 + 1/9), use the first three terms of your expansion to find an estimate for √2.5, giving your answer to 4 significant figures.
(3)
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Answer
Mark
Notes
Part (a)
1 + ½x + [(½)(−½)/2]x2 + [(½)(−½)(−3/2)/6]x3
M1
= 1 + ½x − ⅛x2
A1
+ (1/16)x3
A1
Part (b)
1 + 1/4 − 1/32 + 1/128 = 157/128
B1
157/128 = 1.2266, within 0.002 of √1.5 = 1.2247, as the circle at n = 3 in Figure 1 shows
B1
Part (c)
the sums swing above and below √2.5, with swings that get larger, so they do not settle on √2.5
B1
the expansion is valid only for |x| < 1, and 1.5 is outside this range
B1
Part (d)
√2.5 = (3/2)(1 + 1/9)½
M1
(3/2)(1 + 1/18 − 1/648)
M1
= 683/432 = 1.581
A1
Exam tip: When the number you want is outside the range of validity, rewrite it as (a perfect square) × (1 + small): here 2.5 = (9/4)(1 + 1/9) brings the value inside |x| < 1.
Figure 1 shows part of the curve with equation y = (1 + kx)n, where k and n are non-zero constants. The curve passes through the points A(−1.5, 2) and B(6, 0.5).
(a)
Show that (1 − 1.5k)(1 + 6k) = 1, and hence find the value of k.
(4)
(b)
Find the value of n.
(2)
(c)
Find the binomial expansion of y in ascending powers of x, up to and including the term in x3, simplifying each coefficient.
(3)
(d)
Explain why this expansion cannot be used to estimate the y-coordinate of B.
(1)
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Answer
Mark
Notes
Part (a)
(1 − 1.5k)n = 2 and (1 + 6k)n = 0.5
B1
multiplying: [(1 − 1.5k)(1 + 6k)]n = 1 and n ≠ 0, so (1 − 1.5k)(1 + 6k) = 1
it is valid only for |x/2| < 1, i.e. |x| < 2, and B has x = 6
B1
Exam tip: With an unknown power, multiplying or dividing the two equations can remove it: here the product of the right-hand sides is 1, so the n disappears.
Figure 1 shows the curve C with equation y = (8 + x)⅓ and the straight line L1, which passes through the points Q(−24, 0) and P(0, 2) and touches C at P.
(a)
Find the binomial expansion of (8 + x)⅓ in ascending powers of x, up to and including the term in x2. Give each coefficient as a simplified fraction.
(4)
(b)
Use Figure 1 to find an equation of L1, and show that it is given by the first two terms of your expansion.
(2)
(c)
Use your expansion with x = 1 to find an estimate for ∛9, giving your answer to 3 decimal places.
(2)
(d)
State the range of values of x for which the expansion is valid.
(1)
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Answer
Mark
Notes
Part (a)
(8 + x)⅓ = 8⅓(1 + x/8)⅓ = 2(1 + x/8)⅓
B1
(1 + x/8)⅓ = 1 + (⅓)(x/8) + [(⅓)(−⅔)/2](x/8)2 + …
M1
= 2 + x/12
A1
− x2/288
A1
Part (b)
gradient = (2 − 0)/(0 − (−24)) = 1/12
M1
y = 2 + x/12, which is the first two terms
A1
Part (c)
2 + 1/12 − 1/288 = 599/288
M1
∛9 ≈ 2.080
A1
Part (d)
|x/8| < 1, so |x| < 8
B1
Common error: Writing (8 + x)⅓ = 8(1 + x/8)⅓. The 8 taken out must also be cube-rooted, giving 2, not 8.
Figure 1 shows a circle with centre O and radius r. The chord AB has midpoint M, and AM = MB = c. The line through O and M meets the minor arc AB at N, and MN = s.
(a)
Show that s = r − r(1 − c2/r2)½.
(2)
(b)
Hence show that, when c is small compared with r, s ≈ c2/(2r) + c4/(8r3).
(3)
(c)
The cross-section of a curved roof is an arc of a circle of radius 25 m, and the chord AB has length 14 m. Use the approximation in part (b) to estimate s, and find the percentage error compared with the exact value of s.
(3)
(d)
For a different arc, c = 5 m and s = 0.5 m. Use s ≈ c2/(2r) to estimate r, and compare with the exact value of r.
(2)
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Answer
Mark
Notes
Part (a)
OM is perpendicular to AB, so OM2 = r2 − c2
B1
s = ON − OM = r − √(r2 − c2) = r − r(1 − c2/r2)½
A1
Part (b)
(1 − u)½ = 1 − ½u − ⅛u2 + … with u = c2/r2
M1
s ≈ r[½u + ⅛u2]
M1
= c2/(2r) + c4/(8r3)
A1
Part (c)
c = 7: 49/50 + 2401/125000
M1
s ≈ 0.9992 m
A1
exact s = 25 − √576 = 1 m, so percentage error = 0.0792% (accept 0.079%)
A1
Part (d)
r ≈ 25/(2 × 0.5) = 25 m
B1
exact: r2 = c2 + (r − s)2 gives r = (c2 + s2)/(2s) = 25.25 m, so the estimate is about 1% too small
B1
Exam tip: Treat c2/r2 as one quantity u while you expand, and only substitute back at the end. Then (1 − u)½ is a standard expansion.
