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Written exams of Mathematical Analysis I 2018 - 2019 1

Exam of January 30, 2019 - - Ist sitting . . . . . . . . . . . . . . . . . . . . . . . . . nd

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Exam of January 30, 2019 - - II

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Exam of February 12, 2019 - - Ist sitting . . . . . . . . . . . . . . . . . . . . . . . . . 11

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Exam of February 12, 2019 - - IInd sitting . . . . . . . . . . . . . . . . . . . . . . . . 15

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Exam of June 28, 2019 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

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Exam of September 10, 2019 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

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Exam of January 30, 2019 - - Ist sitting

Exercise 1, version A. (8 points) Consider the function   2x + 2 1/(x+1) e . f (x) = x+4 (a) Find domain, limits at the boundary of the domain, and asymptotes (if any) of f . (b) Study the differentiability of f on its domain and compute the first derivative f ′(x). (c) Find the monotonicity intervals and the extremum points of f , specifying whether they are local or global. (d) Plot a qualitative graph of f . (e) Study the sign of the solution of the Cauchy problem n y′ (x) = 2 − f (x) y(1/2) = 0 in a right and in a left neighbourhood of 1/2. Exercise 2, version A. (5 points) (a) Let A ⊆ R be a bounded from below set. Give the definition of greatest lower bound of A. (b) Determine if the sequence an = arctan



2n + 3 n+1



(n ∈ N).

is monotone; if it is, specify whether it’s increasing or decreasing. (c) Given the set A = {an : n ∈ N}, find, if they exist, minimum or greatest lower bound, maximum or least upper bound of A. Motivate your answer.

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SOLUTION Exercise 1. a) domf = R \ {−4, −1}.

1 1 2x + 2 > 0 in a neighbour= −∞, limx→−1+ = +∞, and x+4 x+1 x+1 hood of x = −1. Thus, if we set s = 1/(x + 1), we get

Remark that

lim

x→−1−

lim f (x) = 0,

x→−1−

lim + f (x) =

x→−1

2 2/s es = +∞. es = lim s→+∞ 1 + 3s s→+∞ (1/s) + 3 lim

Moreover lim f (x) = lim f (x) = 2.

x→+∞

x→−∞

Since f (x) > 0 in a left neighbourhood of x = −4 and f (x) < 0 in a right neighbourhood of x = −4, we have lim f (x) = +∞

x→4−

and

lim f (x) = −∞.

x→4+

Therefore, (a) x = −1 is the equation of the vertical right asymptote of f . (b) x = −4 is the equation of the vertical bilateral asymptote of f . (c) y = 2 is the equation of the horizontal bilateral asymptote of f . b) f is a composition of differentiable functions, hence it’s differentiable for all x ∈ domf . Moreover ′

f (x) =

 2(x + 1) −1 2(x + 4) − 2(x + 1) + · e1/(x+1) = (x + 4)2 x+4 (x + 1)2   2 1 2x − 1 1/(x+1) 3 2 e . e1/(x+1) = − · 2 x+1 (x + 4) x+4 x+4 x+1



=

c) In order to find the extremum points and the monotonicity intervals of f we find first of all its critical points (that is the points where f ′ (x) = 0) and then the sign of f ′. Remark that f ′(x) = 0 if and only if x = 1/2. Moreover • f ′ (x) > 0

⇐⇒

x ∈ (−∞, −4) ∪ (−4, −1) ∪ (1/2, +∞)

• f ′ (x) < 0

⇐⇒

x ∈ (−1, 1/2)

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Written exams of Mathematical Analysis I 2018 - 2019

By the Mean Value Theorem, we know that a function is monotone on every interval where the sign of its derivative is constant; moreover, f is continuous at x = 1/2. Hence: • f is strictly increasing in (−∞, −4), in (−4, −1) and in [1/2, +∞) (but not necessarily in the union of any couple of those intervals), • f is strictly decreasing in (−1, 1/2]. Since f is continuous at x = 1/2, it yields that x = 1/2 is a local minimum point of f , which has no other extremum points. Moreover, observe that f (1/2) =

