Title: Diagnostic questions
Course: Demonstrating STACK (v4.3.7)

Questions written as part of the Diagnostic testing project, 2010.

Differentiate [value] with respect to [value]
.
Feedback: Recall: \[\frac{d}{d[value]}\left(\frac{1}{[value]^{\frac{1}[value]}}+[value]\right)=\frac{d}{d[value]}\left(\frac{1}{[value]^{\frac{1}[value]}}\right)+\frac{d}{d[value]}\left([value]\right).\] Now remember that the differential of a constant is \(0\). So we need to find: \[\frac{d}{d[value]}\left(\frac{1}{[value]^{\frac{1}[value]}}\right).\] Using the laws of indices this is: \[\frac{d}{d[value]}\left([value]^{-\frac{1}[value]}\right).\] We can now apply the standard rules of differentiation to get: \[\frac{d}{d[value]}\left([value]^{-\frac{1}[value]}\right)=-\frac{1}[value][value]^[value]=-\frac{1}{[value]\cdot[value]^[value]}.\]

The derivative of the function \(y\) satisfying

\[[value] = [value]\]

is of the form

\[\frac{dy}{dx}=\frac[value][value]\]

where \(u\left(x\right)=\),

\(v\left(x\right)=\),

and \(w\left(x\right)=\).
Feedback: Differentiating both sides of the equation with respect to \(x\) we obtain \[[value] \frac{dy}{dx} + [value] + [value] \frac{dy}{dx} = [value]\mbox{.}\] Collecting the terms in \(\frac{dy}{dx}\) on one side we have that \[\frac{dy}{dx}\left([value]\right) = [value]\] and so \[\frac{dy}{dx} = [value]\mbox{.}\] We see that \[\begin{array}{rcl}\displaystyle u&\displaystyle =&\displaystyle [value]\mbox{,}\\ v&\displaystyle =&\displaystyle [value]\mbox{,}\\ w&\displaystyle =&\displaystyle [value]\mbox{.} \end{array}\]

Find the value of \(x\) for which the function [value] is a minimum.
Feedback: The function [value] has a stationary point at the value of \(x\) where its derivative with respect to \(x\) vanishes. Its derivative is [value] which is zero at \([value]\). The second derivative is [value], which is positive. Therefore the point is a minimum.

Differentiate \[[value]\] with respect to [value].
Feedback: In order to solve this problem we will have to use the linearity rules for differentition. calc_diff_linearity_rule \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left[ [value]\right] &\displaystyle =&\displaystyle \frac{d}{d[value]}\left[[value]\right] +\frac{d}{d[value]}\left[[value]\right]+\frac{d}{d[value]}\left[[value]\right], \\ &\displaystyle =&\displaystyle [value]\cdot \frac{d}{d[value]}[[value]]+[value]\cdot \frac{d}{d[value]}\left[[value]\right] +[value]\cdot \frac{d}{d[value]}\left[[value]\right], \\ &\displaystyle =&\displaystyle [value]\cdot [[value]]+[value]\cdot [[value]] +[value]\cdot \left[[value]\right], \\ &\displaystyle =&\displaystyle [value]. \end{array}\]

Find the following derivative:
\[ \frac{d}{d[value]}[[value]]. \]
Feedback: In order to solve this problem we will have to use the product rule calc_product_rule as follows: \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \left[\frac{d}{d[value]}[[value]]\right]\cdot [value] + [value]\cdot \left[\frac{d}{d[value]}[[value]]\right], \\ &\displaystyle =&\displaystyle [value]\cdot [value] + [value]\cdot [[value]], \\ &\displaystyle =&\displaystyle [value]. \end{array}\]

Find the derivative with respect to \(x\) of

\[[value].\]
Feedback: calc_quotient_rule Here \(f(x)=[value]\) and \(g(x)=[value]\), so the derivative of \(f(x)\) is [value] and the derivative of \(g(x)\) is [value]. The derivative of the quotient is \[ \frac[value]{x^[value]}. \]

Given that \([value]-[value]\geq [value]\) then \(p \leq \).
Feedback: First, we rearrange so on one side we have multiples of \(p\) and constants on the other, i.e.: \[\begin{array}{rcl}\displaystyle [value]-[value] &\displaystyle \geq&\displaystyle [value],\\ -[value] &\displaystyle \geq&\displaystyle [value].\\ \end{array}\] Now recall that if you multiple an inequality by minus \(1\) you have to swap the inequality sign: \[\begin{array}{rcl}\displaystyle -[value] &\displaystyle \ge&\displaystyle [value],\\ [value] &\displaystyle \le&\displaystyle [value]. \end{array}\] We can now divide through to get the values of \(p\): \[p \leq [value].\]

Given \([value] = [value]\) then \(x=\).
Feedback: We need to rearrange the equation to make \(x\) the subject: \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value], \\ x &\displaystyle =&\displaystyle \frac{[value] - [value]}[value],\\ &\displaystyle =&\displaystyle \frac[value][value],\\ &\displaystyle =&\displaystyle [value]. \end{array}\]

If \( [value]\circ [value] \) is defined to mean [value], what must [value] equal for \( [value]\circ [value] = [value]\circ [value] \) to be true?

