Calculus 1 Final 2021 RUG - University of Groningen Final Exam (solutions) 5/Nov/ [0] 1. Prove that - Studeersnel (2024)

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University of Groningen Final Exam (solutions) 5/Nov/

[0] 1. Prove that (−1)

1 + (−1)

2 + (−1)

3 + · · · + (−1)

n = (

−1)

n − 1

2

, for any n ∈ N.

• Base step: (−1)

1 = −1 = (

−1)

1 − 1

2

✓

• Induction step: assume that

k∑

j=

(−1)

j = (

−1)

k − 1

2

for some k ∈ N. Then:

k∑

j=

(−1)

j + (−1)

k+ = (

−1)

k − 1

2

+ (−1)

k+

= (

−1)

k − 1 + 2(−1) k+

2

= (

−1)

k − 1 + (−1) k+ + (−1) k+

2

= (

−1)

k − 1 − (−1) k + (−1) k+

2

= (

−1)

k+ − 1

2

.

• Conclusion: it follows from the principle of mathematical induction that (−1)

1 + (−1)

2 + (−1)

3 + · · · +

(−1)

n = (

−1)

n − 1

2

, for any n ∈ N.

[0] 2. Find all complex numbers z ∈ C that satisfy the equation

∣

∣

∣

∣

z + 1 +

ı

1 − ı

∣

∣

∣

∣

=

∣

∣

∣

∣

z −

1 + ı

1 − ı

∣

∣

∣

∣

and plot them in the Argand

plane.

First we see that:

1 + ı

1 − ı

= 1 +

ı

1 − ı

·

1 + ı

1 + ı

= ı.

Therefore:

∣

∣

∣

∣

z + 1 +

ı

1 − ı

∣

∣

∣

∣

=

∣

∣

∣

∣

z −

1 + ı

1 − ı

∣

∣

∣

∣

⇐⇒ |z + ı| = |z − ı|

|a + (b + 1)ı| = |a + (b − 1)ı|

a

2 + (b + 1)

2 = a

2 + (b − 1)

2

a

2 + b

2 + 2b + 1 = a

2 + b

2 − 2 b + 1

4 b = 0

b = 0.

Therefore, the solution is z = a, with a ∈ R, or all real numbers. The plot is simply the real line.

University of Groningen Final Exam (solutions) 5/Nov/

[0] 3. Prove rigorously, using the (ε, δ)-definition of limit, that lim x→ 0

x

2 − 2 x + 1

x − 1

= −1.

• Preliminary analysis: First we notice that, due to continuity and together with lim x→c

f (x)

f (x)

= 1 for

arbitrary functions, lim x→ 0

x

2 − 2 x + 1

x − 1

= − 1 ⇐⇒ lim x→ 0

(x − 1)

2

x − 1

= − 1 ⇐⇒ lim x→ 0

(x − 1) = −1. This

means that it suffices to prove lim x→ 0

(x − 1) = −1. Now it is straightforward to choose δ = ε.

Proof. For all ε > 0, let δ > 0. If 0 < |x| < δ, then |x − 1 + 1| = |x − 1 − (−1)| < δ = ε.

Therefore, from the (ε, δ)-definition of limit it follows that lim x→ 0

(x − 1) = −1 and therefore

limx→ 0

x 2 − 2 x + 1

x − 1

= − 1.

[0] 4. Evaluate the limit lim x→ 0

sin x − 2 cos x − x + 2

x 2

without using L’Hospital’s rule.

We know from the lectures that

sin x = x −

x

3

3! +

x

5

5! +

· · ·

cos x = 1 −

x 2

2! +

x 4

4! +

· · ·

Therefore:

limx→ 0

sin x − 2 cos x − x + 2

x 2

= lim x→ 0

(x −

x 3

3!

+

x 5

5!

+ · · · ) − 2(1 −

x 2

2!

+

x 4

4! + · · · ) − x + 2

x 2

= lim x→ 0

2

x 2

2!

−

x 3

3!

+ · · ·

x 2

= 1.

[0] 5. Let x ∈ R and k ∈ N. Prove that for every k ≥ 1, the equation

x

2 k+ + a 2 kx

2 k + a 2 k− 1 x

2 k− 1 + · · · + a 2 x

2 + a 1 x + a 0 = 0,

where all the coefficients ai, i = 0, 1 ,... , 2 k, are real numbers, has at least one real solution.

Let p(x) = x

2 k+ + a 2 kx

2 k + a 2 k− 1 x

2 k− 1 + · · · + a 2 x

2 + a 1 x + a 0. Since p(x) is an odd polynomial, there

exist α < 0 and β > 0 such that p(α) < 0 and p(β) > 0.

It follows from the intermediate value theorem that there exists a c ∈ (α, β) such that p(c) = 0.

University of Groningen Final Exam (solutions) 5/Nov/

[0] 10. Evaluate the following integral

∫ π 2

− π 2

(

x

3 cos x sin

2 x + x sin x

)

dx.

Since x

3 cos x sin

2 x is an odd function, it immediately follows that

∫ π 2

− π 2

(

x

3 cos x sin

2 x + x sin x

)

dx =

∫ π 2

− π 2

(x sin x) dx.

Integrating by parts we have that

∫

x sin xdx = −x cos x + sin x. Therefore:

∫ π 2

− π 2

(

x

3 cos x sin

2 x + x sin x

)

dx =

∫ π 2

− π 2

(x sin x) dx = (−x cos x + sin x)|

π/ 2−π/ 2

= 2.

[0] 11. Find the average of the function f (t) =

1

√

2 − t

on the interval [0, 2].

