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

Reply 40

16

Original post

by _gcx

Might be a bit tricky but doable:

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

)

Spoiler

Is this right?

Reply 41

Loci Pi

16

Original post

by notnek

Yes that's correct :smile:

Yay

Reply 42

_gcx

21

Original post

by Loci Pi

x

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

Reply 43

Loci Pi

16

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?

Reply 44

B0redBrioche

12

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:

$\displaystyle \frac{3}{3},\frac{20}{15}, \frac{49}{35}, \frac{90}{63}, \frac{143}{99}$

The $n$ ^th term of the sequence is $u_n$ . Find the limiting value of $u_n$ as $n\rightarrow \infty$ .

Spoiler

Reply 45

_gcx

21

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:

Reply 46

Loci Pi

16

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.

Reply 47

3317752

13

I got one for y'all.

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

What is $cosx = ?$

Reply 48

B0redBrioche

12

Original post

by Pretish

I got one for y'all.

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

What is $cosx = ?$

Spoiler

Reply 49

3317752

13

Original post

by B0redBrioche

Spoiler

ye.

Reply 50

Loci Pi

16

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.

Reply 51

B0redBrioche

12

Original post

by Pretish

ye.

That means the question should read square root 11 not square root 1

Reply 52

Loci Pi

16

Original post

by Pretish

I got one for y'all.

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

What is $cosx = ?$

Is this right?

Spoiler

Reply 53

_gcx

21

Original post

by Loci Pi

Is this right?

Spoiler

Reply 54

Loci Pi

16

Original post

by _gcx

Spoiler

Good method :smile:

Reply 55

Pastelx

18

(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

$\displaystyle \frac{\sin \theta}{\cos \theta + 1} + \frac{\cos \theta}{\sin \theta + 1} = 1$

for $0^o \leq \theta \leq 360^o$ .

Reply 56

Notnek

OP

21

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.

Reply 57

thekidwhogames

18

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:

$\displaystyle \frac{3}{3},\frac{20}{15}, \frac{49}{35}, \frac{90}{63}, \frac{143}{99}$

The $n$ ^th term of the sequence is $u_n$ . Find the limiting value of $u_n$ as $n\rightarrow \infty$ .

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.

Reply 58

thekidwhogames

18

Original post

by _gcx

Might be a bit tricky but doable:

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

)