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Level 2 Further Maths - Post some hard questions (Includes unofficial practice paper)

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Original post by _gcx
Might be a bit tricky but doable:

Given that f(x)=x3kx213x+5kf(x) = x^3 - kx^2 - 13x + 5k and f(5)=0f(5) = 0, find the other two solutions to f(x)=0f(x) = 0. (no trial and error :wink:)

Spoiler



Is this right?
(edited 6 years ago)
Original post by notnek
Yes that's correct :smile:

Yay :smile:
Original post by Loci Pi
x


Yes but I would put the solution in a spoiler so other people can have a go :smile:
Original post by _gcx
Yes but I would put the solution in a spoiler so other people can have a go :smile:


Oops sorry! Is it possible to go back and edit a comment?
Original post by notnek
Sequences. It's definitely grade A^ and I've seen a question like this but easier in textbooks but not in an exam (yet).

Here are the first five terms of a sequence:

33,2015,4935,9063,14399\displaystyle \frac{3}{3},\frac{20}{15}, \frac{49}{35}, \frac{90}{63}, \frac{143}{99}

The nnth term of the sequence is unu_n. Find the limiting value of unu_n as nn\rightarrow \infty.


Spoiler

Original post by Loci Pi
Oops sorry! Is it possible to go back and edit a comment?


Yep, just click the "edit" button, to the left of reply :smile:
Original post by _gcx
Yep, just click the "edit" button, to the left of reply :smile:


Done it lol. The edit button didn't load yesterday so I was wondering if I had done something wrong.
I got one for y'all.

sinx=116 sinx = \frac{\sqrt{11}}{6} and it's obtuse

What is cosx=? cosx = ?
(edited 6 years ago)
Original post by Pretish
I got one for y'all.

sinx=16 sinx = \frac{\sqrt{1}}{6} and it's obtuse

What is cosx=? cosx = ?


Spoiler

(edited 6 years ago)
Original post by B0redBrioche

Spoiler




ye.
Good luck for tomorrow everyone :smile:

Does anyone have any difficult sine/cosine rule questions? I feel like I could do with more practice on those.
Original post by Pretish
ye.


That means the question should read square root 11 not square root 1
Original post by Pretish
I got one for y'all.

sinx=116 sinx = \frac{\sqrt{11}}{6} and it's obtuse

What is cosx=? cosx = ?


Is this right?

Spoiler

Original post by Loci Pi
Is this right?

Spoiler




Spoiler

Original post by _gcx

Spoiler




Good method :smile:
(Update: read the previous comments and made the same error as someone else! Didn't know the thing about the denominators- any way of easily being able to tell in an exam if you need to worry about that or not?)

Haven't inserted Theta symbols so I hope the working is still understandable (also don't know how to do the "spoiler" thing on my phone sorry!)

Sin^2 + Sin + Cos^2 + Cos
= CosSin + Sin + Cos + 1

Sin^2 + Cos^2 CosSin 1 = 0
1 Cos^2 + Cos^2 CosSin –1 = 0
CosSin =0

If Cos = 0
Theta = 90°, 270°

If Sin = 0
Theta = 0°, 180°, 360°

Original post by notnek
Here's my first one. There's a very tricky part in this question that you need to watch out for (most A Level students would miss it) but apart from that it's good practice for Level 2 FM.


Solve the equation


sinθcosθ+1+cosθsinθ+1=1\displaystyle \frac{\sin \theta}{\cos \theta + 1} + \frac{\cos \theta}{\sin \theta + 1} = 1


for 0oθ360o0^o \leq \theta \leq 360^o .
(edited 6 years ago)
Reply 56
Original post by Pastelx
(Update: read the previous comments and made the same error as someone else! Didn't know the thing about the denominators- any way of easily being able to tell in an exam if you need to worry about that or not?)

I highly doubt that this part of the question could occur in the Level 2 FM exams. But it's always a good idea to substitute solutions back into the original equation to check they're correct (if you have time of course). If you did that then you'd have noticed that two of the solutions don't work.
Original post by notnek
Sequences. It's definitely grade A^ and I've seen a question like this but easier in textbooks but not in an exam (yet).

Here are the first five terms of a sequence:

33,2015,4935,9063,14399\displaystyle \frac{3}{3},\frac{20}{15}, \frac{49}{35}, \frac{90}{63}, \frac{143}{99}

The nnth term of the sequence is unu_n. Find the limiting value of unu_n as nn\rightarrow \infty.


Find nth term of numerator and denominator so 6n2-n-2/4n2-1. Apply l'hopitals rule to get the limit to infinity to get 3/2.
Original post by _gcx
Might be a bit tricky but doable:

Given that f(x)=x3kx213x+5kf(x) = x^3 - kx^2 - 13x + 5k and f(5)=0f(5) = 0, find the other two solutions to f(x)=0f(x) = 0. (no trial and error :wink:)


Plug in 5 to get k=3. Use inspection or long division to get the quadratic factor then factor this giving you

(X-5)(x+3)(x-1)
Reply 59
ABC is a straight line where BC is 20% of AC. In the diagram below, the coordinates of A, B and C are shown. pp and qq are positive numbers.



Find the value of pp and qq.
(edited 6 years ago)

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