1 + 2 2/3 2 2/3 e = e > 0. 4 + 1/2 3

d) A qualitative graph of f is y

2 −4

−1

1 2

x

e) Consider the Cauchy problem: n y′ (x) = 2 − f (x). y(1/2) = 0 Since f (1/2) ∈ (0, 2), 2 − f (1/2) > 0. The function g (x) = 2 − f (x) is continuous on its domain: by the Sign Permanence Theorem, there exists a neighbourhood I of x = 1/2 where  c 2019 Politecnico di Torino

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g(x) > 0. Hence the solution y(x) of the given Cauchy problem is such that y′ (x) = g (x) > 0 in I. Therefore the solution is strictly increasing in I . Remind that y(1/2) = 0; therefore, y(x) < 0, ∀x ∈ I ∩ (−∞, 1/2) and y(x) > 0, ∀x ∈ I ∩ (1/2, +∞).

Exercise 2, version A. a) We call greatest lower bound of a set A bounded from below the maximum of its lower bounds, that is the real number ℓ such that – ∀x ∈ A, ℓ ≤ x – ∀ε > 0, ∃x ∈ A such that ℓ ≤ x < ℓ + ε. b) Consider the sequence an = arctan



2n + 3 n+1



(n ∈ N).

Remark that bn =

2n + 3 2(n + 1) + 1 1 . = =2+ n + 1 n+1 n+1

Since, for all n ≥ 0 bn+1 = 2 +

1 1 < bn = 2 + , n+2 n+1

the sequence bn is strictly decreasing. The function arctan(·) is strictly increasing in R, therefore the sequence an = arctan(bn ) is strictly decreasing. In order to prove that the sequence is decreasing, we may use a different method as well. Actually, an is the restriction to the set of natural numbers of the function   2x + 3 f (x) = arctan x+1 defined in R \ {−1}. Since f ′ (x) = 1+

1 

2x+3 x+1

2 ·

−1 0. Prove that

Z

3

g(x) dx > 0.

−3

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Written exams of Mathematical Analysis I 2018 - 2019

SOLUTION Exercise 1. a)

• ∀k ∈ R, domf = R • ∀k ∈ R, f is continuous if x 6= 0, since it’s a composition of continuous functions. •

lim f (x) = +∞; the order of infinity of f with respect to x is greater than 1: therefore

x→+∞

f has no right oblique asymptote. •

lim f (x) = 2; therefore f has horizontal left asymptote, with equation y = 2.

x→−∞

• lim f (x) = 2 + e0 = 3, therefore f is continuous at x = 0 if and only if k = f (0) = 3. x→0

b) We set k = 3. The function f is differentiable, since it’s a composition of differentiable functions, for all x 6= 0. Moreover ′

(x ln |x|−x)

f (x) = e



 1 ln |x| + x · − 1 = (ln |x|) e(x ln |x|−x) . x

f is continuous at x = 0 and lim f ′ (x) = −∞. By a consequence of de l’Hopital Theorem, x→0

we have that f is not differentiable at x = 0, which is a point with vertical tangent.

c) In order to find the extremum points and the monotonicity intervals of f we find first of all its critical points (that is the points where f ′ (x) = 0) and then the sign of f ′. f ′ (x) = 0

⇐⇒

ln |x| = 0

⇐⇒

x = ±1.

Moreover • f ′ (x) < 0

⇐⇒

x ∈ (−1, 0) ∪ (0, 1).

• f ′ (x) > 0

⇐⇒

x ∈ (−∞, −1) ∪ (1, +∞).

By the Mean Value Theorem, we know that a function is monotone on every interval where the sign of its derivative is constant. Moreover, f is continuous in R, therefore, • f is strictly increasing in [−1, 1], • f is strictly decreasing in (−∞, −1] and in [1, +∞) (but not necessarily in the union of the two intervals). Hence x = −1 is the unique local maximum point of f and x = 1 is the unique local minimum

point of f . Moreover, f (1) = 2 + e−1 > 2.

