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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)

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

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`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)

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`text(See Worked Solutions)`

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`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)`