Title: STACK Used Around the World
Course: Demonstrating STACK (v4.3.7)

Give an example of a non-trivial homogeneous system of three equations in \(x, y, z\) that has infinitely many solutions. (Non-trivial means at least one entry in the left hand matrix must be non-zero!)

\( \displaystyle \left[ \begin{array}{c}x\\y\\z\end{array}\right] = \)
Feedback: One way to solve this problem is to choose two equations which have a linear dependence. For example \(x+y=0\), \(x+y=0\) and \(z=0\). These have any solution \(x=-y\).

Give an example of a matrix with eigenvalues [value].
Feedback: We want the roots of \[ \mbox{det}\left( [value]-t[value]\right)=[value] =0\] to be [value]. There are lots of ways to arrange this, but the simplest is to take \(b_1=b_2=0\). We are left with \[ t^2-(a_1+a_2)t+a_1a_2=0\] which has as roots \(a_1,a_2\), and these must be \([value]\). This gives one choice as [value]. There are many others. Can you find an example where the matrix has no zero entries?

Find the vector form of the equation of the line in \( \mathbb{R^{2}} \) that passes through the point \(P=[value]\) and is perpendicular to the line with general equation \([value]\).

\(\mathbf{x}\) = + \(t\)
Feedback: To find the vector form of the equation of the line in \( \mathbb{R^{2}} \) we need to find two things (1) a point on the line and (2) the direction vector of the line. (1) We are actually given a point on the line in the question \(P=[value]\). (2) We need to find a direction perpendicular to the line with general equation \([value]\). Note that \[ [value] = [value]\cdot [value].\] Therefore we can turn the general form of the equation into the normal form \( \mathbf{n}\cdot \mathbf{x} = \mathbf{n}\cdot \mathbf{p}\) where \(\mathbf{p}\) is a point on the line. The vector \(\mathbf{n}\) is perpendicular to the line, so this will give us the required direction. We can therefore write \[ \mathbf{x} = [value] + t [value].\]

In each case below, give an example of an arithmetic sequence with the stated property, by entering an expression for the general term. If it is not possible, enter none.

(a) Increasing

      \( u_n=\)

(b) Decreasing

      \( u_n=\)

(c) Bounded above

      \( u_n=\)

(d) Decreasing and bounded below

      \( u_n=\)
Feedback: (a) To be increasing, each term must be strictly larger than the last. Since \(u_{n+1}-u_n=d\), that means that the common difference \(d\) must be stricly larger than \(0\). So, for example, \(u_n=n\). (b) Similarly, to be decreasing we need \(d<0\). So an example is \(u_n=-n\). (c) To be bounded above, we need the sequence to be non-increasing -- if we had \(d>0\) then the terms would keep increasing and eventually be bigger than any proposed upper bound. So we can take \(u_n=-n\) again. Another example would be a constant sequence, e.g. \(u_n=1\). (d) It is not possible for an arithmetic sequence to be decreasing and bounded below. If \(u_n\) is decreasing then the common difference is \(d<0\), and if we add enough multiples of \(d\) then we will eventually get a value smaller than any proposed lower bound.

Find the Maclaurin series of \(f(x)=[value]\).

(a) The first five nonzero terms are:

(b) The general term is:
Feedback: We compute the derivatives and evaluate them at \(x=0\): \[ \begin{align*} f(x)&=[value] &\quad f(0)&=[value] \\ f'(x)&=[value] &\quad f'(0)&=[value] \\ f''(x)&=[value] &\quad f''(0)&=[value] \\ f^{(3)}(x)&=[value] &\quad f^{(3)}(0)&=[value] \\ f^{(4)}(x)&=[value] &\quad f^{(4)}(0)&=[value]\end{align*} \] From the computed values and the pattern, we have \[\begin{align*} f(x) &=[value] \\ &=\sum_{n=1}^\infty [value]. \end{align*} \]

(a) Find the equation of the parabola shown in the diagram.

      \(y=\)

(b) The line with equation \(y= [value]\) is tangent to the parabola. Find the value of \(k.\)

      \(k=\)
Feedback: (a) The parabola has roots when \(x=[value]\) and \(x=[value]\) so its equation will be of the form \(y=[value]\). Since the parabola passes through \(([value],[value])\) we have \( [value]=[value] \) thus \(c=[value]\). So the parabola has equation \(y=[value]\). (b) The line \(y= [value]\) is tangent to the parabola \( y = [value] \) when these two intersect at one point. So we solve \[ \begin{align*} [value] &= [value] \\ [value] &= 0 \end{align*}\] To intersect at one point means that the discriminant of this quadratic equation should be \(0\). We have \(a=[value]\), \(b=[value]\) and \(c=[value]\), so \[ \begin{align*} b^2-4ac = ([value])^2 - 4([value])([value]) &= 0 \\ [value] &= 0 \\ k &= [value]\end{align*} \]

Find, without using a calculator, the value of \( \log_{[value]}[value]\).

