smc-1205-30-Other

Functions, EXT1 F2 2023 HSC 14b

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

Show that the \(x\)-coordinates of any points of intersection of the hyperbola and circle are zeros of the polynomial \(P(x)=x^4-2 c x^3+1\). (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
The graphs of \(y=x^4-2 c x^3+1\) 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 \(y=\dfrac{1}{x}\) and the circle \((x-c)^2+y^2=c^2\) intersect at only one point. (3 marks)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

\(\text{See Worked Solutions}\)
\(\sqrt[4]{\dfrac{16}{27}}\approx 0.877\)

Show Worked Solution

i. \(y=\dfrac{1}{x}\ …\ (1) \)

\((x-c)^2+y^2=c^2\ …\ (2) \)

\(\text{Substitute (1) into (2):}\)

\((x-c)^2+\Big{(}\dfrac{1}{x}\Big{)}^2 \)	\(=c^2\)
\(x^2-2cx+c^2+\dfrac{1}{x^2}\)	\(=c^2\)
\(x^4-2cx^3+1\)	\(=0\)

Mean mark (i) 53%.

ii. \(\text{The two graphs show that for some value of}\ \ 0.8 \leq c \leq 1,\)

\(P(x)\ \text{has a minimum that touches the}\ x\text{-axis once.}\)

\(P(x)\)	\(=x^4-2cx^3+1\)
\(P^{′}(x)\)	\(=4x^3-6cx^2\)

\(\text{Find}\ x\ \text{when}\ P^{′}(x)=0: \)

\(4x^3-6cx^2\)	\(=0\)
\(2x^2(2x-3c)\)	\(=0\)
\(x\)	\(=\dfrac{3c}{2}\ \ (x \neq 0)\)

\(\text{Find}\ c\ \text{when}\ P(\frac{3c}{2})=0: \)

\(\Big{(} \dfrac{3c}{2} \Big{)}^4-2c\Big{(} \dfrac{3c}{2} \Big{)}^3+1 \)	\(=0\)
\(\dfrac{81c^4}{16}-\dfrac{54c^4}{8}+1\)	\(=0\)
\(\dfrac{(108-81)c^4}{16}\)	\(=1\)
\(\dfrac{27c^4}{16}\)	\(=1\)
\(c^4\)	\(=\dfrac{16}{27}\)
\(c\)	\(=\sqrt[4]{\dfrac{16}{27}} \)
	\(\approx 0.877\)

Mean mark (ii) 19%.

Functions, EXT1′ F2 2019 HSC 16ai

Consider the equation `x^3 - px + q = 0`, where `p` and `q` are real numbers and `p > 0`.

Let `r = sqrt((4p)/3)` and `cos 3 theta = (-4q)/r^3`.

Show that `r cos theta` is a root of `x^3 - px + q = 0`.

You may use the result `4 cos^3 theta - 3 cos theta = cos 3 theta`. (Do NOT prove this.) (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ text{(See Worked Solutions)}`

Show Worked Solution

`x^3 – px + q = 0\ …\ (1)`

♦ Mean mark 49%.

`r = sqrt((4p)/3) \ => \ p=(3r^2)/4`

`cos 3 theta = (-4q)/r^3 \ => \ q = (-r^3 cos 3 theta)/4`

`text(Given)\ \ 4 cos^3 theta – 3 cos theta = cos 3 theta\ …\ (2)`

`text(Substitute)\ \ x = r cos theta\ \ text{into (1)}`

`r^3 cos^3 theta – (3r^2)/4 r cos theta – (r^3 cos 3 theta)/4`	`= 0`
`r^3/4 underbrace((4 cos^3 theta – 3 cos theta – cos 3 theta))_(=\ 0\ text{(see (2) above)})`	`= 0`

`:.r cos theta\ \ text(is a root).`

Functions, EXT1 F2 2019 HSC 14b

The diagram shows the graph of `y = 1/(x - k)`, where `k` is a positive real number.

By considering the graphs of `y = x^2` and `y = 1/(x - k)`, explain why the function `f(x) = x^3 - kx^2 - 1` has exactly one real zero. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Draw)\ \ y = x^2\ \ text(on the diagram:)`

`text(One point of intersection occurs when)`

`x^2`	`= 1/(x – k)`
`x^3 – kx^2`	`= 1`
`x^3 – kx^2 – 1`	`= 0`

`text(S)text(ince only 1 point of intersection)`

`=> x^3 – kx^2 – 1 = 0\ \ text(has exactly 1 zero)`

Functions, EXT1 F2 2016 HSC 10 MC

Consider the polynomial `p(x) = ax^3 + bx^2 + cx - 6` with `a` and `b` positive.

Which graph could represent `p(x)`?

Show Answers Only

`A`

Show Worked Solution

`p(x)`	`= ax^3 + bx^2 + cx – 6`
`p prime (x)`	`= 3ax^2 + 2bx + c`
`p ″ (x)`	`= 6ax + 2b`

`text(S) text(ince)\ \ a, b > 0,`

`text(As)\ \ x -> oo,\ \ p(x) -> oo`

`:.\ text(Eliminate)\ \ C and D.`

`text(POI occurs when)\ \ p ″ (x) = 0,`

`6 ax + 2b`	`= 0`
`x`	`= -b/(3a) < 0,\ \ \ (a, b > 0)`

`:.\ text(Eliminate)\ \ B`

`=> A`