[value]=
Feedback: We know: \[ [value]\circ [value] = [value] \mbox{ and } [value]\circ [value] = [value]. \] So the question asks when are these are equal, that is when does: \[ [value]=[value]. \] Putting similar terms on the same side, and simplifying we get: \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value],\\ [value][value]-[value] &\displaystyle =&\displaystyle [value][value]-[value],\\ [value][value] &\displaystyle =&\displaystyle [value][value],\\ [value] &\displaystyle =&\displaystyle [value]. \end{array}\]

If \( [value]=[value] \) then what does \( [value]+[value]= \) .
Feedback: We have no information about [value], so we can only add it to the right and left hand side of the equation, and leave our answer in that form. So \[\begin{array}{rcl}\displaystyle [value]+[value] &\displaystyle =&\displaystyle [value],\\ [value]+[value]+[value] &\displaystyle =&\displaystyle [value]+[value]. \end{array}\]

If \(y = [value]x + [value]\) then what is \(x\) equal to?

\(x =\)
Feedback: Rearranging purely algebraic equations is no different to numerical questions: \[\begin{array}{rcl}\displaystyle y &\displaystyle =&\displaystyle [value]x + [value], \qquad\mbox{first take [value] from each side} \\ y - [value] &\displaystyle =&\displaystyle [value]x, \qquad\mbox{now divide through by [value]} \\ \frac{y - [value]}[value] &\displaystyle =&\displaystyle x, \qquad\mbox{i.e.} \\ x &\displaystyle =&\displaystyle [value]. \end{array}\]

If \([value] = [value]\) then [value] \(=\)
Feedback: We need to rearrange the equation to get [value], on its own, on one side. \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value], \\ 2[value] &\displaystyle =&\displaystyle [value], \\ [value] &\displaystyle =&\displaystyle [value]. \end{array}\]

What is the positive solution of the quadratic equation [value] \( =0\)?
Feedback: First recall the quadratic formula: \[\begin{array}{rcl}\displaystyle \mbox{For} \quad a\cdot [value]^2 + b\cdot [value] + c &\displaystyle =&\displaystyle 0, \\ \mbox{We have} \quad [value] &\displaystyle =&\displaystyle \frac{-b \pm \sqrt{b^2 - 4\cdot a \cdot c}}{2a}. \end{array}\] So our question is to find the positive solution to \([value]^2 - 2\cdot [value] - 7 =0\) that is we substitute in the formula: \[\begin{array}{rcl}\displaystyle a &\displaystyle =&\displaystyle 1, \\ b &\displaystyle =&\displaystyle -2, \\ c &\displaystyle =&\displaystyle -7. \end{array}\] Giving us: \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle \frac{2 \pm \sqrt{4 - (-28)}}{2}, \\ &\displaystyle =&\displaystyle \frac{2 \pm \sqrt{32}}{2}, \\ &\displaystyle =&\displaystyle \frac{2 \pm \sqrt{16\cdot 2}}{2}, \\ &\displaystyle =&\displaystyle \frac{2 \pm 4\sqrt{2}}{2}, \\ &\displaystyle =&\displaystyle 1 \pm 2\sqrt{2}. \end{array}\] We then see that there are two solutions here (as there is a \(=\pm\) sign) we want the positive solution, and given the other terms are positive that is with the \(+\) sign. So the positive solution is: \[ 1 + 2\sqrt{2}. \]

In a railway journey of \(90km\) an increase of \(5\) kilometers per hour in the speed decreases the time taken by \(15\) minutes.

Write a system of equations (one equation per line) to represent this situation using \(v\) as the speed of the train and \(t\) as the time.

Eliminate \(t\) from your equations, writing a single equation only in \(v\) which represents this situation.

What is the velocity?
Feedback: Let \(v\) be the speed of the train, \(d\) the distance travelled and \(t\) time. If \(t\) is the time (in hours) then \(15\) minutes is \(\frac{1}{4}\) hour. We need to use the equation \(d=v\times t\) twice. For the original journey we have \[ 90 = v\times t.\] For the faster journey we have \[ 90 = (v+5)(t-1/4).\]

Given \[\begin{array}{rcl}\displaystyle [value]&\displaystyle =&\displaystyle [value]\mbox{,}\\ [value]&\displaystyle =&\displaystyle [value]\mbox{,} \end{array}\] then \(x=\)

and \(y=\).
Feedback: We are given \[\begin{array}{rcl}\displaystyle [value]&\displaystyle =&\displaystyle [value]\mbox{,}\\ [value]&\displaystyle =&\displaystyle [value]\mbox{.} \end{array}\] Multiplying the first equation by [value] and the second equation by [value] we obtain \[\begin{array}{rcl}\displaystyle [value]&\displaystyle =&\displaystyle [value]\mbox{,}\\ [value]&\displaystyle =&\displaystyle [value]\mbox{.} \end{array}\] Subtracting the latter equation from the former one we find \[ x=\frac{[value] - [value]}{[value] - [value]} = \frac[value][value] = [value] \] and then substituting \(x=[value]\) in the former equation we see \[y = \frac{[value] - [value] \times [value]}[value] = \frac[value][value] = [value]\mbox{.}\]

