Edexcel P2

hi, i don't have answers to this edexcel mock paprt 6672/01

can someone pls help.....

The curve C has equation y=4x^(3/2) - ln(5x), where x>0. The tangent at the point on C where x=1 meets the x-axis at the point A.

Prove that the x-coordinate of A is 1/5 ln (5e)

Thanks.

well first you need to differentiate y to get an expression for dy/dx.

so dy/dx = 6x^(1/2) - 1/x, and at x = 1, dy/dx = 5

so using y - y1 = (dy/dx) (x - x1), the equation of the tangent is
y - (4 - ln 5) = 5 (x - 1)

tidy that up and u get
y + ln 5 + 1 = 5x

At point A the y coordinate is 0 (since on the x - axis), so u get that

5x = 1 + ln 5
and x = 1/5 (1 + ln 5)

slightly different from ur answer, but i dont think ur answer is correct.

x = 1/5 (1 + ln 5)

slightly different from ur answer, but i dont think ur answer is correct.

His answer is correct.

ln e = 1

Therefore

x = 1/5 (1 + ln 5) = 1/5 (ln e + ln 5) = 1/5 ln (5e)

how do u put 4ln4 - 4ln1/2 in the form bln2? where b is a rational constant? the actual answer is 12ln2, but i don't know how to work it out. thanx

4ln4 - 4ln(1/2) = 4( ln4 - ln(1/2) ) = 4( ln(2^2) - ln(1/2) )

Using the rules: ln(a^b) = blna, ln(a/b) = lna - lnb and ln1 = 0 we have:

4( 2ln2 -(ln1 - ln2) ) = 4( 2ln2 - (-ln2) ) = 4(3ln2) = 12ln2

omg thank u .........how do u know all this!? as u can tell.......i've just started some revision and i haven't been listening during my lessons, lol. i take it u do maths too.

Yes, I do a lot of maths because I do further maths as well. I have found that the more I have done maths the more internalised it has become.

The more you practice maths, the more it becomes second nature. That is why I would recommend just doing loads of past papers to revise for any maths exam.

yeah thanks, that wot i'm doing.....i'm not planning to study my textbooks and in all my other maths exams i usually just go through past papers too, except this time its more difficult cos i can't actually figure the answers out.......but i'm gonna keep going!!