Exam tip: In a 'show that' substitution, set out both sides: what the expression becomes (√5/2) and what the series gives (181/162), then connect them.
Given that f(x) = A/(1 + x) + B/(1 + x)2 + C/(3 − x), find the values of the constants A, B and C.
(4)
(b)
Hence find the series expansion of f(x) in ascending powers of x, up to and including the term in x2, giving each coefficient as a simplified fraction.
(6)
(c)
State, with a reason, the range of values of x for which your expansion is valid.
f(x) = (1 + kx)(1 − 3x)−⅓, where k is a constant and |x| < 1/3.
(a)
Find the binomial expansion of (1 − 3x)−⅓ in ascending powers of x, up to and including the term in x3.
(4)
(b)
In the expansion of f(x), the coefficient of x is zero. Find the value of k.
(2)
(c)
Using this value of k, show that f(x) ≈ 1 + x2 + (8/3)x3.
(2)
(d)
By substituting x = 1/24 into both the exact form of f(x) and the approximation in part (c), find an estimate for ∛7, giving your answer to 3 decimal places.
This question looks for a formula for every coefficient in a power series. The function 3/[(1 − x)(1 + 2x)] is called f(x), and its series in ascending powers of x is to be found using partial fractions.
(a)
Express f(x) in partial fractions.
(2)
(b)
Find the series expansion of f(x) in ascending powers of x, up to and including the term in x3.
(4)
(c)
State the range of values of x for which the expansion is valid.
(1)
(d)
Show that the coefficient of xr in the expansion of f(x) is 1 + (−1)r2r + 1.
(3)
(e)
Find the smallest value of r for which the coefficient of xr is greater than 1000.
(2)
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Answer
Mark
Notes
Part (a)
3 ≡ A(1 + 2x) + B(1 − x), substituting x = 1 and x = −½
M1
A = 1, B = 2: f(x) = 1/(1 − x) + 2/(1 + 2x)
A1
Part (b)
(1 − x)−1 = 1 + x + x2 + x3
B1
2(1 + 2x)−1 = 2[1 + (−1)(2x) + (2x)2 − (2x)3]
M1
= 2 − 4x + 8x2 − 16x3
A1
f(x) = 3 − 3x + 9x2 − 15x3
A1
Part (c)
|x| < ½ (the narrower of |x| < 1 and |2x| < 1)
B1
Part (d)
the xr term of (1 − x)−1 is xr, coefficient 1
B1
the xr term of 2(1 + 2x)−1 is 2(−2x)r
M1
2(−2)r = (−1)r2r + 1, so the coefficient is 1 + (−1)r2r + 1
A1
Part (e)
for odd r the coefficient is negative, so r is even and 1 + 2r + 1 > 1000, i.e. 2r + 1 > 999
M1
r = 8 gives 513 and r = 10 gives 2049, so r = 10
A1
Common error: Forgetting that (−2x)r alternates in sign. The coefficients of odd powers are negative, so they can never exceed 1000.
Figure 1 shows the curve with equation y = (1 + kx)n, where k > 0 and 0 < n < 1, together with (dashed) the parabola y = 1 + nkx + ½n(n − 1)k2x2, formed by the first three terms of the binomial expansion of (1 + kx)n. The vertex of the parabola is at V(0.5, 1.25).
(a)
Show that the x-coordinate of V is 1/[(1 − n)k] and that its y-coordinate is 1 + n/[2(1 − n)].
(4)
(b)
Use the coordinates of V to find the value of n and the value of k.
(3)
(c)
Find the coefficient of x3 in the binomial expansion of (1 + kx)n.
(2)
(d)
State the range of values of x for which the expansion is valid, and use it to explain why the two graphs in Figure 1 are far apart when x = 1.
(2)
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Answer
Mark
Notes
Part (a)
dy/dx = nk + n(n − 1)k2x = 0 at the vertex
M1
x = −nk/[n(n − 1)k2] = 1/[(1 − n)k]
A1
y = 1 + n/(1 − n) + ½n(n − 1)/(1 − n)2
M1
= 1 + n/(1 − n) − n/[2(1 − n)] = 1 + n/[2(1 − n)]
A1
Part (b)
1 + n/[2(1 − n)] = 1.25, so 4n = 2(1 − n)
M1
n = ⅓
A1
1/[(⅔)k] = 0.5, so k = 3
A1
Part (c)
[(⅓)(−⅔)(−5/3)/6](3x)3 = (5/81)(27x3)
M1
coefficient = 5/3
A1
Part (d)
|3x| < 1, so |x| < ⅓
B1
x = 1 is outside this range, so the terms of the series do not get smaller and three terms need not be close to (1 + 3x)⅓
B1
Exam tip: The vertex of a quadratic a + bx + cx2 is at x = −b/(2c). Work with n and k as letters until the end; the numbers only go in when you use V.
g(x) = (a + bx)/(1 − x)2, where a and b are constants. Figure 1 shows the coefficients of xr, for r = 0, 1, …, 5, in the series expansion of g(x) in ascending powers of x.