Given \(f\left(x\right) = [value]\) and \(g\left(y\right) = [value]\) then the composite function \(g\left(f\left(x\right)\right)=\).
Feedback: Substituting \(f\left(x\right)=[value]\) for \(y\) in the expression \(g\left(y\right) = [value]\) we obtain the composite function: \[g\left(f\left(x\right)\right)=[value]=[value].\]

Given \(f\left(x\right) = [value]\) then \(f\left([value]\right)=\) .
Feedback: We must substitute [value] in place of \(x\) in the definition of the function: \[\begin{array}{rcl}\displaystyle f\left([value]\right) &\displaystyle =&\displaystyle [value]\mbox{,}\\ &\displaystyle =&\displaystyle [value] \cdot ([value])^2 + [value] \cdot ([value]) + [value]\mbox{,}\\ &\displaystyle =&\displaystyle [value]\mbox{.} \end{array}\]

Given that \[f\left(x\right) = [value],\] then find the inverse function \(f^{-1}\left(x\right)=\).
Feedback: Given that \[f\left(x\right) = [value],\] then \[ f^{-1}(x) = [value]. \] You can confirm the answer by checking that \(f\left(f^{-1}\left(x\right)\right) = f^{-1}\left(f\left(x\right)\right) = x\).

Given \(f\left(x\right) =  [value] \cdot \mid [value] + [value] \mid + [value]\) then \(f\left([value]\right)=\).
Feedback: \[f\left(x\right) = [value] \cdot \mid [value] + [value] \mid + [value],\] We now substitute [value] in place of \(x\): \[\begin{array}{rcl} \displaystyle f\left([value]\right) &\displaystyle =&\displaystyle [value] \cdot \mid [value] + [value] \mid + [value],\\ &\displaystyle =&\displaystyle [value] \cdot \mid [value] \mid + [value],\\ &\displaystyle =&\displaystyle [value] \cdot [value] + [value],\\ &\displaystyle =&\displaystyle [value] + [value],\\ &\displaystyle =&\displaystyle [value]. \end{array}\]

To how many degrees do \(\frac{\pi}[value]\) radians correspond?

An angle of \(\frac{\pi}[value]\) radians can also be described as being degrees.
Feedback: There are [value] radians in a revolution; \(\pi\) radians correspond to \(180\) degrees. The number of degrees equivalent to \(\frac{\pi}[value]\) radians is \(\frac{\pi}[value]\) multiplied by \(\frac{180}{\pi}\), which is [value] degrees.

To how many radians do [value] degrees correspond?

An angle of [value] degrees can also be described as being radians.
Feedback: There are [value] radians in a revolution; \(\pi\) radians correspond to \(180\) degrees. We calculate that: \[\frac[value]{180} = [value],\] and so [value] degrees correspond to [value] radians.

A circle has a diameter of [value]cm. What is the area of the circle?
Feedback: The formula for calculating the area of a circle of radius \(r\)cm is: \[ Area\, = \pi r^2. \] So for a circle of radius [value]cm the area is: \( \pi \times [value]^2 = [value]. \)

Find the equation of a straight line in the \((x,y)\) plane which passes through \(([value],[value])\) and has gradient \([value]\).
Feedback: Recall that the equation of a straight line passing through the point \((x_1,y_1)\) with gradient \(m\) is \[y-y_1=m(x-x_1).\] So substituting in the values we get: \[y-[value]=[value](x-[value]),\] Which gives the equation of the line: \[\begin{array}{rcl}\displaystyle y - [value] &\displaystyle =&\displaystyle [value](x-[value]),\\ y - [value] &\displaystyle =&\displaystyle [value]x - [value],\\ y &\displaystyle =&\displaystyle [value]x + [value]. \end{array}\]

Given that \(\theta\) lies in the range \[[value]<\theta<[value]\mbox{,}\] and that \(\tan \left(\theta\right) = 1\), then \(\theta=\).

You may enter \(\pi\) as pi or %pi.
Feedback: Given that \(\tan \left(\theta\right) = 1\) then \(\theta= [value] + [value]\). Setting \(m=[value]\) we see that \(\theta=[value]+[value]=[value]\) lies in the range \([value]<\theta<[value]\).

