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Musterklausur 19

Take Musterklausur 19 and compare your solution. From the course Mathematik I at Humboldt-Universität zu Berlin (HU Berlin).

Section MAIN-53a73233-9d76-4c66-8ff5-29a643847b0a

Mixed
Funktion f(x)
1 P

The following function is considered: f(x)=4(x−2)(x−1)2f(x) = \frac{4(x-2)}{(x − 1)²}f(x)=(x−1)24(x−2)​


a
1 P

Give the domain of f(x)f(x)f(x) and examine the function for zeros, poles and extreme values.

Your answer:

b
1 P

Examine the behavior of f(x)f(x)f(x) for x→+∞x → +∞x→+∞ and x→∞x → ∞x→∞ with the rules of l'Hospital.

Your answer:

c
1 P

Sketch the graph of the function on the interval [−2,4] (rough sketch).

Your answer:

d
1 P

Calculate the Taylor polynomial of the 2nd order with respect to the point x0=3x_0 = 3x0​=3.

Your answer:

e
1 P

Show that for x>1x > 1x>1 the function F(x)=4x−1+4ln(x−1)F(x) = \frac{4}{x-1} + 4 ln(x − 1)F(x)=x−14​+4ln(x−1) is an antiderivative of f(x)f(x)f(x) and determine ∫2e5+1f(x)dx\int_2^{e^5+1} f(x)dx∫2e5+1​f(x)dx.

Your answer:

Section MAIN-5592da82-ed63-4bfb-83c2-d6cac1de99d7

Mixed
Continuous payment stream
1 P

The following continuous payment stream with the flow rate R(t)=200—e0,8tR(t) = 200 — e^{0,8t}R(t)=200—e0,8t [€/year], for t≥0t≥ 0t≥0 is considered. The annual interest rate is p=4p = 4p=4.


a
1 P

Determine the interval [0,T], in which R(t)≥0R(t) ≥ 0R(t)≥0 for t=[0,T]t = [0,T]t=[0,T] applies.

Your answer:

b
1 P

Determine the continuous interest rate ppp equivalent to ppp.

Your answer:

c
1 P

Determine the present value of the payment stream, which flows from t=0t = 0t=0 to t=Tt = Tt=T, where TTT is the size determined under (a).

Your answer:

Section MAIN-f95a2f1a-ecbc-4924-a110-85aba9237677

Mixed
Isoquant
1 P

Given is the function z=f(x,y)=x2y2+xyz = f(x, y) = x²y² + xyz=f(x,y)=x2y2+xy, x,y>0x, y >0x,y>0. The equation ƒ(x,y)=6ƒ(x, y) = 6ƒ(x,y)=6 implicitly defines a function y=h(x)y = h(x)y=h(x) and thus the isoquant to the level z=6z = 6z=6 is defined.


a
1 P

Show that the point (4,½)(4, ½)(4,½) lies on the isoquant to the level z=6z = 6z=6.

Your answer:

b
1 P

Show by implicit differentiation that h′(x)=−2xy2−y2x2y+xh'(x) = \frac{-2xy^2 - y}{2x^2y + x}h′(x)=2x2y+x−2xy2−y​ and calculate h′(4)h'(4)h′(4).

Your answer:

c
1 P

Calculate the tangent that lies on the graph of the isoquant to the level z=6z = 6z=6 at the point (4,1)(4, 1)(4,1).

Your answer:

Section MAIN-fd912d41-3bbb-4388-a378-dfdfab372e77

Mixed
Local Extrema
1 P

Given is the function f(x,y)=x2+(x−2)y+y2+2f(x, y) = x² + (x − 2)y + y²+2f(x,y)=x2+(x−2)y+y2+2, x,y∈Rx, y∈ Rx,y∈R.


a
1 P

Determine the points where fff takes on local extrema using the method of Lagrange multipliers, where x,yx, yx,y must satisfy the constraint 2x−y=22x - y = 22x−y=2. (The proof of the sufficient condition is omitted.)

Your answer:

b
1 P

By how much does the optimal objective function value change approximately if the right-hand side of the constraint in (a) is reduced from 2 to 1.8 by 0.2? (The Lagrange multiplier is to be used!)

Your answer:

Section MAIN-c8cdb870-5cb0-4cf3-bca9-3574a7c40863

Mixed
Function f(x, y)
1 P

The function f(x,y)=(y−2)e−2x−x(y−2)2f(x, y) = (y-2)e^{-2x} - x(y-2)²f(x,y)=(y−2)e−2x−x(y−2)2, x,y∈Rx, y∈Rx,y∈R is considered.


a
1 P

Determine the gradient of f(x,y)f(x, y)f(x,y) at the point (xº,yº)=(0,3)(xº, yº) = (0,3)(xº,yº)=(0,3).

Your answer:

b
1 P

Determine the directional derivative of f(x,y)f(x, y)f(x,y) at the point (xº,yº)=(0,3)(xº, yº) = (0,3)(xº,yº)=(0,3) for the direction vector r=(2 1)r = \begin{pmatrix} 2 \ 1 \end{pmatrix}r=(2 1​).

Your answer:

c
1 P

Determine the total differential f(x,y)f(x, y)f(x,y) at the point (xº,yº)=(0,3)(xº, yº) = (0,3)(xº,yº)=(0,3) for dx=0,1dx = 0, 1dx=0,1 and dy=−0,01dy = −0, 01dy=−0,01.

Your answer:

d
1 P

Determine the partial elasticity ef,y(0,3)e_{f,y} (0,3)ef,y​(0,3) of ƒƒƒ with respect to yyy at the point (x,y)=(0,3)(x, y) = (0,3)(x,y)=(0,3).

Your answer:

e
1 P

Determine the points where f(x,y)f(x, y)f(x,y) has local extrema or saddle points, and determine what is the case.

Your answer:

f
1 P

By 0=f(x,y)–10 = f(x, y) – 10=f(x,y)–1 in a sufficiently small neighborhood of x=0x = 0x=0 a function y=h(x)y = h(x)y=h(x) is implicitly defined, for which h(0)=3h(0) = 3h(0)=3 applies. Determine the derivative h′(0)h'(0)h′(0).

Your answer:

Section MAIN-a8603b23-4234-442b-a392-556f62d3de64

Mixed
Difference Equation
1 P

The observation of a time-dependent economic characteristic yty_tyt​ showed that it satisfies the following difference equation: 13yt+2−13yt+1−13yt=3t+20\frac{1}{3}y_{t+2} - \frac{1}{3}y_{t+1} - \frac{1}{3}y_t = 3t+2031​yt+2​−31​yt+1​−31​yt​=3t+20, t=0,1,2,...t = 0, 1, 2, . . .t=0,1,2,...


a
1 P

Determine the solution of (1) for the initial conditions y0=10y_0 = 10y0​=10 and y1=17y_1 = 17y1​=17.

Your answer:

b
1 P

For the general solution yty_tyt​ of (1) determine the limit limt→∞ytlim_{t→∞} y_tlimt→∞​yt​.

Your answer:
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