** Easy mistake**: an axis can’t have 2 directions.

Several our students persistently started to sketch graphs from the diagram above. By definition, an *axis* is an infinite straight line with one direction, so we need to know a direction of increase of the variable along the axis. In computer programming the axes have different directions, because we usually type on a screen from top to bottom starting from the top left corner (the origin).

So the vertical \(y\)-axis is oriented downwards and the parabola \(y=(x-8)^2\) has the reversed shape on such a diagram. However, if two directions are specified on the line of the \(y\)-coordinate, then we can not know how to draw even the simple parabola \(y=(x-8)^2\). By the standard convention in mathematics the horizontal \(x\)-axis is oriented to the right hand side and the vertical \(y\)-axis is oriented upwards.

**Common mistake**: corners/cusps vs smooth points.

Many students often draw the graph \(y=2\sin^2 x\) with corners and cusps at the points

\(x=\frac{\pi}{2}+\pi n\) for any integer n, where the graph meets the \(x\)-axis. Starting from the familiar graph \(y=\sin x\) and before scaling it by factor 2, the students probably reflect the negative parts to the upper half-plane and get the following picture with cusps (“acute corners”). This reflection should give right-angled (not acute) corners, because \(y=\sin x\) meets the \(x\)-axis at angles \(\pm\frac{\pi}{4}\) since the gradient is \(y'(x)=\cos x=\pm 1\) at \(x=\frac{\pi}{2}+\pi n\).

However, \(y=2\sin^2 x\) has neither corners nor cusps, because the derivative \(y'(x)=2\sin x \cos x =\sin 2x\) is well-defined everywhere. Indeed, \(y'(x)=sin 2x=0\) at \(x=\frac{\pi}{2}+\pi n\), so the graph the \(x\)-axis smoothly touches \(y=2\sin^2 x\) at all the points \(x=\frac{\pi}{2}+\pi n\) as in the correct picture below.

Actually, best students know the formula \(2\sin^2 x=1-\cos 2x\) and sketch the simpler graphs \(y=\cos x, y=\cos 2x, y=1-\cos 2x\).

**Unexpected mistake**: Q3(v) in the MAT paper 2011.

Q3 in the MAT paper 2011 is about the cubic parabola \(y(x)=x^3-x\) and its tangent line \(y=m(x-a)\) having a slope \(m>0\) and meeting the \(x\)-axis at a point \(x=a\leq 1\).

Through any point (a,0) for \(a<-1\) we can draw 3 tangent lines to \(y=x^3-x\). One the them touches the cubic parabola at a point \(0<x<1\) and has a negative slope. However, there are two different tangent lines with a positive slope. Only one of them was shown in the original problem as in the picture above. No restrictions apart from \(m>0, a\leq -1\) were given in the problem. Hence we may have the second tangent line for \(b<a<-1\) in our picture below.

Part (ii) asks to prove that \(a=\frac{2b^3}{3b^2-1}\). This formula also works for both tangent lines: if \(b=-2\), then \(a=-\frac{16}{11}>b=-2\).

Part (iii) asks to find an approximate value of b when \(a=-10^6\). The examiners’ solution of part (iii) considers only one possibility when \(a=\frac{2b^3}{3b^2-1}\) has a large (absolute) value because of a small denominator, which implies that \(b\approx -\frac{1}{\sqrt{3}}\) (if negative). However, the second possibility is that b also has a large (absolute) value. For instance, if we set \(b=\frac{3}{2}a\), then \(\frac{2b^3}{3b^2-1}=\frac{\frac{27}{4}a^3}{\frac{27}{4}a^2-1}=a+\frac{a}{\frac{27}{4}a^2-1}\) is approximately equal to \(a\) when \(a\) has a large absolute value. So another tangent line at \(b\approx -\frac{3}{2}10^6\) meets the x-axis at the same point \(a=-10^6\).