Give an expression of the form [value] which is equivalent to \(\sin\left([value] + x\right)\).
Feedback: Standard trigonometric identities are given below. trig_standard_identities The appropriate identity in this case is the angle addition formula \[\sin\left(A + B\right)=[value] + [value],\] with \(A=[value]\) and \(B=x\). We substitute \(\cos\left([value]\right) = 0\) and \(\sin\left([value]\right) = [value]\) and find that \[\sin\left([value] + x\right)=[value] + [value]=[value].\]

Give an expression of the form [value] which is equivalent to \(\cos\left([value] + x\right)\).
Feedback: Standard trigonometric identities are given below. trig_standard_identities The appropriate identity in this case is the angle addition formula: \[\cos\left(A + B\right)=[value] - [value],\] with \(A=[value]\) and \(B=x\). We substitute \(\cos\left([value]\right) = 0\) and \(\sin\left([value]\right) = [value]\) and find that \(\cos\left([value] + x\right)=[value] - [value]=[value]\).

Work out and give the area bounded by the lines \(x=[value]\), \(x=[value]\), the \(x\) axis and the polynomial \([value]\).
Feedback: Integrate in the usual way, between two limits: \[\begin{array}{rcl}\displaystyle \int_{x=[value]}^{x=[value]} \left([value]\right)\;dx&\displaystyle =&\displaystyle \left[[value]\right]_{x=[value]}^{x=[value]}\mbox{,}\\ &\displaystyle =&\displaystyle \frac{[value]\left([value]^[value] - [value]^[value]\right)}[value] + \frac{[value]\left([value]^[value] - [value]^[value]\right)}[value]\mbox{,}\\ &\displaystyle =&\displaystyle [value]\mbox{.} \end{array}\]

Find the integral: \[\int [value] \,dx \]
Feedback: Below is the formula for integration by parts. calc_int_methods_parts From the product rule for differentiating \(u\left(x\right)v\left(x\right)\) we can derive the rule for integration by parts, namely: \[\int u\left(x\right)v'\left(x\right)\, dx= u\left(x\right)v\left(x\right)- \int u'\left(x\right)v\left(x\right)\, dx.\] We set \(u\left(x\right)=[value]\) and \(v'\left(x\right)=[value]\), so \(u'\left(x\right)=[value]\) and \(v\left(x\right)=[value]\). Then \[\begin{array}{rcl}\displaystyle \int [value]\,dx&\displaystyle =&\displaystyle [value]-\int [value] \,dx,\\ &\displaystyle =&\displaystyle [value]-\frac[value][value]+c. \end{array}\] Where \(c\) is an integral constant.

Integrate [value] with respect to [value].

\(\int [value]\,d[value] = \) .
Feedback: For this question we will use integration by parts. Remembering that the formula is: \[\int u\,\frac{dv}{dt}\,dt = uv - \int v\,\frac{du}{dt}\,dt.\] Now we want to find \(\int [value]\,d[value]\). So, we let \[ u = [value]^2\quad\Rightarrow\quad \frac{du}{d[value]} = 2[value],\] and \[ \frac{dv}{dt} = [value]\quad\Rightarrow\quad v = [value].\] We can now substitute these into the formula to find that: \[ \int [value]\,d[value] = [value] - \int [value]\,d[value]. \] We now have to perform an integration by parts again on what we have found. This time: \[ u = 2[value]\quad\Rightarrow\quad \frac{du}{d[value]}=2,\] and \[\frac{dv}{dt} = [value]\quad\Rightarrow\quad v=[value]. \] So we get: \[\begin{array}{rcl}\displaystyle \int [value]\,d[value] &\displaystyle =&\displaystyle [value] - \int [value]\,d[value], \\ &\displaystyle =&\displaystyle [value] - \left([value] - 2\int [value]\,d[value]\right). \end{array}\] Which means: \[\begin{array}{rcl}\displaystyle \int [value]\,d[value] &\displaystyle =&\displaystyle [value] - \left([value] - ([value])\right),\\ &\displaystyle =&\displaystyle [value] + constant . \end{array}\]

Using the substitution \(u = \cos([value])\) integrate [value] with respect to [value].
Feedback: We are told to use the substitution \(u=\cos([value])\), so we therefore get \(du = -\sin([value])\,dv\). Hence: \[ dv = -\frac{du}{\sin([value])}. \] So the integral becomes: \[\begin{array}{rcl}\displaystyle \int [value]\,dv &\displaystyle =&\displaystyle \int e^u \cdot \sin([value]) \cdot -\frac{du}{\sin([value])}, \\ &\displaystyle =&\displaystyle - \int e^u \,du, \\ &\displaystyle =&\displaystyle -e^u + constant ,\\ &\displaystyle =&\displaystyle -e^{\cos([value])} + c. \end{array}\]

Integrate [value] with respect to [value].
Feedback: This just uses the standard rules of integration: \[\int [value]\,d[value] = [value]+c. \] Don't forget to add a constant of integration.