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d) A qualitative graph of f is y

3 2

−1

O

1

x

e) Observe that, for all x > 0 x  f (x) = 2 + e(x ln x − x) = 2 + eln x · e−x = 2 + xx · e−x . (i) The first statement is false. Actually, we have that f (x) ∼ xx · e−x as x → +∞, hence the order of infinite of f is lower than xx .

(ii) The second statement is true. Actually: lim

x→+∞

f (x) xx · e−x = lim e−x = 0, = lim xx xx x→+∞ x→+∞

therefore f (x) = o(xx ), as x → +∞. (iii) The statement: f (x) = xx + o(1) as x → +∞ is false. If not, lim [f (x) − xx ] should be zero, while x→+∞

lim (f (x) − xx ) = lim (2 + xx e−x − xx ) = 2 + lim xx (e−x − 1) = −∞.

x→+∞

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x→+∞

x→+∞

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Written exams of Mathematical Analysis I 2018 - 2019

Exercise 2, Version A. a) Consider a function f , integrable on a closed and bounded interval I. For all a, b, c ∈ I : Z

b

f (x) dx = a

Z

c

Z

f (x) dx +

b

f (x) dx.

c

a

b) Consider a function f , continuous and even on [−3, 3]. Then the function g(x) = f (x) sin x is odd, since g(−x) = f (−x) sin(−x) = −f (x) sin x, for all x ∈ [−3, 3]. Now consider Z

3

g(x) dx =

0

Z

g(x) dx +

g(x) dx = I1 + I2 .

0

−3

−3

3

Z

We set t = −x; then, Z Z 0 Z 0 [−g (−t)] dt = g(x) dx = I1 = Therefore, Z

g(t) dt = −

3

3

−3

0

3

f (x) sin x dx =

−3

Z

Z

3 0

g(t) dt = −I2 .

3 −3

g(x) dx = −I2 + I2 = 0.

c) By assumption, g is continuous and non-negative in [−3, 3]. Moreover g(0) > 0. By the Sign Permanence Theorem, there exists a neighbourhood (−δ, δ) of 0 such that ∀x ∈ (−δ, δ), g(x) > g(0) > 0. 2 The properties of the definite integral ensure that Z b – If g(x) ≥ 0 on [a, b], then g(x) dx ≥ 0. a

– If g(x) >

g(0) on (−δ, δ), then 2

Z

δ

−δ

g(x) dx ≥

Z

δ

−δ

g(0) g(0) = δg (0). dx = 2δ · 2 2

Using the property of additivity with respect to the domain, we have, Z

3

−3

g(x) dx =

Z

−δ −3

g(x) dx +

Z

δ

g(x) dx + −δ

Z

3

g(x) dx

δ

≥ 0 + δg(0) + 0 ≥ δg (0) > 0.

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Exam of February 12, 2019 - - Ist sitting

Exercise 1. (8 points) Consider the function f (x) = |1 − log3 (x − 1)|. (a) Determine domain, zeroes, limits at the boundary of the domain and asymptotes (if any) of f. (b) Study the differentiability of f on its domain, and determine the type of its non-differentiability points. Compute the first derivative f ′(x). (c) Find the monotonicity intervals and the extremum points of f , specifying whether they are local or global. (d) Plot a qualitative graph of f . (e) Find out, with suitable motivations, if it’s true that lim [f (x)] = [ lim f (x)]

x→2+

x→2+

where [·] denotes the integer part function. Exercise 2. (5 points) (a) Write the definition of oblique asymptote, as x → +∞, of a function f (x). (b) Consider the following statement: If f : R → R is an infinite of order 1/2 with respect to u(x) = x2 + 1, as x → +∞, then f

has an oblique asymptote as x → +∞.

Is it true or false? If it’s true, prove it; if it’s false, find a counterexample. (c) Determine the principal part the and order of infinite of f (x) =

p

3 + (x +

√ 2 x) with respect

to u(x) = x, as x → +∞; then find out if f has an oblique asymptote as x → +∞.

