Question 1
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 1 (Total: 20 marks)
Forx∈R, let
f(x) = x
1 +x2.
(a) (i) Find the limits of f(x) as x→±∞, all zeros and all stationary points. Classify each
stationary point. [5 marks]
(ii) Find all points of inflection and sketch the graph of f, marking the stationary and
inflection points. [5 marks]
x
f(x)
0
(b) State precisely what it means for the improper integral
∫∞
−∞f(x)dx to exist. Determine whether
this integral exists. [4 marks]
(c) Let p∈R.
(i) Determine all p for which ∫ ∞
0
xp
(1 +x2)2dx
converges. [3 marks]
(ii) Determine all p for which
∞∑
n=1
np
(1 +n2)2
converges. [3 marks]
2
Worked solution and marking guidance
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 1 - Complete solution (Total: 20 marks)
(a) (i) Since f(x)∼1/x as|x|→∞, both end limits are 0. The only zero is x = 0. Also
f′(x) = 1−x2
(1 +x2)2.
Thusf is increasing on (−1, 1) and decreasing on (−∞,−1) and (1,∞). Hence (−1,−1/2) is the global
minimum and (1, 1/2) is the global maximum. [5 marks] Suggested level: A/B
(ii)
f′′(x) = 2x(x2−3)
(1 +x2)3 .
The sign changes at x =−
√
3, 0,
√
3, so the inflection points are
(
−
√
3,−
√
3
4
)
, (0, 0),
(√
3,
√
3
4
)
.
The graph is odd and has horizontal asymptote y = 0.
[5 marks] Suggested level: B
(b) For any fixed c∈R, existence means that both
∫ c
−∞
f(x)dx and
∫ ∞
c
f(x)dx
exist as finite limits. Taking c = 0,
∫ M
0
x
1 +x2 dx = 1
2 log(1 +M 2)−→∞.
Therefore the improper integral does not exist. Odd symmetry only gives a Cauchy principal value of 0; it
does not make the two improper halves finite. [4 marks] Suggested level: C
(c) (i) Near 0 the integrand behaves like xp, so integrability requires p> −1. At infinity it behaves like xp−4,
so integrability requires p−4<−1, i.e. p< 3. Therefore
−1<p< 3.
[3 marks] Suggested level: A/C
(ii) The general term is asymptotic to np−4. By the p-series test the series converges exactly when
p−4<−1, namely
p< 3.
There is no lower-end condition because the sum starts at n = 1. [3 marks] Suggested level: A/C
2
Question 2
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 2 (Total: 20 marks)
Use the Fourier transform convention
ˆf(ω) =
∫ ∞
−∞
f(x)e−iωxdx, f (x) = 1
2π
∫ ∞
−∞
ˆf(ω)eiωxdω.
Let a> 0 and b∈R.
(a) Find the Fourier transform of
f(x) =e−a|x−b|.
[5 marks]
(b) Put h(x) =e−a|x|. Compute the convolution (h∗h)(x) explicitly. [5 marks]
(c) Hence obtain
F−1
{ 1
(a2 +ω2)2
}
(x).
[4 marks]
(d) Use the Fourier energy theorem to evaluate
∫ ∞
−∞
dω
(a2 +ω2)2.
[4 marks]
(e) Use Fourier duality to find the Fourier transform of ( a2 +x2)−2. [2 marks]
3
Worked solution and marking guidance
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 2 - Complete solution (Total: 20 marks)
(a) Put u =x−b. Then
ˆf(ω) =e−iωb
∫ ∞
−∞
e−a|u|e−iωudu =e−iωb
( 1
a−iω+ 1
a +iω
)
.
Hence
ˆf(ω) = 2ae−iωb
a2 +ω2 .
[5 marks] Suggested level: A/B
(b) The convolution is even, so take x≥0 and split at u = 0 and u =x:
(h∗h)(x) =
∫ 0
−∞
e−a(x−u)eaudu +
∫ x
0
e−a(x−u)e−audu +
∫ ∞
x
ea(x−u)e−audu
= e−ax
2a +xe−ax + e−ax
2a .
Thus, for all x,
(h∗h)(x) =
(
|x|+ 1
a
)
e−a|x|.
[5 marks] Suggested level: B
(c) Since ˆh = 2a/(a2 +ω2),
F{h∗h}= 4a2
(a2 +ω2)2.
Therefore
F−1
{ 1
(a2 +ω2)2
}
(x) = 1 +a|x|
4a3 e−a|x|.
[4 marks] Suggested level: C
(d) Parseval gives
1
2π
∫ ∞
−∞
4a2
(a2 +ω2)2 dω=
∫ ∞
−∞
e−2a|x|dx = 1
a.