The area between the curve [value], the \(y\)-axis and the lines \(x=[value]\) and \(x=[value]\) is revolved once around the \(x\)-axis. Find the volume of revolution of the solid swept out.
Feedback: Recall that the volume swept out by a curve \(y(x)\) revolved once around the \(x\) axis, and between the lines \(x=a\) and \(x=b\) is given by: \[\mbox{Volume} = \pi \int_a^b y(x)^2\,dx.\] So the volume swept out in this question is: \[\begin{array}{rcl}\displaystyle \pi\int_[value]^[value] \left([value]\right)^2 \, dx&\displaystyle =&\displaystyle [value] \int_[value]^[value] [value] \, dx,\\ &\displaystyle =&\displaystyle [value] \int_[value]^[value] \left([value]\right) \, dx,\\ &\displaystyle =&\displaystyle \left[ [value]\right]_[value]^[value],\\ &\displaystyle =&\displaystyle [value]. \end{array}\]

Given that \(p=[value]\) and \(q=[value]\) give \([value]\) as a function of \(p\) and \(q\).
Feedback: Below are the basic rules of logarithms. alg_logarithms Given \(p=[value]\) and \(q=[value]\) then, \[[value] = \ln \left([value]\right)+[value] = [value] = [value].\]

Given \[x=\frac{e^[value]}{y^[value]},\] then find \(z\).

We find that \(z=\).
Feedback: Given \(x=\frac{e^[value]}{y^[value]}\) then \(e^[value] = [value]\) and \(z = [value]\).

Imagine you have a deck of four cards in which you know every card has one letter on one side and one number on the other. You can see the following:

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

You must choose the fewest cards to turn over to establish the truth of the following statement.

"Every card which has a [value] on one side has a [value] on the other."

Which cards will you turn over? Your answer should be a set of cards, e.g. {?, ...}.
Feedback: Turning over all the cards is not the right answer. While it does enable you to establish whether this is true, it is not the fewest possible cards. One way to think about this statement is to re-phrase it to "If we see a [value] on one side then we have a [value] on the other." Keep in mind that every card has a number on one side and a letter on the other! Do we turn over the [value]? Yes, if we see a [value] then we must check if we have a [value]. This kind of if-then statement is known as implication. Do we turn over the [value]? Yes, if we have a [value] and a [value] then the statement is false. Hence we must check for this possibility. We can re-phrase the statement to "If we do not have a [value] on one side then we do not have [value] on the other." This form is known as the contrapositive. It is equivalent to the original statement we are trying to check. Do we turn over the [value]? No, since we have not satisfied the hypothesis that we have one [value]. The other side must be a number but we don't care if it is a [value] or not. Any situation does not affect the truth of the statement we are trying to check. Do we turn over the [value]? No, since we could easily have [value] on one side and a [value] on the other. The statement is unaffected. Similarly if we don't have a [value] on the other side the statement is unaffected.

What is the value of: \[\frac[value][value] \div \frac[value][value],\] .
Feedback: Recall that to divide by a fraction you can flip one of the fractions (swap its denominator and numerator), change the divide sign to a multiply and proceed like you usually would. So we get: \[\begin{array}{rcl}\displaystyle \frac[value][value] \div \frac[value][value] &\displaystyle =&\displaystyle \frac[value][value] \times \frac[value][value], \\\\ &\displaystyle =&\displaystyle \frac{[value]\times [value]}{[value]\times [value]}, \\\\ &\displaystyle =&\displaystyle [value]. \end{array}\]

Write \[\frac{1}[value]+\frac{1}{[value]^2}+\frac{1}{[value]^3}.\] as a single fraction.
Feedback: \[\begin{array}{rcl}\displaystyle \frac{1}[value] + \frac{1}{[value]^2} + \frac{1}{[value]^3} &\displaystyle =&\displaystyle \left(\frac{1}[value]\cdot\frac{[value]^2}{[value]^2}\right)+ \left(\frac{1}{[value]^2}\cdot\frac[value][value]\right) + \frac{1}{[value]^3},\quad(\mbox{make the denominator the same})\\ &\displaystyle =&\displaystyle \frac{[value]^2}{[value]^3}+\frac[value]{[value]^3}+\frac{1}{[value]^3},\\ &\displaystyle =&\displaystyle \frac{[value]^2 + [value] + 1}{[value]^3}.\qquad(\mbox{collect terms}) \end{array}\]

Write \[[value],\] as a single fraction.
Feedback: We must add fractions in the usual way to find the answer here: \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle \frac{([value]\times [value]) + [value]}{[value][value]}, \\ &\displaystyle =&\displaystyle \frac{[value] + [value]}{[value][value]}. \end{array}\]

Given \(\left(y^[value]\right)^[value] = y^n\) then \(n=\).

Enter your answer as an integer (whole number) with no decimal point.
Feedback: Recalling the laws of indices we multiply the two powers to get: \[ \left(y^[value]\right)^[value] = y^{[value] \times [value]} = y^[value]. \]

Given that \(y^[value] \div y^[value] = y^n\) then \(n=\).

