2005CSSA Q1 - (b) (ii) (1 Viewer)

[Stefano]

2005 CSSA: Question 1 (b) (ii)

The line y=mx is a tangent to the curve y=3-[1/x^2]. Show that the equation mx^3 - 3x^2 + 1=0 has a double root and hence find any values of m.

I got this during the exam, but looking back at it now I have no idea how to approach it. When I look at the solutions I cann follow all the steps, yet cannot understand WHY they are doing them. Could someone help? Thanks.

 

[robbo_145]

since its a tangent we know that at the point of intersection that the x,y values are equal
thus
mx = 3 - 1/x^2
mx^3 - 3x^2 + 1 = 0 (A)

and the gradients are equal, thus the solution of (A) will be shared with its derivitive
hence

3mx^2 - 6x = 0
3x(mx-2) = 0
hence mx - 2 = 0 is the correct solution as x = 0 is not defined (since 1/x^2 is a term in the original graph)
mx = 2

thus x = 2/m into (A)
m(8/m^3) - 3 (4/m^2) + 1 = 0
thus 8/m^2 - 12/m^2 +1 = 0
4/m^2 =1
m^2=4
m = +- 2

there are no doubt other (quicker) ways to do this question

 

[Stefano]

Nicely done.
Thanks man!

The tricky part, for me, was:

and the gradients are equal, thus the solution of (A) will be shared with its derivitive

 

[christoph]

Still, the question says to SHOW that\

mx^3 - 3x^2 + 1 = 0

has a double root. Is your working really sufficient?

 

[robbo_145]

like i said there are other ways of doing it

a more verbose way of saying it would be

let α be the intercept of the two curves

thus α is a solution to mx^3 - 3x^2 + 1 = 0

but at α the gradient of each line will be equal
thus dy/dx of each curve is equal at x = α
2/α = m
α = 2/m

then prove alpha is a root of the derivitive etc.
so you could do it this way as well

 

[Stefano]

Roger that!

Thanks robbo.