Mathematical Methods — Question 19

QCAA 2020, Paper 1 · 6 marks

Q19 · 2020 · Technology-freeComplex unfamiliar6 marks

QUESTION 19 (6 marks)

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

has only one horizontal point of inflection.

WORKED SOLUTION

6 marks

ANSWER

k=\pm2

Worked solution

One accepted method is:

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

Stationary points satisfy

x^2+(2-k^2)x+1=0.

Requiring only one stationary point gives discriminant

0

, so

(2-k^2)^2-4=0

. The valid values are

k=\pm2

(while

k=0

is invalid). Substituting either value gives the stationary point

x=1

. For

k=2

f''(x)=-\frac{1}{2x^2}+\frac{4}{(x+1)^3},

and for

k=-2

f''(x)=\frac{1}{2x^2}-\frac{4}{(x+1)^3}.

In each case

f''(1)=0

, so

x=1

is a horizontal point of inflection. Hence

k=\pm2

correctly determines the first derivative

[1 mark]

correctly determines the quadratic equation to identify the stationary point/s

[1 mark]

determines valid and non-valid solutions of

k

[1 mark]

determines

x

-ordinate of stationary point

[1 mark]

determines values of second derivative for both values of

k

[1 mark]

shows logical organisation communicating key steps

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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