Enter your answer as an integer (whole number) with no decimal point.
Feedback: \[ y^[value] \div y^[value] = y^{[value] - [value]} = [value] \]

What is the value of \(2^{-[value]}\)?
Feedback: First recall that: \[ x^{-y}=\frac{1}{x^y}. \] So, \[2^{-[value]}=\frac{1}{2^[value]}.\] Then we perform the calculation to find see that: \[ \frac{1}{2^[value]} = [value]. \] Hence \[2^{-[value]} = [value]. \]

If £ [value] is divided between two people \(A\) and \(B\) in the ratio \(5:7\), then \(A\) gets how much? £
Feedback: We first work out the total number of parts that are to be shared out. We know the ratio is \(5:7\), so the total number of parts is \(5+7=12\). So we divide [value] by 12 to get [value], which is the value of each part. To work out how much \(A\) gets we multiply [value] by \(5\). So \(A\) gets £ [value].

Two blocks, \(A\) and \(B\) are the same shape but different sizes. If the measurements of block \(A\) are all [value] the lengths of those for block \(B\), and block \(B\) is [value]cm tall, how tall is block \(A\) in cm?
Feedback: Block \(A\) is [value] times the height of block \(B\). i.e. block \(A\) is \([value]\times [value]\mbox{cm} = [value]\)cm tall.

Let term \(n\) of an arithmetic progression be denoted \(u_{n}\).

Given that \(u_[value] = [value]\) and that the common difference is \(d=[value]\) then \(u_[value]=\).
Feedback: Term \(n\) of an arithmetic progression with common difference \(d\) and first term \(a\) is [value]. So \(u_[value] = a + [value] = [value]\) and \(u_[value] = a + [value]\). Eliminating \(a\) we obtain the equation \(u_[value] = [value] + [value]\) and substituting \(d=[value]\) we find that \(u_[value]=[value]\).

What is the sum of the first [value] terms of the arithmetic progression \([value]+[value]+[value]+[value]+[value]+\ldots\)?
Feedback: The sum \(S_{n}\) of an arithmetic progression of \(n\) terms, first term \(a\) and common difference \(d\), is \[\frac[value]{2}.\] Here \(a=[value]\) and \(d=[value]\) so \(S_[value]=[value]\).

Evaluate the coefficient of the term in \(x^3\) in the binomial expansion of \(\left([value]\right)^{\frac[value]{2}}\,\,\) (for \(\mid x \mid < [value]\)).
Feedback: For \(n\) a negative integer, or in general any rational number, and \(\mid x \mid<1\), then \[ \left(1 + x\right)^n = 1 + [value] + \frac[value]{2!}x^2 + \frac[value]{3!}x^3 +\ldots. \] The expansion is finite if \(n\) is a positive integer but is in general infinite otherwise. In this case we have that \(n = \frac[value]{2}\) and the expansion of \(\left([value]\right)^{\frac[value]{2}}\) up to terms in \(x^3\) is \[[value]\mbox{.} \] Therefore the coefficient in \(x^3\) is [value]. This expansion is valid provided \(\mid x \mid< [value]\).

Let term \(n\) of a geometric progression be denoted \(u_{n}\). Given that \(u_[value] = [value]\) and that the common ratio is \(r=[value]\) then \(u_[value]=\).
Feedback: Term \(u_{n}\) of a geometric progression with common ratio \(r\) and first term \(a\) is [value]. So \(u_[value] = [value] = [value]\) and \(u_[value] = [value]\). Dividing \(u_[value]\) by \(u_[value]\) we eliminate \(a\) and obtain the equation \(u_[value] = [value]\). Substituting \(r=[value]\) we find that \(u_[value]=[value]\).

What is the sum of the first [value] terms of the geometric progression \([value] + [value] + [value] + [value] + \ldots\)?
Feedback: The partial sum \(S_{n}\) of the first \(n\) terms of a geometric progression, first term \(a\) and common ratio \(r\), is given by the equation \[ S_{n}=\frac{a(1-r^n)}{1 - r}. \] Here \(a = [value]\) and \(r = [value]\) so \(S_[value] = [value]\).

What is the sum to infinity of the geometric progression \([value] + [value] + [value] + [value] + \ldots\)?
Feedback: The sum to infinity \(S_{\infty}\) of a geometric progression, first term \(a\) and common ratio \(r\), is defined provided \(\mid r \mid<1\), in which case it is given by the equation \[ S_{\infty}=\frac{a}{1 - r}. \] Here \(a = [value]\) and \(r = [value]\) so \(S_{\infty} = [value]\).

Given that \[u_n = \frac{1}[value] [value],\] then \(u_[value]=\).
Feedback: Substituting \(n=[value]\) we see that \(u_[value]=[value]\).