By definition:

favg = 1 2

∫ 2

1

√

2 − t

dt,

We notice that

∫

2

1

√

2 − t

dt is improper. Therefore:

favg = 1 2

limx→ 2 −

∫ x

1

√

2 − t

dt

= 1

2

limx→ 2 −

(

− 2

√

2 − t

) ∣∣

∣

x

= lim x→ 2 −

(

−√ 2 − x +

√

2

)

=

√

2.

University of Groningen Final Exam (solutions) 5/Nov/

[0] 12. Let a and b be positive numbers. Find the length of the shortest line segment that is cut off by the first quadrant

and passes through the point (a, b).

x

y

(a, b)

The line that passes through (a, b) is given by y = m(x − a) + b, where m is the slope of the line. From this

we know that the length of the line segment is a function of m.

Of course there is an infinite number of lines that pass through (a, b) and that are cut-off by the first

quadrant. However, from the sketch we observe that if the line is going to have finite length, then m < 0.

Notice that if y = 0, then x =

ma − b

m

, while when x = 0, then y = b − ma. This means that the length of

the line that is cut off by the first quadrant and passes through the point (a, b) is:

l(m) =

√(

ma − b

m

) 2

• (b − ma)

2 .

Minimizing l(m) is equivalent to minimizing L(m) = l(m)

2 =

(

ma − b

m

) 2

+(b − ma)

2 = (ma−b)

2

(

1 +

1

m 2

)

,

so now we only focus on minimizing L(m). So:

L

′ (m) = 2a(ma − b)

(

1 +

1

m 2

)

−

2

m 3

(ma − b)

2 = 2(ma − b)

(

a +

a

m 2

−

a

m 2

+

b

m 3

)

= 2(ma − b)

(

a +

b

m 3

)

= 2(

ma − b)

m 3

(am

3 + b).

Notice at this point that since

ma − b

m 3

> 0. That implies that since m < 0, the only solution for L

′ (m) = 0

is m = −

(

b

a

) 1 / 3

. Moreover, the sign of L

′ (m) is exclusively given by the sign of (am

3 + b).

Next, we notice that L

′ (m) < 0 for m < −

(

b

a

) 1 / 3

and L

′ (m) > 0 for m > −

(

b

a

) 1 / 3

, which means that

the value of m that we have found corresponds to a minimum of L(m).

Finally, let us define m

∗ = −

(

b

a

) 1 / 3

. Thus, the length of the shortest line segment that is cut off by the

first quadrant and passes through the point (a, b) is l(m

∗ ), that is:

l(m

∗ ) =

√(

m ∗ a − b

m ∗

) 2

• (b − m ∗ a)

2 =

√

(m ∗ a − b) 2 + (m ∗ ) 2 (b − m ∗ a) 2

(m ∗ ) 2

=

√

(m ∗ a − b) 2 (1 + (m ∗ ) 2 )

(m ∗ ) 2

=

m

∗ a − b

m ∗

√

1 + (m ∗ ) 2 =

b

1 / 3 a

2 / 3 + b

b 1 / 3 a − 1 / 3

√

1 +

b 2 / 3

a 2 / 3

=

b

1 / 3 a

2 / 3 + b

b 1 / 3

√

a 2 / 3 + b 2 / 3

= (a

2 / 3 + b

2 / 3 )

√

a 2 / 3 + b 2 / 3 = (a

2 / 3 + b

2 / 3 )

3 / 2

University of Groningen Final Exam (solutions) 5/Nov/

[0] 15. Solve the differential equation xy

′ + 2y = x

2 − x + 2, and state for which values of x is the solution well-defined.

(That is, state the domain of the solution)

First we write the given equation as a linear first order differential equation:

y

′ + 2 x y

= x − 1 + 2 x ,

and compute that the integrating factor is

I(x) = x

2 .

Then

(x

2 y)

′ = x

3 − x

2 + 2x

x

2 y =

x

4

4

−

x

3

3

• x

2 + C

y =

x 2

4

−

x

3 + 1 +

C

x 2

.

For arbitrary constant C, we notice that the solution is well defined for all x ̸= 0. Only in the particular

case where C = 0 is the solution defined for all x ∈ R.

University of Groningen Final Exam (solutions) 5/Nov/

[1 bonus] 16. Consider a fixed population of N > 0 individuals. In epidemiology, the so-called SIS model is given by the

equations:

dS

dt

= −

β

N SI

• γI

dI

dt

=

β

N SI

− γI,

(1)

where S = S(t) denotes the number of Susceptible individuals at time t, and I = I(t) denotes the number of

Infected individuals at time t. Because the total population N is fixed, the conservation law S(t) + I(t) = N

holds for all t. Moreover, for the model one considers β > 0 and γ > 0.

a) Using the conservation law and (1), obtain a single differential equation that models the evolution of the

Infected individuals over time. [Hint: this equation should be of the form d

I

dt

= f (I, N, β, γ)]

b) Solve the equation obtained in item a). That is, find the function I(t) that satisfies the differential equation

of item a).

c) Prove, using the solution obtained in b) that: if

β

γ <

1, then lim t→∞

I(t) = 0. Give an interpretation of this

result.

d) Assuming

β

γ >

1, what is lim t→∞

I(t)? Give an interpretation of this result.

a) We substitute S = N − I in (1) and obtain:

dI

dt

=

β

N

(N − I)I − γI

= (β − γ)I −

β

N I

2

b) To simplify notation, let a = β − γ and b =

β

N

. Then we have:

dI

dt

= aI − bI

2

= (a − bI)I

dI

I(a − bI) = d

t.

Using partial fractions we find that

1

I(a − bI) =

1

aI

+

b

a(a − bI)

.