EXAM STRUCTURE
(preliminary version, the number of questions/problems/points may vary)
Exam consists of a theoretical part and a practical part.
THEORETICAL PART
Any auxiliaries (incl. notes and calculators) are prohibited!
Duration: 40 min.
Content:
I) 5 true/false questions; 1 point for each correct answer, 5 points in total.
Example question: Decide whether the following statements are true. Argument if yes,
give a counterexample if not.
1. The product of two strictly decreasing functions is a strictly decreasing function.
1
Answer: False. Counterexample: f1 : (−∞; 0) → (−∞; 0), f1 (x) = , f2 : (−∞; 0) →
x
(0, +∞), f2 (x) = x2 . f1 , f2 are strictly decreasing on (−∞; 0), (f1 · f2 )(x) = x is
strictly increasing on (−∞; 0).
2. Any quadratic polynomial has two complex roots.
Answer: True. Justification: fundamental theorem of algebra.
II) 4 questions on formulating definitions and statements, giving examples; 3–5 points for
each correct answer, 12–20 points in total.
Example questions:
1. Formulate the definition of a strictly increasing function.
Answer: Let f : D → R be a function, D ⊆ R, x1 and x2 be any two points in D. The
function f is strictly increasing in D if f (x2 ) > f (x1 ) whenever x1 < x2 .
2. Formulate the Fundamental theorem of algebra.
Answer: Fundamental theorem of algebra: every single-variable polynomial with
complex coefficients
p(x) = xn + an−1 xn−1 + · · · + a1 x + a0 ,
a0 , a1 , . . . , an−1 ∈ C, n ≥ 1,
has at least one complex root.
PRACTICAL PART
Notes are allowed, handwritten, max. 1 A4 sheet (two-sided). Calculators are prohibited!
Duration: 70 min.
Content: 5 problems; 5–10 points for each correct answer (with justified solution). 25–40
points in total.
Example problems:
1
c
an−1 +
for all n ∈ N, where c > 0
1. Consider the sequence {an }n∈N with an =
2
an−1
and a0 = M > 0.
√
1.1) Prove that an ≥ c ∀n ∈ N. (2 point)
1
Hint: use the fact that t + ≥ 2 ∀t ≥ 0.
t
1.2) Using 1.2), prove that {an }n∈N monotonically decreases. (2 points)
1
1.3) Determine whether the sequence {an }n∈N is bounded or unbounded. (1 point)
1.4) Prove that {an }n∈N has a finite limit. (1 point)
1.5) Prove that lim an =
n→∞
√
c. (2 points).
Solution:
1.1) By mathematical induction, an > 0 ∀n∈ N. Indeed,
a0 = M > 0. For an arbitrary
c
1
ak +
> 0 because ak > 0 and c > 0. By
k ∈ N, assume that ak > 0. Then ak+1 =
2
ak
the principle of mathematical induction, an > 0 ∀n ∈ N.
1
Applying the fact t + ≥ 2 ∀t ≥ 0, we get
t
√ ! √
√
√
k an
k
k
1
k
√ +
an+1 =
an +
=
≥
2 = k, ∀n ∈ N.
2
an
2
an
2
k
√
k
an
k
an
k
1
an +
=
1+ 2 ≤
1+
= an ,
1.2) Since an ≥ k ∀n ∈ N, an+1 =
2
an
2
an
2
k
∀n ∈ N. Thus, an+1 ≥ an ∀n ∈ N, so {an }n∈N monotonically decreases.
1.3) Since {an }n∈N monotonically decreases, we have an ≤ a0 = M ∀n ∈ N.
1.4) The sequence {an }n∈N is bounded from below and monotonically decreasing, therefore, it has a finite limit (by the Weierstrass theorem).
1.5) Denote lim an = x. Then lim an−1 = x, and
n→∞
n→∞
1
k
1
k
x = lim an = lim
an−1 +
=
x+
.
n→∞
n→∞ 2
an−1
2
x
Solving the above equation for x ≥ 0 (because of an ≥ 0∀n ∈ N), we get x =
2. Compute
1/2
R
√
k.
ex sin(πx)dx (4 points).
0
Solution:
(
Z1/2
ex sin(πx)dx =
0
u = ex
du = ex dx
1
dv = sin(πx)dx v = − cos(πx)
π
e1/2
e0
1
=−
cos(π/2) + cos(0) +
π
π
π
|
{z
} | {z }
=0
=1/π
1
ex
= + 2 sin(πx)
π
π
Denote A :=
1/2
R
1/2
0
1
− 2
π
Z1/2
(
x
e cos(πx)dx =
0
0
1/2
0
1
+
π
u = ex
A=
√
1
e
A
+ 2 − 2,
π
π
π
√
π+ e
A= 2
.
π +1
2
Z1/2
ex cos(πx)dx
0
du = ex dx
1
dv = cos(πx)dx v = sin(πx)
π
0
0
√
π+ e
Answer: A = 2
.
π +1
ex
= − cos(πx)
π
√
Z1/2
Z1/2
1
e
1
x
e sin(πx)dx = + 2 − 2
ex sin(πx)dx.
π
π
π
ex sin(πx)dx. Then
therefore,
)
)