Type the intermediary steps in your answer to show your working. The syntax for logarithms is as follows: \( \log_{a} b\) is entered as lg(b,a).

Your last line will be taken as your final answer.
Feedback: \( \log_{[value]}[value] = \log_{[value]} [value]^{[value]} = [value]\cdot \log_{[value]} [value] = [value] \cdot 1 = [value]\).

Use the Squeeze Theorem to determine the value of \(\displaystyle \lim_{x \rightarrow [value]} [value]\).
Feedback: We need to find two functions that bound \(f(x) = [value]\) from above and below and tend to the same value when \(x\) tends to [value]. Considering that \(-1 \leq [value] \leq 1\) and \([value] \geq 0\) for all values of \(x\), it follows that \([value] \leq [value] \leq [value]\). We now consider the limits of both bounding functions: \(\displaystyle \lim_{x\rightarrow [value]} [value] = 0\) \(\displaystyle \lim_{x\rightarrow [value]} [value] = 0\) Therefore, \(\displaystyle \lim_{x\rightarrow [value]} [value] = 0\)

Determine the value of the constant \(A\) for which the function

\[ f(x) = \begin{cases} [value] & x\neq [value] \\ A & x = [value]  \end{cases} \]

is continuous.

\(A=\)
Feedback: For \(f(x)\) to be continuous, we need \(\displaystyle \lim_{x\rightarrow [value]^-} f(x) = \lim_{x\rightarrow [value]^+} f(x) = f([value])\). \(\displaystyle \lim_{x\rightarrow [value]^-} f(x) = \lim_{x\rightarrow [value]^-} [value] = \lim_{x\rightarrow [value]^-} \frac{[value]}{[value]} = [value]\) Similarly, \(\displaystyle \lim_{x\rightarrow [value]^+} f(x) = \lim_{x\rightarrow [value]^+} [value] = \lim_{x\rightarrow [value]^+} \frac{[value]}{[value]} = [value]\) Since \(A\) is a contsant, we require \(A = [value]\) for continuity.

Complete the following table of standard derivatives. Try to do this without checking your notes -- which of the most important standard derivatives can you remember?

Function | Derivative
\(x^n\) |

Past QMB Exam Question (modified)

 | 2013 | 2014 | 2015 | 2016 | 2017
January | \(245.8\) | [value] | [value] | \(258.8\) | \(265.5\)
February | \(247.6\) | [value] | [value] | \(260.0\) | \(268.4\)
March | \(248.7\) | [value] | [value] | \(261.1\) | \(269.3\)
April | \(249.5\) | [value] | [value] | \(261.4\) | \(270.6\)
May | \(250.0\) | [value] | [value] | \(262.1\) | \(271.7\)
June | \(249.7\) | [value] | [value] | \(263.1\) | \(272.3\)
July | \(249.7\) | \(256.0\) | \(258.6\) | \(263.4\) | \(272.9\)
August | \(251.0\) | \(257.0\) | \(259.8\) | \(264.4\) | \(274.7\)
September | \(251.9\) | \(257.6\) | \(259.6\) | \(264.9\) | \(275.1\)
October | \(251.9\) | [value] | \(259.5\) | \(264.8\) |
November | \(253.4\) | [value] | \(259.8\) | \(265.5\) |

(Partially taken from https://www.ons.gov.uk/economy/inflationandpriceindices/timeseries/chaw/mm23)

(a) Use the values of the RPI given above to calculate

inflation in the year to January 2015: %

inflation in the year to February 2015: %

inflation in the year to March 2015: %

inflation in the year to April 2015: %

inflation in the year to May 2015: %

inflation in the year to June 2015: %

(b) What was the rate of inflation for the month of November 2014?

%

(c) If an item cost £60 in January 2014, what would be the equivalent cost (i.e. adjusted cost for inflation) in January 2015?

£

(d) If the RPI has been rebased so that the index was 100 in June 2014, what would be the values of the index for the first six months of 2014? You should give your answers to 3 decimal places.

 | Old index | New index
January | [value] |
February | [value] |
March | [value] |
April | [value] |
May | [value] |
June | [value] |
Feedback: FEEDBACK (a) Inflation in year to Jan 2015 \(= ([value] - [value])/[value] \times 100 = [value]\)%. Jan: | [value] | Apr: | [value] Feb: | [value] | May: | [value] Mar: | [value] | Jun: | [value] (b) Inflation for November \(= ([value]-[value])/ [value] \times 100 = [value]\)%. The minus sign above indicates that overall prices dropped in November by [value]%. (c) In January 2015 the price would be £\(60 \times [value]/[value] = [value].\) (d) | Old index | New index January | [value] | [value] February | [value] | [value] March | [value] | [value] April | [value] | [value] May | [value] | [value] June | [value] | [value]

For the following function find \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).

\[f(x,y) = [value].\]

\(\dfrac{\partial f}{\partial x} =\)

\(\dfrac{\partial f}{\partial y} =\)

For the above function also find \(f_x([value], [value])\), \(f_x([value],[value])\), \(f_y([value],[value])\) and \(f_y([value],[value])\).

