Delft University of Technology

Calculus (CSE1200)
Test 1, 10-12-2019, 9:00–11:00
Remarks:
3
No calculators allowed, grade = 1 + 10
Score.

SHORT ANSWER QUESTIONS

Only the answers will be graded.
p
2pt 1. Let f (x) = 2 − ln(x).
Find the maximal domain of f .

Answer:
There are two conditions: x > 0 because of the logarithm, and 2 − ln(x) ≥ 0
because of the square root. The latter is equivalent to x ≤ e2 . We find (0, e2 ]
as maximal domain.

2pt 2. Write the following expressions without (inverse) trigonometric functions.

a. tan(arcsin( √17 ))

Answer:
Draw a right-angled triangle with side opposite to angle α equal to 1 and
√ √
hypotenuse 7. Then the adjacent side is 6. Therefore,
1 1
tan α = tan(arcsin( √ )) = √ .
7 6

b. arccos(cos( 53 π))

Answer:
Note that the answer x must satisfy cos(x) = cos( 53 π) AND 0 ≤ x ≤ π.
This holds for x = 13 π.
√ √ √ 
2pt 3. Find lim x · x−2− x+5 .
x→∞
If it does not exist, indicate whether
it is ∞, −∞ or neither.
 Answer:
Using the square root trick we can rewrite the function:
√ √ √  √ −7
x x − 2 − x + 5 = x√ √
x−2+ x+5
−7
=q q
1 − x2 + 1 + x5
7
→− as x → ∞.
2
1
2pt 4. Find lim (1 − 3x) x .
x→0
If it does not exist, indicate whether
it is ∞, −∞ or neither.

Answer:
The function can be rewritten as follows:
1 ln(1−3x)
(1 − 3x) x = e x .

Using l’Hospital, we find that

ln(1 − 3x) −3
lim = lim = −3.
x→0 x x→0 1 − 3x

By the substitution rule we find:

1
lim (1 − 3x) x = e−3 .
x→0

ln(4 − x)
2pt 5. Let f (x) = .
x2 − 3x
Find all x-values at which f has
a vertical asymptote.

Answer:
Values of x to investigate:

• x = 4: here the argument of the logarithm is 0, and the denominator is

finite, so the function has a vertical asymptote.
• x = 0: here the denominator is zero, but the numerator is not: the
function has a vertical asymptote.
 • x = 3: here both numerator and denominator vanish. We use l’Hospital
to find the limit:
ln(4 − x) −1/(4 − x) 1
lim 2
= lim =− .
x→3 x − 3x x→3 2x − 3 3
This is finite, so there is no vertical asymptote at x = 3.

2pt 6. Consider the curve defined by the equation

2x2 − x = y 3 − 7y.
dy
Find dx at the point (2, −1).

Answer:
We use implicit differentiation w.r.t. x:
dy
4x − 1 = (3y 2 − 7) .
dx
dy
Plugging in x = 2 and y = −1 and rewriting we get dx = − 74 .
√
2pt 7. Let f (x) = 13 − x2 .
Find the linearization of f near x = 2.
L(x) =

Answer:
We know that the linearization L of f near x = a is given by:

L(x) = f (a) + f 0 (a)(x − a).

Here a = 2 and f 0 (x) = √ −x . We find:

13−x2

2
L(x) = 3 − (x − 2).
3

2pt 8. Suppose x = 7 ± 0.2.

Using a linear approximation,
estimate the absolute error in arctan(x).

Answer:
Write y = arctan(x). The uncertainty in y can be estimated by

1 0.2
|dy| = |y 0 (x)dx| = dx = = 0.004.
1 + x2 50
2pt 9. Rewrite the following definite integral using the
√
substitution
Z 4 √ u = x:
cos( x)
dx
2 x2
Do not evaluate the integral!

Answer:
1
We get du = √
2 x
dx. We find that the integral equals:
Z 2
cos(u)
√
2 du.
2 u3

2pt 10. Consider

Z 2 the integral
an =
In = xn e3x dx for n ≥ 1.
0
Find coefficients an and bn such that bn =
In = an + bn In−1 .

Answer:
We use integration by parts to evaluate the integral:
 2 Z 2
1 n 3x n n−1 3x
In = x e − x e dx
3 0 0 3
1 n
= 2n e6 − In−1 .
3 3

OPEN QUESTIONS
Show your calculations and explanations.
Z ∞
ln(x)
4pt 11. Evaluate, if possible, the integral dx.
1 x3
In case of divergence, indicate whether it is ∞, −∞ or neither.
Explain all your steps!

Answer:
It is convenient to first find an antiderivative. We use integration by parts:
Z Z
1 1 1 1 1
3
ln(x) dx = − 2 ln(x) − − 3 dx
x 2x 2x
Z
1 1 1
= − 2 ln(x) + dx
2x 2 x3
1 1
= − 2 ln(x) − 2
2x 4x
 Now we evaluate the improper integral:
Z ∞ Z t
ln(x) ln(x)
3
dx = lim dx
1 x t→∞ 1 x3
 t
1 1
= lim − 2 ln(x) − 2
t→∞ 2x 4x 1
1 1 1
= lim − 2 ln(t) − 2 +
t→∞ 2t 4t 4
The first term goes to 0 (l’Hospital or “power function pulls harder than
logarithm”), the second goes to 0 (standard limit), hence we find:
Z ∞
1 1
3
ln(x) dx = .
1 x 4

a0 = 2,
12. Consider the sequence {an }∞
n=0 recursively defined as:
2
an+1 = 1 + an .
5

4pt a. Show that for all integer n ≥ 0 we have 1 ≤ an+1 ≤ an .

Explain all your steps!

Answer:
We use induction. Let P (n) be the statement 1 ≤ an+1 ≤ an . Note that
1 ≤ a1 = 95 ≤ a0 = 2, so the base case P (0) holds.
Assume that P (k) holds for some k: 1 ≤ ak+1 ≤ ak . Then 1 ≤ a2k+1 ≤ a2k
a2 a2
(since all are positive), hence 65 ≤ 1 + k+1
5
≤ 1 + 5k . Using the recursion
relation and the fact that 1 ≤ 65 , we find that 1 ≤ ak+2 ≤ ak+1 . Therefore,
P (k + 1) holds. By induction, it follows that P (n) holds for all integer n.
2pt b. Find lim an .
n→∞
No explanation needed.

Answer:
Since the sequence is decreasing and bounded from below, we know that it
2
converges to some L. Furthermore, we know that L = 1 + L5 , or equivalently,
√
L2 − 5L + 5 = 0. It follows that L = 12 (5 ± 5). Since a0 = 2 and the
sequence is decreasing, we must have that L ≤ 2, hence it follows that
√
L = 12 (5 − 5).

Exam created by B. van den Dries, checked by CSE1200 team.