[hello]

I have a few questions about P6, from the Review Exercise in the Heinemann Edexcel book.

I'm not sure how to search this forum, but I'm sure they've been answered before, so I'll list them first then write them! Thanks a lot.

75, 98, 107b/c, 108, 116b

Q. 75

Sequence, u_nu_n+1 + 2u_n+1 - 4u_n - 3 = 0, and u₁=2.

Prove by induction that u_n>0 and u_n<3, and hence prove the sequence is increasing.

Q. 98

Show that 7ⁿ - 6n - 1 is divisible by 36 and 5ⁿ - 4n - 1 is divisible by 16.

Q. 107

z⁵ - 1 = 0. Given θ is the complex root of this equation, show the roots are 1, θ, θ², θ³, θ⁴.

Show θ⁴ = θ*, and hence obtain (z⁵ - 1) as a product of real linear and quadratic factors giving coefficients in terms of integers and cosines.

Show also that z⁵ - 1 = (z - 1)(z⁴ + z³ + z² + z + 1) and hence find cos(2pi/5) in terms of surds.

Q. 108

Show y = 1 + ∫ e^t² dt is the solution of dy/dx = e^x² for which y=1 at x=0.

Q. 116

r₁ = (-2i + 5j - 11k) + s(3i + j + 3k)
r₂ = (8i + 9j) + s(4i + 2j + 5k)

Show these lines intersect at P, stating P.

Show that the plane θ containing these lines is parallel to the plane Φ with equation r . (i + 3j - 2k) = 7.

Find the distance between the planes θ and Φ.

[dvs]

75. What have you tried here?

98. http://www.thestudentroom.co.uk/t125126.html

107. You want to find the fifth roots of unity. The things the question asks for are properties of those roots. As for the last part, divide z^5-1 by z-1.

108. dy/dx = d(1 + ∫ e^(t²) dt)/dx = d(∫ e^(t²) dt)/dx = e^(x²) -- assuming that the limits are 0->x. This is because of the first fundamental theorem of calculus (do a google search).

116. Find a normal to plane θ, and show that this normal is parallel (i + 3j - 2k) (by using the cross product, for example).

[lgs98jonee]

Originally Posted by hello

Q. 98

Show that 7ⁿ - 6n - 1 is divisible by 36

When n=1, 7-6-1=0

Assume true for n=k

If n=k+1, then 7^(k+1)-6(k+1)-1=7.7^k-6k-7

Mulitiply this by 5 to get 35.7^k-30k-35

now add the equation for when n=k, which we have assumed to be divisible by 36

and we get 36.7^k-36k-36=36(7^k-k-1) which is clearly divisible by 36.

[hello]

Originally Posted by dvs

75. What have you tried here?

98. http://www.thestudentroom.co.uk/t125126.html

107. You want to find the fifth roots of unity. The things the question asks for are properties of those roots. As for the last part, divide z^5-1 by z-1.

108. dy/dx = d(1 + ∫ e^(t²) dt)/dx = d(∫ e^(t²) dt)/dx = e^(x²) -- assuming that the limits are 0->x. This is because of the first fundamental theorem of calculus (do a google search).

116. Find a normal to plane θ, and show that this normal is parallel (i + 3j - 2k) (by using the cross product, for example).

75: Basically I don't have a clue - is this induction or what?

98: Cheers!

107: I'm fine with finding the roots, I just can't express z⁵ - 1 as a product of real linear and quadratic factors.

116: That part's okay, it's finding the distance between the planes that I can't do (I wrote the whole question to save on any confusion on what's already been found).

[Gaz031]

Sequence, u_nu_n+1 + 2u_n+1 - 4u_n - 3 = 0, and u₁=2.

Prove by induction that u_n>0 and u_n<3, and hence prove the sequence is increasing.

[Mimetex cannot convert this formula]

Assume true for [Mimetex cannot convert this formula], prove true for [Mimetex cannot convert this formula] and so on..

[dvs]

75. It's induction.
Assume that for some n=k we have u{k}>0. Then we also have:
4u{k} + 3 > 0, and
u{k} + 2 > 0

So:
(4u{k} + 3)/(u{k} + 2) > 0

But (4u{k} + 3)/(u{k} + 2) = u{k+1}. So we managed to get u{k+1} > 0 iff u{k} > 0. So by checking this for n=1 we complete the induction.

We can prove that u{n}<3 in a similar way I think. And for the final part, you want to prove that u{n} < u{n+1}.
-

107. You have the roots of z^5 = 1, i.e. of z^5 - 1 = 0. Two are complex conjugates and one is 1.

You can think of the complex conjugates as being the roots of a quadratic equation. For example, let's take cos(2pi/5)±i.sin(2pi/5). Now:
The product of these two roots (which is the difference between 2 squares) = cos²(2pi/5) - i².sin²(2pi/5) = cos²(2pi/5) + sin²(2pi/5) = 1
The sum of these two roots = 2cos(2pi/5)

Therefore, the quadratic equation with those as its roots is:
z² - [2cos(2pi/5)]z + 1 = 0
-

116. If the two planes are parallel, then you want to find the perpendicular distance from any point on θ to Φ.
-

I hope that isn't too difficult to read.

[RichE]

Originally Posted by hello

Q. 108

Show y = 1 + ∫ e^t² dt is the solution of dy/dx = e^x² for which y=1 at x=0.

This doesn't make sense without limits. Surely you mean show

[Mimetex cannot convert this formula]

is the solution.

[hello]

Originally Posted by RichE

This doesn't make sense without limits. Surely you mean show

[Mimetex cannot convert this formula]

is the solution.

Yes, sorry, I omitted the limits in error.

[Fermat]

Originally Posted by hello

...

116: That part's okay, it's finding the distance between the planes that I can't do (I wrote the whole question to save on any confusion on what's already been found).

the distance between a point P(x₀,y₀,z₀) and a plane given by ax + by + cz + d = 0, with a normal n [= (a,b,c)], is given by,

d = |ax₀ + by₀ + cz₀ + d| / |n|

now all you have to do is find a point on the second plane.