2023HSCShort answer

Q14 (b) (ii)

3 marks

Question

Consider the hyperbola $y=\dfrac{1}{x}$ and the circle $(x-c)^2+y^2=c^2$ , where $c$ is a constant.

(ii) The graphs of $y=x^4-2cx^3+1$ for $c=0.8$ and $c=1$ are shown.

By considering the given graphs, or otherwise, find the exact value of
$c>0$ such that the hyperbola $y=\frac{1}{x}$ and the circle
$(x-c)^2+y^2=c^2$ intersect at only one point.

Uncropped source page 14 for question 14.b.ii; requires human review and cropping — Reviewed visual · source page 14

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	3
Finds the $x$ -coordinate of the double root in terms of $c$ , or equivalent merit	2
Recognises that the circle and hyperbola are tangent when the polynomial has a double root and so is tangent to the $x$ -axis, or equivalent merit	1

Sample answer:

Let $f(x)=x^4-2cx^3+1$ .

The hyperbola and the circle have exactly one point of intersection if there exists a value of $x$ such that $f(x)=0$ and $f'(x)=0$ .

\begin{aligned} f'(x)&=4x^3-6cx^2\\ &=2x^2(2x-3c)\\ f'(x)=0&\iff x=0\text{ or }x=\frac{3c}{2} \end{aligned}

Note that $f(0)\ne0$ .

\begin{aligned} f\left(\frac{3c}{2}\right)=0&\iff\left(\frac{3c}{2}\right)^4-2c\left(\frac{3c}{2}\right)^3+1=0\\ &\iff3^4c^4-3^3\times2^2c^4+2^4=0\\ &\iff-27c^4=-16\\ &\iff c=\frac{2}{\sqrt[4]{27}} \end{aligned}

Supplied marking guidance · human reviewed

Compare this with your working, not only your final sentence.

Syllabus alignment

Functions

PolynomialsUse polynomial division, the factor and remainder theorems, zeros and graphs to solve polynomial problems.

Spot a problem? Send feedback