\(f_x([value], [value]) = \)

\(f_x([value], [value]) = \)

\(f_y([value], [value]) = \)

\(f_y([value], [value]) = \)
Feedback: FEEDBACK For a function of two variables, \(f(x,y)\), then \(\frac{\partial f}{\partial x}\) is the partial derivative of \(f\) with respect to \(x\). It is obtained by differentiating \(f(x,y)\) with respect to \(x\) in the usual way but treating the \(y\)-variable as if it were a constant. For the case of \(f(x,y)=[value]\) we have \(\frac{\partial f}{\partial x} = [value]\). Similarly, \(\frac{\partial f}{\partial y}\) is the partial derivative of \(f\) with respect to \(y\). It is obtained by differentiating \(f(x,y)\) with respect to \(y\) in the usual way but treating the \(x\)-variable as if it were a constant. For the case of \(f(x,y)=[value]\) we have \(\frac{\partial f}{\partial y} = [value]\). To find the values for \(f_x([value], [value])\), \(f_x([value],[value])\), \(f_y([value],[value])\) and \(f_y([value],[value])\) we simply plug the corresponding values of \(x\) and \(y\) into \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\). Recall that \(f_x\) and \(f_y\) are simply different ways of writing \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) respectively. This gives us \(f_x([value], [value]) = [value] \), \(f_x([value], [value]) = [value]\), \(f_y([value], [value]) = [value]\), and \(f_y([value], [value]) = [value]\).

A fishing company measured the lengths of octopuses caught after a trawl of the North Sea. The results are shown below in cm.

[value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value]

Summarise these data in the grouped frequency table below.

Length (cm) | Frequency
0-9.9 |
10-19.9 |
20-29.9 |
30-39.9 |
40-49.9 |
50-59.9 |

Calculate a five point summary for this data set. Enter your answers to 1 d.p.

Minimum:  cm

25th percentile (1st quarter):  cm

50th percentile (median):  cm

75th percentile (3rd quarter):  cm

Maximum:  cm
Feedback: FEEDBACK First, sort the data by length. [value] | [value] | [value] | [value] | [value] [value] | [value] | [value] | [value] | [value] [value] | [value] | [value] | [value] | [value] [value] | [value] | [value] | [value] | [value] [value] | [value] | [value] | [value] | [value] From this, we can read off that the frequency of octopuses with a length \(0-9.9\) is \([value]\), with a length \(10-19.9\) is \([value]\), and so on. Five point summary We can see that the first (minimum) and last (maximum) lengths are \([value]\)cm and \([value]\)cm respectively. The median is the middle length, \([value]\)cm. To calculate the \(P\)-th percentile for \(n\) observations we use \(\dfrac{P}{100}\times(n+1)\). For \(P=25\) we have \(\dfrac{25}{100}\times([value]+1)=[value]\). This tells us which observation corresponds to the \(25\)th percentile. We take a weighted average for this, \([value]\), between the \([value]\)th and \([value]\)th observations. So the \(25\)th percentile is \([value] + [value] \times ([value] - [value]) = [value]\). Similarly, for \(P=75\) we have \(\dfrac{75}{100}\times([value]+1)=[value]\)cm. This tells us which observation corresponds to the \(75\)th percentile. We take a weighted average for this, \([value]\), between the \([value]\)th and \([value]\)th observations. So the \(75\)th percentile is \([value] + [value] \times ([value] - [value]) = [value]\).

Geben Sie eine Funktion \(f:\mathbb{R}\to\mathbb{R}\) mit den Nullstellen \(x_1=\)[value], \(x_2=\)[value] und \(x_3=\)[value] an!

\(f(x)=\)
Feedback: Lösung: Unendlich viele Funktionen erfüllen die Bedingungen aus der Aufgabenstellung. Eine mögliche Funktion lautet: \(f(x)=\)[value] Erläuterung: Wir suchen eine Funktion \(f(x)\), die die Bedingung \(f([value])=f([value])=f([value])=0\) erfüllt. Eine Funktion mit diesen Eigenschaften ist beispielsweise: \(f(x)=[value]=[value]\)

Kürzen Sie den Bruch für \(x^2\neq [value]\) vollständig!

\(\dfrac{[value]\cdot ([value])^2}{[value]\cdot ([value])}=\  \)
Feedback: Lösung: \(\dfrac{-[value]\cdot (x+[value])^2}{[value]\cdot ([value])}= \)\(\dfrac{[value]}{[value]}\) Erläuterung: Zunächst faktorisieren wir Zähler und Nenner mit Hilfe der binomischen Formel und kürzen den Faktor [value] \(\dfrac[value][value]=-\dfrac[value][value]\) Nach Kürzen von [value] erhalten wir den Bruch \(-\dfrac[value][value]\) Dieser Ausdruck lässt sich nicht weiter vereinfachen.