Hence ∫ ∞
−∞
dω
(a2 +ω2)2 = π
2a3 .
[4 marks] Suggested level: B/C
(e) Fourier duality gives F{G(x)}(ω) = 2πF−1{G}(−ω). Using part (c),
F
{ 1
(a2 +x2)2
}
(ω) = π
2a3 (1 +a|ω|)e−a|ω|.
[2 marks] Suggested level: D
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Question 3
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 3 (Total: 20 marks)
(a) For t≥0, consider
y′′+ 2y′+y = 2e−tH(1−t), y (0) = 0, y ′(0) = 0,
where H denotes the Heaviside function.
(i) Find the complementary function and prove that the two basis functions used are linearly
independent. [4 marks]
(ii) Find the solution for 0 ≤t< 1. [5 marks]
(iii) Continue the solution to t> 1 by imposing continuity of y and y′at t = 1. [4 marks]
(b) For x> 0, consider the Euler–Cauchy equation
x2y′′+xy′−y = logx +x.
(i) Using z = logx, find the complementary function. [3 marks]
(ii) Find a particular integral and hence the general solution. [4 marks]
4
Worked solution and marking guidance
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 3 - Complete solution (Total: 20 marks)
(a) (i) The characteristic equation is ( λ+ 1)2 = 0, so
yc = (C1 +C2t)e−t.
Fory1 =e−t and y2 =te−t,
W =y1y′
2−y′
1y2 =e−2t̸= 0,
so they are linearly independent. [4 marks] Suggested level: A
(ii) For 0≤t< 1, write y =e−tu(t). Then (D + 1)2y =e−tu′′, hence u′′= 2. The initial conditions give
u(0) =u′(0) = 0, so u =t2 and
y(t) =t2e−t, 0≤t< 1.
[5 marks] Suggested level: B
(iii) At t = 1,
y(1−) =e−1, y ′(1−) =e−1.
Fort> 1, y = (C +Dt)e−t. Continuity gives
C +D = 1, −C = 1,
so C =−1, D = 2. Therefore
y(t) = (2t−1)e−t, t> 1.
Both y and y′are continuous at 1; the forcing jump permits a jump in y′′.
[4 marks] Suggested level: C
(b) Put Y (z) =y(ez). Then xy′=Y′and x2y′′=Y′′−Y′, so the equation becomes
Y′′−Y =z +ez.
(i) The complementary solution is Yc =C1ez +C2e−z, hence
yc =C1x + C2
x .
[3 marks] Suggested level: A/B
(ii) For the term z, take Yp1 =−z. Since ez is resonant, take Yp2 = 1
2zez. Thus
y(x) =C1x + C2
x −logx + 1
2x logx.
[4 marks] Suggested level: C/D
4
Question 4
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 4 (Total: 20 marks)
Consider
u(x,y ) = (x2 +y2−4)(y−x).
(a) Find the complete zero contour u(x,y ) = 0. [3 marks]
(b) Find all stationary points of u. [6 marks]
(c) Compute the Hessian and classify every stationary point. [6 marks]
(d) Sketch representative contours, clearly marking the zero contour, stationary points and regions
where u is positive or negative. Explain why the sketch is consistent with the classifications in
part (c). [5 marks]
x
y
0
5
Worked solution and marking guidance
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 4 - Complete solution (Total: 20 marks)
(a)
u = 0 ⇐⇒x2 +y2 = 4 or y =x.
Thus the zero contour is the circle of radius 2 together with the line y =x. [3 marks] Suggested level: A
(b)
ux =−3x2 + 2xy−y2 + 4, u y =x2−2xy + 3y2−4.
Adding gives 2(y2−x2) = 0, so y =±x. If y = x, then x2 = 2. If y =−x, then x2 = 2/3. Hence the
stationary points are
(±
√
2,±
√
2),
(
−
√
6
3 ,
√
6
3
)
,
(√
6
3 ,−
√
6
3
)
.
[6 marks] Suggested level: B
(c)
H =
(−6x + 2y 2x−2y
2x−2y −2x + 6y
)
.
At (±
√
2,±
√
2), detH =−32 < 0, so both are saddles. Let a =
√
6/3. At (−a,a ), detH = 32 > 0 and
trH >0, so it is a strict minimum. At ( a,−a), detH = 32 > 0 and trH <0, so it is a strict maximum.
[6 marks] Suggested level: C
(d) The sign is the product of the signs of x2 +y2−4 and y−x: outside the circle it agrees with y−x, and
inside the circle it is reversed. The minimum lies in the negative region and the maximum in the positive
region; the two intersections of the zero curves are saddles.
[5 marks] Suggested level: C/D
5