(a)
Write down the binomial expansion of (1 − x)−2 up to and including the term in x2, and use Figure 1 to find the values of a and b.
(3)
(b)
Show that your values agree with the coefficient of x2 in Figure 1.
(1)
(c)
Express g(x) in the form A/(1 − x) + B/(1 − x)2.
(3)
(d)
Hence show that the coefficient of xr in the expansion of g(x) is 3r + 2.
(3)
(e)
Find the sum of the coefficients of x0, x1, …, x19 in the expansion of g(x).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
(1 − x)−2 = 1 + 2x + 3x2 + …
B1
constant term: a = 2
B1
coefficient of x: 2a + b = 5, so b = 1
A1
Part (b)
3a + 2b = 6 + 2 = 8, as shown
B1
Part (c)
2 + x ≡ A(1 − x) + B
M1
x = 1: B = 3
A1
x = 0: 2 = A + 3, so A = −1
A1
Part (d)
the coefficient of xr in (1 − x)−1 is 1
B1
in (1 − x)−2 it is [(−2)(−3)…(−r − 1)/r!](−1)r = (r + 1)!/r! = r + 1
M1
3(r + 1) − 1 = 3r + 2
A1
Part (e)
arithmetic series with first term 2, common difference 3 and 20 terms: ½ × 20 × (2 × 2 + 19 × 3)
M1
= 610
A1
Common error: Reading off the coefficients of (1 − x)−2 with alternating signs. Both the power and the term −x are negative, so every coefficient is positive: 1, 2, 3, 4, …
Figure 1 shows four points A, O, B and P, in that order, on a straight line, with AO = OB = a and OP = d, where d > a > 0. A quantity Q is defined by Q = 1/PB2 − 1/PA2.
(a)
Show that Q = (1/d2)[(1 − a/d)−2 − (1 + a/d)−2].
(2)
(b)
Use binomial expansions to show that Q ≈ 4a/d3 + 8a3/d5.
(4)
(c)
Explain why these expansions are valid for every position of P allowed in Figure 1.
(1)
(d)
Given that a = 1 and d = 5, find the exact value of Q, the value of the approximation in part (b), and the percentage error in the approximation.
(3)
(e)
With a = 1, the simpler approximation Q ≈ 4/d3 is to be within 1% of Q. Using the ratio of the second term to the first term in part (b), find the smallest integer value of d for which this is true.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
PB = d − a and PA = d + a
B1
(d − a)−2 = d−2(1 − a/d)−2, and similarly for PA, giving the result
A1
Part (b)
(1 − u)−2 = 1 + 2u + 3u2 + 4u3 + …
M1
with u = a/d
A1
(1 + u)−2 = 1 − 2u + 3u2 − 4u3 + …
A1
difference = 4u + 8u3, so Q ≈ 4a/d3 + 8a3/d5
A1
Part (c)
they need |a/d| < 1, which holds because d > a > 0 (P is beyond B)
Exam tip: Expansions of (1 − u)n and (1 + u)n differ only in the signs of the odd powers. When you subtract them, the even powers cancel and the odd powers double.
f(x) = 3/[(1 + x)(2 − x)]. Figure 1 shows part of the curve y = f(x), which has vertical asymptotes x = −1 and x = 2. The shaded region R is bounded by the curve, the coordinate axes and the line x = 0.4.
(a)
Express f(x) in partial fractions.
(2)
(b)
Hence find the series expansion of f(x) in ascending powers of x, up to and including the term in x3. Give each coefficient as a simplified fraction.
(5)
(c)
State the range of values of x for which this expansion is valid, and explain why it may be used for every x-value in R.
(1)
(d)
Use your answer to part (b) to estimate the area of R, giving your answer to 3 decimal places.
(3)
(e)
Use your answer to part (a) to find the exact area of R in the form ln k, and hence find the percentage error in your estimate.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
3 ≡ A(2 − x) + B(1 + x); substitutes x = 2 and x = −1
M1
A = 1, B = 1: f(x) = 1/(1 + x) + 1/(2 − x)
A1
Part (b)
(1 + x)−1 = 1 − x + x2 − x3
B1
(2 − x)−1 = ½(1 − x/2)−1
M1
= ½ + x/4 + x2/8 + x3/16
A1
adds the two series term by term
M1
f(x) ≈ 3/2 − 3x/4 + 9x2/8 − 15x3/16
A1
Part (c)
|x| < 1; R has 0 ≤ x ≤ 0.4, which lies inside this range
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