2020 Paper 1, Question 19

Technology-free Engine: inversion [6 marks]

The question, as it appeared

A horizontal point of inflection is a point of inflection that is also a stationary point.

Determine the value/s of $k$ for which the graph of $f(x)=\dfrac{\ln(x)}{k}-\dfrac{kx}{x+1}$ has only one horizontal point of inflection.

First, watch it happen

Turn the dial on $k$ and watch a flat spot arrive

One family, one parameter. For small $k$ the curve climbs without ever levelling. At one value it pauses for an instant. Then that pause splits into a peak and a trough.

k = Slope at x = 1:

No flat spot yet, so the curve climbs the whole way.

At $k=2$ the tangent at $x=1$ is level, and the curve keeps rising through it.

Past $k=2$ the flat spot splits into a maximum and a minimum.

$k$ sweeps from 1.45 to 2.9 and back. The orange bar is the tangent at $x=1$. Heights are rescaled each frame, so read the tilt, not the size.

Then the mathematics

One flat spot means a repeated root

$f'(x)=0$ tidies into the quadratic $x^2+(2-k^2)x+1=0$. Two roots give two turning points. No roots give none. One root only happens when the discriminant is zero.

The curve

$f(x)=\frac{\ln(x)}{k}-\frac{kx}{x+1}$

The quadratic whose roots are those points

$y=x^2+(2-k^2)x+1$

Size of $k$

One stationary point at $x=1$, and $f''(1)=0$, so it is a horizontal point of inflection.

Discriminant negative: $f'$ never reaches zero, so there is nothing flat to inflect.

Two different stationary points, a maximum and a minimum. Neither is a point of inflection.

Discriminant

$(2-k^2)^2-4$

Stationary points

Second derivative at 1

Zero only when $k^2=4$

Before you read the solution

What does “only one” translate to?

Close, but it is the other way around. $f''(x)=0$ has one solution for lots of values of $k$, and most of them give an ordinary point of inflection where the curve is still climbing. The stationary point is the condition you have to force.

Correct. One stationary point means $x^2+(2-k^2)x+1$ has a repeated root, so its discriminant is zero. A repeated root of $f'$ is exactly a horizontal point of inflection.

No. Nothing in the question says $k$ is unique. Two values of $k$ are left, and the mark for step 5 is for checking both of them.

One mark at a time

QCAA marking guide

Six steps, one mark each. Try the question on paper before you reveal the first one.

Step 1: differentiate

[1 mark]

$$f'(x)=\frac{1}{kx}-\left[\frac{k(x+1)-kx}{(x+1)^2}\right]=\frac{1}{kx}-\frac{k}{(x+1)^2}$$

Marker: correctly determines the first derivative. Note $k$ is a constant, so the $\ln$ term differentiates to $\frac{1}{kx}$ and the second term needs the quotient rule.

Step 2: the quadratic

[1 mark]

$$\frac{1}{kx}=\frac{k}{(x+1)^2}\ \Rightarrow\ (x+1)^2=k^2x\ \Rightarrow\ x^2+(2-k^2)x+1=0$$

Marker: correctly determines the quadratic equation to identify the stationary point/s. Equivalent forms accepted, e.g. $0=(x+1)^2-k^2x$ or $k^2=\frac{(x+1)^2}{x}$.

Step 3: one root, and the value you throw away

[1 mark]

$$(2-k^2)^2-4=0\ \Rightarrow\ 2-k^2=\pm 2\ \Rightarrow\ k^2=0\ \text{or}\ k^2=4$$

$k=0$ is not valid, because $f$ divides by $k$, so $k=0$ has no function to speak of. That leaves $k=2$ and $k=-2$.

Marker: determines valid and non-valid solutions of $k$. FT marks allowed for errors in prior working.

Step 4: where the flat spot is

[1 mark]

$$k^2=4:\quad x^2-2x+1=0\ \Rightarrow\ (x-1)^2=0\ \Rightarrow\ x=1$$

Marker: determines the $x$-ordinate of the stationary point. The repeated root is the single stationary point, and $x=1$ is in the domain $x>0$.

Step 5: check both values, not one

[1 mark]

$$f''(x)=-\frac{1}{kx^2}+\frac{2k}{(x+1)^3}$$

$$k=2:\ f''(1)=-\tfrac12+\tfrac48=0 \qquad k=-2:\ f''(1)=\tfrac12-\tfrac48=0$$

Marker: determines values of the second derivative for both values of $k$. For each $k$, $x=1$ is the $x$-ordinate of both a stationary point and a point of inflection.

Step 6: the answer, and the communication mark

[1 mark]

Answer There is a horizontal point of inflection at $x=1$ when $k=\pm 2$.

Marker: shows logical organisation communicating key steps, labelling and combining equations when solving simultaneously, use of substitution, use of the discriminant when determining the number of solutions of a quadratic, and concluding statements.

Putting it all together

Two routes to $k=\pm 2$

Method 1: discriminant first

Force the quadratic to have one root, solve for $k$, then confirm with $f''$. This is the route the marking guide leads with, and the one the marks are written for.

Method 2: solve both conditions together

$f'=0$ gives $k^2=\frac{(x+1)^2}{x}$, and $f''=0$ gives $k^2=\frac{(x+1)^3}{2x^2}$. Equate the two: $2x=x+1$, so $x=1$ and $k^2=4$. It is shorter, and it still earns full marks.

Both routes rely on one fact worth remembering: a repeated root of $f'$ is a horizontal point of inflection. $f'$ touches zero and comes straight back. It never changes sign, so the curve never turns. That is why the discriminant, a Year 10 tool, settles a Year 12 inflection question.

What makes this complex unfamiliar

The question defines the term for you in the stem, then hands you a function with two unknowns in it. The step you need to see is that you never need $f''$ to find $k$. “Only one” tells you how many roots a quadratic has, so the discriminant does the work and $f''$ only confirms it. There are two more marks in the tidying up. You must reject $k=0$, which the algebra gives you but the function does not allow, and you must check both values of $k$ that are left. Plenty of students stop at $k=2$ and lose that mark.

Keep going

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Question wording and marking-guide steps are from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.