Past Exam Questions (1 Viewer)

[allGenreGamer]

Just doing some past exam papers in preparation for the half-yearly. I have some questions:

1) I've proven that the normal to the parabola x^2 = 4ay at the point P(2ap, ap^2) has equation x + py = 2ap + ap ^ 3, the question now reads "Hence show that there's only one normal to the parabola passing through the focus". How?

2) Given a = 3(x - x^-1) and b =6(x = x^-1) write an equation connecting a and b which is independent of x.

3) Solve for x: (0.3)^x < 4/3

4) Solve 2log(3x+1) - log(x+1) + log(7x+4)

My test is on Monday and these are my final questions - please, help me solve these problems so that I can be confident in the exams. Even if you don't answer them all it is fine. Thanks in advance.

 

[Estel]

1) Can't do 'cause I'm in Yr 11

2)
a=3(x - x^-1) and b=6(x - x^-1) [I presume the = is a minus]
2a=b

3)(0.3)^x < 4/3
xlog(3/10)<log(4/3)
x>(log4/3)/log(3/10)
x>[log4-log3]/[log3-1]

4)When you say solve, what is it meant to be equal to?

 

[CM_Tutor]

1) I've proven that the normal to the parabola x^2 = 4ay at the point P(2ap, ap^2) has equation x + py = 2ap + ap ^ 3, the question now reads "Hence show that there's only one normal to the parabola passing through the focus". How?

You have the general form of the normal, and you require it to pass through the focus (0, a). When you substitute this into the equation of the normal, you will get a polynomial in p, whose solution(s) are all parameter(s) p corresponding to point(s) where the normal(s) pass through the focus. Since you are asked to show there is only one such normal, and hence only one such point (and one such parameter), you want to show the polynomial has only one real solution - note that, by symmetry, that means the only solution must be p = 0.

So, put (0, a) into x + py = 2ap + ap³: 0 + p(a) = 2ap + ap³
0 = ap + ap³ = ap(1 + p²)
p = 0 is clearly the only solution to this equation, as 1 + p² > 0 for all real p, and so the result is proven.

For question 2, assuming that you typed the question correctly, apart from the error Estel pointed out, then Estel's solution is correct - however, this is really easy, so I'm wondering if you have mistyped the question. Could it have been a = 3^{x - x^-1} and b = 6^{x - x^-1}?

3) Solve for x: (0.3)^x < 4/3

Expanding on Estel's solution, and taking log x to mean a base 10 logarithm:
0.3^x < 4 / 3
log 0.3^x < log (4 / 3)
x * log 0.3 < log 4 - log 3
x > (log 4 - log 3) / log 0.3, as log 0.3 < 0
x > (log 4 - log 3) / (log 3 - 1), as log 0.3 = log (3 / 10) = log 3 - log 10 - log 3 - 1, as log 10 = 1

4) Solve 2log(3x+1) - log(x+1) + log(7x+4)

I am going to take this to mean 2log(3x + 1) = log(x + 1) + log(7x + 4)
log(3x + 1)² = log(x + 1)(7x + 4)
(3x + 1)² = (x + 1)(7x + 4)
9x² + 6x + 1 = 7x² + 11x + 4
2x² - 5x - 3 = 0
(2x + 1)(x - 3) = 0
So, x = -1 / 2 or x = 3

BUT, the solutions must satisfy the original equation, and so (3x +1), (x + 1) and (7x + 4) must all be positive.
This restricts the solutions to the domain x > -1 / 3, and so x = -1 / 2 is not a valid solution.

So, x = 3

 

[allGenreGamer]

Indeed I have typed the question wrong:

2) Given a = 3(x - x^-1) and b =6(x + x^-1) write an equation connecting a and b which is independent of x.

Thanks for being so helpful, please finish off ur charity

 

[CM_Tutor]

Rewriting the equations, we have x - x^-1 = a / 3 _____ (1) and x + x^-1 = b / 6 _____ (2)

(1) + (2): (a / 3) + (b / 6) = 2x, and so x = (2a + b) / 12

Putting this back into (1), we get: (2a + b) / 12 - 12 / (2a + b) = a / 3
Multiply by 12(2a + b): (2a + b)² - 12² = 4a(2a + b)
4a² + 4ab + b² - 144 = 8a² + 4ab
b² - 4a² = 144