Question

How to do these question?

Answer 1

a) `p=2/3` and `q=-4/27`

b) `a != 2 => a in RR \\\ {2}`

c) `(1/3(8-a), 4/27(2-a)^3)`

d) `a-2`

Explanation:

Let `g:RR rarr RR`, where `g(x)=(x-2)^2(x-a)` where `a in RR`

a) The coordinates of the stationary points of `g` are `P(2,0)` and `Q(p(a+1),q(a-2)^3)`, where `p` and `q` are rational numbers. Fin the values of `p` and `q`

b) Show that the gradient of the tangent to the graph of `y=g(x)` at the point `(a,0)` is positive for `a in RR \\\ {2}`

c) Find the coordinate of another point where the tangent to the graph of `y=g(x)` is parallel to the tangent at `x=a`

d) Hence, find the distance between this point and `Q` when `a gt 0`

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Part (a):

We have

`g(x) = (x-2)^2(x-a)`

Differentiating wrt `x` using the product rule and the chain rule, we get the first derivative:

`g'(x) = (x-2)^2(1) + 2(x-2)(x-a)`
`\ \ \ \ \ \ \ \ = (x-2){(x-2) + 2(x-a)}`
`\ \ \ \ \ \ \ \ = (x-2)(x-2 + 2x-2a)`
`\ \ \ \ \ \ \ \ = (x-2)(3x-2 -2a)`

At a stationary point (or a critical point), the first derivative vanishes, thus we gave `g'(x)=0` which requires that:

`(x-2){3x-2 -2a} = 0`

Yielding two solutions:

`x_1 = 2` and `x_2 = (2a+2)/3 = 2/3(a+1)`

Consider the case `x_1=0`

`g(x_1) = 0`, so this corresponds to coordinate `P`

Consider the case `x_2=0`

`g(x_2) = ((2a+2)/3-2)^2((2a+2)/3-a)`
`\ \ \ \ \ \ \ \ = ((2a+2-6)/3)^2((2a+2-3a)/3)`
`\ \ \ \ \ \ \ \ = ((2a-4)/3)^2((2-a)/3)`
`\ \ \ \ \ \ \ \ = ((2(a-2))/3)^2(-(a-2)/3)`
`\ \ \ \ \ \ \ \ = (4/9)(a-2)^2(-1/3)(a-2)`
`\ \ \ \ \ \ \ \ = -4/27(a-2)^3`

Comparing with the given form `Q(p(a+1),q(a-2)^3)`, we conclude that:

`p=2/3` and `q=-4/27`

Part (b):

The gradient of the tangent to a curve at any particular point is given by the value of the derivative of the function at that point.

So, at the given coordinate `(a,0)`, the gradient of the tangent is given by:

`g'(a) = (a-2)(3a-2 -2a}`
`\ \ \ \ \ \ \ \ = (a-2)(a-2)`
`\ \ \ \ \ \ \ \ = (a-2)^2`

Which is positive provided `a != 2`, hence provided `a in RR \\\ {2}` QED

Part (c):

The tangents are parallel, providing they have the same gradient. We have already established that the gradenite of the tangent at `x=a` is given by `g'(a)`, so we seek coordinates on the initial curve satisfying `g'(x)=g'(a)`, which requires that:

`(x-2)(3x-2 -2a) = (a-2)^2`
`:. x(3x-2 -2a) -2(3x-2 -2a) = (a-2)^2`
`:. 3x^2-2x -2ax -6x+4 +4a = a^2-4a+4`
`:. 3x^2-2x -2ax -6x +8a - a^2 =0`
`:. 3x^2-(2 +2a +6) + (8a - a^2) =0`

This is quadratic in `x` and so has `2` solutions, we already know that `x=a` is a solution (part b), so we can use the sum of roots property, that is `alpha+beta=-b/a` so that with `alpha=a`, then our second solution `x=beta` is given by:

`a + beta = (2 +2a +6)/3`
`:. beta = (8+2a)/3 - a`
`\ \ \ \ \ \ \ = (8+2a-3a)/3`
`\ \ \ \ \ \ \ = 1/3(8-a)`

And with this value of `x` we calculate the `y`-coordinate:

`g(x) = (1/3(8-a)-2)^2(1/3(8-a)-a)`
`\ \ \ \ \ \ \ = (1/3(8-a-6))^2(1/3(8-a-3a))`
`\ \ \ \ \ \ \ = (1/3(2-a))^2(1/3(8-4a))`
`\ \ \ \ \ \ \ = (1/3)^2(2-a)^2(1/3)(4)(2-a)`
`\ \ \ \ \ \ \ = 4/27(2-a)^3`

So that the required coordinate is `(1/3(8-a), 4/27(2-a)^3)`

Part (d):

We require the distance, `d`, say, between the coordinates:

`Q(2/3(a+1),-4/27(a-2)^3)` and `(1/3(8-a), 4/27(2-a)^3)`

Which by pythagoras is given by:

`d^2 = {1/3(8-a) - 2/3(a+1)}^2 + {4/27(2-a)^3 -(-4/27(a-2)^3) }^2`

`\ \ \ \ = {1/3(8-a -2a-2))}^2 + {4/27(2-a)^3 -4/27(2-a)^3 }^2`

`\ \ \ \ = {1/3(6-3a))}^2 + 0`

`\ \ \ \ = (2-a)^2`

So that:

`d = +-(2-a)`

We require that `d gt 0` and we are given a restriction `a gt 2 => 2-a lt 0`, thus the distance is:

`d = a-2`