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Written exams of Mathematical Analysis I 2018 - 2019

SOLUTION Exercise 1. a) domf = (1, +∞). Moreover, f (x) = 0

⇐⇒

log 3 (x − 1) = 1

⇐⇒

x−1=e

⇐⇒

⇐⇒

log(x − 1) = 1

x = e + 1.

Hence f has the unique zero x = e + 1. Moreover, •

lim f (x) = +∞. Since the order of infinite of f is lower than 1 with respect to x, f

x→+∞

has no right oblique asymptote. • lim f (x) = +∞. Hence the line with equation x = 1 is a vertical (right) asymptote of x→1+

f. Moreover, since 1 − log3 (x − 1) > 0 if and only if 1 < x < e + 1, we have ( 1 − log3 (x − 1) 1 < x ≤ e + 1 f (x) = log3 (x − 1) − 1 x > e + 1. b) f is continuous on its domain, since it’s a composition of continuous functions. Moreover, f is differentiable, since it’s a composition of differentiable functions, for all x ∈ domf \{e + 1}.

f ′ (x) =

Moreover

 −3 log2 (x − 1) · 1    x−1     3 log2 (x − 1) ·

1 x−1

lim

f ′(x) = −

lim

f ′ (x) =

x→(e+1)− x→(e+1)+

1 < x < e+1

x > e + 1.

3 log2 e 3 =− e e 3 3 log2 e . = e e

Since f is continuous at x = e + 1, by a consequence of de l’Hopital Theorem, we have f−′ (e + 1) = −

3 3 6= f+′ (e + 1) = . e e

Hence f is non-differentiable at x = e + 1, which is a corner point of f .  c 2019 Politecnico di Torino

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c) In order to find the extremum points and the monotonicity intervals of f we find first of all its critical points (that is the points where f ′ (x) = 0) and then the sign of f ′. f ′ (x) = 0 ⇐⇒

3 log 2 (x − 1) = 0 ⇐⇒ log 2 (x − 1) = 0 ⇐⇒ x − 1 = 1 ⇐⇒ x = 2. x−1

Hence f has the unique critical point x = 2. Moreover f ′(x) > 0 if x > e + 1 and f ′ (x) < 0 if 1 < x < e + 1. By the Mean Value Theorem, we know that a function is monotone on every interval where the sign of its derivative is constant. Moreover f is continuous at x = e + 1. Therefore, • f is strictly decreasing in (1, e + 1] and x = 2 is an inflectionary point, with horizontal tangent, of f ; • f is strictly increasing in [e + 1, +∞). Moreover, x = e + 1 is the unique extremum point of f ; it is an global minimum point, since f (e + 1) = 0 ≤ f (x) for all x ∈ dom f . d) A qualitative graph of f is y

O

1

2

e+1

x

e) Note that f (2) = 1, and there exists an interval (2 − δ, 2) where 2 < f (x) < 3 and that 0 < f (x) < 1 in (2, e + 1). Then,

( 1 [f (x)] = 0  c 2019 Politecnico di Torino

if x ∈ (2 − δ, 2) if x ∈ (2, e + 1).

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Written exams of Mathematical Analysis I 2018 - 2019

Hence lim [f (x)] = 0. x→2+

Moreover, lim f (x) = f (2) = 1 and x→2+   lim f (x) = [1] = 1 6= lim [f (x)] = 0. + x→2

x→2+

Therefore the two limits are not equal. Exercise 2. a) If f is defined on a half-line (α, +∞) and there exist a ∈ R \ {0}, b ∈ R such that f (x) = ax + b + o(1),

as x → +∞

we say that f has oblique asymptote, with equation y = ax + b, as x → +∞. Equivalently, if f is defined on a half-line (α, +∞) and f (x) = a ∈ R \ {0}, x→+∞ x • lim [f (x) − ax] = b ∈ R •