Given that \(\mathbf{a} = [value]\,\mathbf{i} + [value]\,\mathbf{j} + [value]\,\mathbf{k}\) and that \(\mathbf{b} = [value]\,\mathbf{i} + [value]\,\mathbf{j} + [value]\,\mathbf{k}\) then

\(\mathbf{a}+\mathbf{b}=\) \(\mathbf{i}\) \(+\) \(\mathbf{j}\) \(+\) \(\mathbf{k}\).
Feedback: \[\begin{array}{rcl}\displaystyle \mathbf{a}+\mathbf{b}&\displaystyle =&\displaystyle \left([value] + [value]\right)\mathbf{i} + \left([value] + [value]\right)\mathbf{j} + \left([value] + [value]\right)\mathbf{k},\\ &\displaystyle =&\displaystyle [value]\,\mathbf{i} + [value]\,\mathbf{j} + [value]\,\mathbf{k}. \end{array}\]

Given that \(A\) is the point \(([value],\,[value],\,[value])\) and that \(B\) is the point \(([value],\,[value],\,[value])\) then the vector \(\mathbf{r}\) from \(A\) to \(B\) is \(\mathbf{i}\) \(+\) \(\mathbf{j}\) \(+\) \(\mathbf{k}\).
Feedback: Given that \(A\) is the point \(([value],\,[value],\,[value])\) and that \(B\) is the point \(([value],\,[value],\,[value])\) then the vector from A to B is \[\begin{array}{rcl}\displaystyle \mathbf{r}&\displaystyle =&\displaystyle \left([value] - [value]\right)\mathbf{i} + \left([value] - [value]\right) \mathbf{j} + \left([value] - [value]\right)\mathbf{k},\\ &\displaystyle =&\displaystyle [value]\,\mathbf{i} + [value]\,\mathbf{j} + [value]\,\mathbf{k}. \end{array}\]

Given that \(\mathbf{a} = [value]\,\mathbf{i} + [value]\,\mathbf{j} + [value]\,\mathbf{k}\) and that \(\mathbf{b} = [value]\,\mathbf{i} + [value]\,\mathbf{j} + [value]\,\mathbf{k}\) then the scalar or "dot" product \(\mathbf{a}.\mathbf{b}=\) .
Feedback: Calculate the dot product in the usual way by multiplying the \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\) components together. \[\begin{array}{rcl}\displaystyle \mathbf{a}.\mathbf{b}&\displaystyle =&\displaystyle [value].[value] + [value].[value] + [value].[value],\\ &\displaystyle =&\displaystyle [value] + [value] + [value],\\ &\displaystyle =&\displaystyle [value]. \end{array}\]

The unit vector in the direction of \(\mathbf{a}=[value]\,\mathbf{i}+[value]\,\mathbf{j}+[value]\,\mathbf{k}\) is

\( \hat{\mathbf{a}}= \) \(\mathbf{i}\) \(+\) \(\mathbf{j}\) \(+\) \(\mathbf{k}\).

Enter your answer exactly, using the sqrt function for square roots if necessary, for example inputting \( \sqrt{3} \) as sqrt(3).
Feedback: The magnitude of \(\mathbf{a}\), denoted \(\mid \mathbf{a} \mid\), is a positive quantity defined as \(\mid \mathbf{a} \mid = \sqrt{[value]^2 + [value]^2 + [value]^2} = [value]\). A unit vector in the direction of \(\mathbf{a}\) is \[\mathbf{\hat{a}}=\frac{\mathbf{a}}{\mid \mathbf{a} \mid} = [value]\,\mathbf{i}+[value]\,\mathbf{j}+ [value]\,\mathbf{k}.\]

A university has [value] times as many students as professors. If \(S \) represents the number of students and \( P \) represents the number of professors, write an equation expressing the relationship between \( S \) and \( P \).
Feedback: There are [value] times as many students as professors. Which written another way is, "the number of students is equal to [value] times the number of professors." This can be written mathematically as: \[[value].\]ans1:exdowncase(ans1);exdowncase(ans1)

Given that the product \(y^[value] \cdot y^[value] = y^n\) then \(n=\).

Enter your answer as an integer (whole number) with no decimal point.
Feedback: Remember the standard laws of indices. \[ y^[value] \cdot y^[value]=y^{[value]+[value]}=y^[value]. \]

Factorise the expression \[[value]\] into the form \[\left(x - a\right)\left(x - b\right) \left(x - c\right)\] given that the expression evaluates to zero when \(x = [value]\). The roots are real.
Feedback: Given that the polynomial [value] is zero when \(x = [value]\), then one of the factors is \(\left([value]\right)\), which is of course zero when \(x = [value]\). The polynomial factorises as \[\begin{array}{rcl}\displaystyle [value]&\displaystyle =&\displaystyle [value]\mbox{,}\\ &\displaystyle =&\displaystyle [value]\mbox{,}\\ &\displaystyle =&\displaystyle [value]\mbox{,}\\ &\displaystyle =&\displaystyle [value]. \end{array}\] Matching coefficients \[\begin{array}{rcl}\displaystyle [value] + m + n&\displaystyle =&\displaystyle [value]\mbox{,}\\ [value]+ [value] + [value]&\displaystyle =&\displaystyle [value]\mbox{,}\\ [value]&\displaystyle =&\displaystyle [value]. \end{array}\] So \[m = [value] - n\mbox{,}\] and \[ [value] = [value]\mbox{.} \] Substituting for \(m\) in terms of \(n\) \[\displaystyle [value]\displaystyle =\displaystyle [value]\mbox{.}\] This can be rearranged as \[[value]\displaystyle =\displaystyle 0\mbox{.}\] Factorising \[[value] = 0\mbox{,}\] and either \(n = [value]\) and so \(m = [value]\) or \(n = [value]\) and \(m = [value]\). The polynomial [value] factorises as \[[value]\mbox{.}\]