Part (iv) asks to show that the tangent line meets the cubic parabola at the second point \(c=-2b\). This formula also works for both tangent lines: if \(b=-2\), then the tangent line \(y=11\Big(x+\frac{16}{11}\Big)=11x+16\) meets \(y(x)=x^3-x\) at \(x=c=4\) where \(y(4)=60=11\cdot 4+16\).

Part (v) asks to find the largest possible area of the region R bounded above by the tangent line and bounded below by \(y=x^3-x\). For both tangent lines, the region R is finite because both

tangent lines eventually intersect the cubic parabola at \(c=-2b\). The examiners’ solution to part (v) says: “We can see that as \(a\) increases then the tangent line rises and so the area of R increases. So the area is greatest when \(a=-1\)“.

This claim holds only for the first tangent line when \(-1\leq b<0\). However, our last picture with the second tangent line above shows a much larger area when \(b<-1\). Actually both tangent lines coincide when \(a=-1\), so one family of tangent lines for \(a<-1<b<0\) is joining another family of tangent lines for \(b<a<-1\). The picture in the problem with only one tangent line for \(a<b\) degenerates in the case \(a=b=-1\) when two points A,B merge and can hardly be used as a reference for part (v).

It is rather surprising that Q3(v) in the MAT paper 2011 and its solution have not been amended in Oxford MAT 2011 solutions and in Imperial MAT 2011 for almost 2 years after the actual exam: checked on 20th September 2013. You can find more details on this problem in our web tutorial Tangent lines and areas bounded by cubic parabolas.

Mathematics is a wonderful subject, because students can check all arguments and find more efficient solutions themselves. That is why we keep learning from our smart students.

**Riddle 6**: does \(x^2+y^2=1\) define a function \(y(x)\) for \(-1\leq x\leq 1\)?**How to submit**: to write your full answer, simply submit a comment.**Hint**: sketch the curve \(x^2+y^2=1\) on the plane, you may try to express \(y(x)\).**Warning**: a function \(y(x)\) should have a single value of \(y\) over \(-1\leq x\leq 1\).**Restriction**: only the first fully correct public answer will be rewarded.**Prize**: free 1-year access to one of our interactive web tutorials.**Update**: Carlo has solved the problem, see the comment.

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Question 3)v) isn’t a mistake. Read the question carefully, it says the point (a, 0) lies in the Y axis. Y must = 0 when X = a.

Nevermind, I read the question again.

Well, the diagram in question 3)v) clearly shows that b is not less than -1?

Dear Alan, thank you for your comment. We agree that the diagram at the beginning of Question 3 MAT 2011 clearly shows that b>-1. Actually, part (v) does not refer to the diagram for a good reason. Indeed, the claimed answer b=-1 (when the points A and B merge) contradicts the diagram. At this critical moment the tangent line from the solution by MAT examiners is transformed to another tangent line at the point B with b< =-1. Without the explicit restriction b>=-1, the new tangent line continues to rise and the area of the region R (bounded by the cubic and its tangent line at the point B) goes to infinity, so it is impossible to find the largest area of R in 3(v). If we assume that the restriction is b>-1 as in the given diagram, not b>=-1 as required, then the area of R is increasing when b approaches the value (-1) on the right and can not achieve a largest value over the open interval b>-1, simply because the largest value of R is at the endpoint b=-1, which is not covered by b>-1.

I’ll try to answer the riddle: no, it doesn’t. Take for example x = 1/2. Then y(1/2) = sqrt(3)/2 and y(1/2) = – sqrt(3)/2 are both valid values, in that they satisfy x^2 + y^2 = 1. However a function can only have at most one output for each input.

Dear Carlo, yes. We would advise to write two solutions in the form x=1/2, y=+/-sqrt(3)/2, not using the notation y(1/2), simply because you are proving that y(x) can’t be defined around 1/2. You could e-mail your choice of a web tutorial to master.maths.tutor@gmail.com to get a 1-year free access.