lim

x→+∞

we say that f has oblique asymptote, with equation y = ax + b, as x → +∞. b) Recall that f is an infinite of order 1/2 with respect to x2 + 1, as x → +∞, if f (x) lim = ℓ ∈ R \ {0}. Since (x2 + 1)1/2 ∼ x, as x → +∞, we have that f is an x→+∞ (x2 + 1)1/2 infinite of order 1 with respect to x and f (x) ∼ ℓx, as x → +∞ . This is one of the two conditions necessary to a function in order to have an oblique asymptote. The second one is that f (x) − ℓx = b + o(1), as x → +∞. It follows that the statement is false. For instance, the function f (x) = x +log x satisfies the given hypothesis, but it has no oblique asymptote, since f (x) − x = log x, which doesn’t go to a constant b, as x → +∞. c) Recall that (1 + t)1/2 = 1 + 12 t + o(t), as t → 0. Then   1/2 q √ √ 3 √ 2 f (x) = 3 + (x + x)2 = (x + x)2 1 + = (x + x)    √ 1 3 1 = |x + x| 1 + · √ √ +o = 2 (x + x)2 (x + x)2 √ = x + x + o(1) as x → +∞. Hence f has order of infinite 1 with respect to x, as x → +∞, but √ x = +∞. lim [f (x) − x] = lim x→+∞

x→+∞

Hence f has no oblique asymptote, as x → +∞.  c 2019 Politecnico di Torino

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Exam of February 12, 2019 - - IInd sitting

Exercise 1. (8 points) Consider the function 2 f (x) = e−1/ log(x − 4) .

(a) Determine domain, symmetries (if any), limits at the boundary of the domain and asymptotes (if any) of f . Explain with suitable motivations if there exists a continuous prolongation of f at some points. ˜ the continuous prolongation of f (x). Denote by f(x) (b) Study the differentiability of ˜f on its domain, and compute the first derivative f˜′(x). (c) Find the monotonicity intervals and the extremum points of ˜f(x), specifying whether they are local or global. ˜ (d) Plot a qualitative graph of f. (e) Explain how it’s possible to study the monotonicity of f˜ found at (c) without using derivatives. Exercise 2. (5 points) (a) Write Taylor expansion, with Peano remainder, of a function y = f (x), stating which are the hypotheses under which the formula is true. (b) The Maclaurin polynomial of degree 2 of a function f (x) is Q2 (x) = 1 − x + 2x2 . Write the Maclaurin polynomial P2 (x) of degree 2 of the function g(x) = x + log f (x). (c) Verify if it’s true that there exists a neighbourhood I of 0 such that the function g(x) defined at point (b) is positive in I \ {0}. Motivate your answer.

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Written exams of Mathematical Analysis I 2018 - 2019

SOLUTION Exercise 1. a) f is defined if and only if x2 − 4 > 0 and log(x2 − 4) 6= 0, that is if and only if √ (x < −2 ∨ x > 2) ∧ x 6= ± 5. √ √ √ √ Hence domf = (−∞, − 5) ∪ (− 5, −2) ∪ (2, 5) ∪ ( 5, +∞). Moreover, f (−x) = f (x), for all x ∈ domf ; hence f is even and its graph is symmetrical with respect to the y-axis. Furthermore, 2 lim e−1/ log(x − 4) = lim

x→2+

lim √

x→ 5

x→−2−

+

2 e−1/ log(x − 4) =

2 lim e−1/ log(x − 4) = √ −

x→ 5

2

lim√

−

2 e−1/ log(x − 4) = 0,

+

2 e−1/ log(x − 4) = +∞,

x→− 5

lim√

x→− 5

lim e−1/ log(x − 4) = lim

x→+∞

Hence x =

2 e−1/ log(x − 4) = 1,

x→−∞

2 e−1/ log(x − 4) = 1.

√ √ 5 is the equation of the left vertical asymptote of f and x = − 5 is the

equation of its right vertical asymptote. Moreover, f has a bilateral horizontal asymptote, with equation y = 1. The function f is continuous on its domain, because it’s a composition of continuous functions. Since its limits as x → ±2 are finite, f can be continuously prolonged at x = ±2, so tha...