Katie's basic wage is £[value] per week.

She is also paid another £[value] for each hour of overtime that she works.

If \(h\) stands for the number of hours of overtime that she works, and if \(W\) stands for her total wage (in £s). Write down an equation relating \(W\) and \(h\):

What would Katie's total wage be if she worked [value] hours of overtime? £
Feedback: To first work out the equation relating \(W\) and \(h\) we look at the information we are given. We know that \(W\) stands for her total wage, so that will be on one side of the equation, on the other we have to put her basic wage and take into account any overtime she may work. Given that she is paid £[value] for every hour overtime and she works \(h\) hours of over time she gets paid an extra \([value] \times h\), and that is on top of the basic wage of £[value]. So the equation is: \[ [value] + [value]h = W. \] For the second part of the question we have to calculate Katie's actual wage given that she worked [value] hours of overtime. So we must replace \(h\) in our equation with the value [value] and simplify, i.e. we get: \[\begin{array}{rcl}\displaystyle W &\displaystyle =&\displaystyle [value] + [value]h, \\ W &\displaystyle =&\displaystyle [value] + ([value] \times [value]), \\ W &\displaystyle =&\displaystyle [value] + [value], \\ W &\displaystyle =&\displaystyle [value]. \end{array}\]

Give the remainder when [value] is divided by [value].
Feedback: By the remainder theorem when a polynomial \(f\left(x\right)\) is divided by \(x - a\), the remainder is \(f\left(a\right)\). So if we divide [value] by [value] the remainder is \[[value] \cdot [value]^2 + [value] \cdot [value] + [value] = [value] \cdot [value] + [value] + [value] = [value].\]

Fully factorize [value].
Feedback: We want to factorize [value]. So we need to find 2 numbers which add to [value] and multiplies to [value]. Since \[ 1 + [value] = [value] \quad\mbox{ and }\quad 1\times [value] = [value], \] the two numbers are \( 1 \) and [value]. So we can then rewrite the equation and factorize: \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value]^2 + [value] + ([value]) + [value],\\ &\displaystyle =&\displaystyle [value],\\ &\displaystyle =&\displaystyle [value]. \end{array}\]

Write \[[value],\] in the form \[a[value]^3 + b[value]^2 + c[value] + d.\]
Feedback: First expand the first two brackets: \[\begin{array}{rcl}\displaystyle [value] & = & ([value])([value]),\quad\quad\quad(\mbox{now multiply out the two remaining brackets}) \\ &\displaystyle =&\displaystyle [value].\\ \end{array}\]

Expand \(\left([value]\right)^5\).
Feedback: For \(n\) a positive integer then by the binomial theorem \[ \left(a + b\right)^n = a^n + \frac{n!}{\left(n-1\right)!\cdot 1!}a^{n-1}\cdot b + \ldots+\frac{n!}{\left(n-r\right)!\cdot r!}a^{n-r}\cdot b^r + \ldots + b^n\mbox{.} \] In this case \(a=[value]\), \(b=[value]\) and \(n=5\); therefore \[\begin{array}{rcl}\displaystyle \left([value]\right)^5 = [value]^5 + 5\cdot [value]^4 \cdot [value] \cdot x + \frac{5\cdot 4}{2!} \cdot [value]^3 \cdot [value]^2 \cdot x^2 + \frac{5 \cdot 4 \cdot 3}{3!} \cdot [value]^2 \cdot [value]^3 \cdot x^3\mbox{,} \\ + \frac{5 \cdot 4 \cdot 3 \cdot 2}{4!} \cdot [value] \cdot [value]^4 \cdot x^4 + [value]^5 \cdot x^5\mbox{,} \end{array}\] where \[\begin{array}{rcl}\displaystyle 2!&\displaystyle =&\displaystyle 2 \cdot 1 = 2\mbox{,}\\ 3!&\displaystyle =&\displaystyle 3 \cdot 2 \cdot 1 = 3 \cdot 2! = 6\mbox{,}\\ 4!&\displaystyle =&\displaystyle 4 \cdot 3 \cdot 2 \cdot 1 = 4 \cdot 3! = 24\mbox{.} \end{array}\] So \[\left([value]\right)^5 = [value]\mbox{.}\]