For 1J, you can substitute each value of n with 1 as the equation holds for n is greater or equal to 1. After doing that you find you get an arithmetic sequence 2+3+4+5...+101. Using the formula gets the answer of (d).

For 1I, you want the greatest possible value of a which means you must have the smallest possible value of b. Setting b = 1 means that it (logb)^2 = 0
So you are left with 4((loga)^2) = 1. Rearranging this leaves you with 10^1/2 which equals the square root of 10/

(edited 7 years ago)

Reply 409

some-student

7

Original post by molly221

Help with 2007 paper? It is question 1: F, I and J. Thank you!

F: Using laws of indices we get:

2^3x + 4 = 2^2x + 2*2*(2^x)
(2^x)^3 + 4 = (2^x)^2 + 4(2^x)

let u = 2^x:

u^3 + 4 = u^2 + 4u

u^3 - u^2 - 4u + 4

factorising we get
(u - 1)(u^2 - 4)
2^x = 1 (Clearly x = 0)

u^2 = 4 => 2^x = +-2

2^x >= 0 and hence 2^x /= -2. We have two real solutions (c).

I: We want the smallest value of (log₁₀ b)^2, which is 0, when b = 1. Therefore 4(log₁₀a)^2 = 1 => log₁₀a = 1/2 (can only take positive root). a = 10^(1/2) = sqrt(10). Answer is (c)

J: let n = 1. Therefore we have an arithmetic sequence first term 2, difference 1 and length 100.
Sum: 50(4 + 99) = 50(103) = 5150.

Therefore 5150 > k, answer (d).

Reply 410

molly221

3

2008 paper help please? It is question 1: C and D. Thanks

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Reply 411

Nyathigiz

7

Original post by molly221

2008 paper help please? It is question 1: C and D. Thanks

For question 1D put x=1 into the equation and you end up with 1+3+5+7+...99
You can then use the sum formulae (can't remember the exact name of he equation) to get 50/2(1+99). It's fifty because it you go from x^0 to x^49 which is 50 numbers. This gives you 2500

Posted from TSR Mobile

Reply 412

some-student

7

Original post by molly221

2008 paper help please? It is question 1: C and D. Thanks

C: Eliminate x and y respectively each time to get
(sin^2(θ) + cos^2(θ))y = y
(sin^2(θ) + cos^2(θ))x = x

Both of which are true for all values of θ as (sin^2(θ) + cos^2(θ)) = 1. Answer (a)

D: Dividing by x - 1 and getting the remainder is the same as substituting in x = 1. This gives us an arithmetic sequence length 50, difference 2, first term 1.
We sum this to get
25(2 + 98) = 25 * 100 = 2500 (b)

Reply 413

molly221

3

Thank you to everyone who helped me with my previous question. Very much appreciated and helped a lot.

These are the questions I was unsure how to do question 1: D, H, I, J in the 2009 paper and I was wondering if anyone could explain how to do these questions?

Thank you so much in advance.

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Reply 414

alfmeister

2

Here's a solution for J.

Posted from TSR Mobile

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Reply 415

some-student

7

Original post by 11234

Ways to approach this?? I'm really hoping no geometry ones come up. Really would love a number theory one tho

(iii) took quite a long time. I had to look at the solutions to do (iv) - I still don't get why the triangles and sectors subtract to get the middle par although the areas weren't too bad to calculate once I was given the sectors and triangles - but even getting 11+/15 in this question on a paper where the average successful score was 62 will really help boost your mark.

I used http://www.webmath.com/gcircle.html to draw my circles.

Hints

Spoiler

Solutions

Spoiler

(edited 7 years ago)

Reply 416

hoakenfull

1

How much time is everyone dedicating to MAT over half-term? Not sure how I should be balancing MAT prep with A Level stuff.

Reply 417

KloppOClock

14

I was feeling sort of confident for this MAT exam... then I did the 2015 paper. :frown:

Reply 418

11234

8

Help pls with q5 of 2002 paper

http://www.mathshelper.co.uk/Oxford%20Admissions%20Test%202002.pdf

Reply 419

some-student

7

Original post by molly221

Thank you to everyone who helped me with my previous question. Very much appreciated and helped a lot.

These are the questions I was unsure how to do question 1: D, H, I, J in the 2009 paper and I was wondering if anyone could explain how to do these questions?

Thank you so much in advance.

D: Assume n is odd. Infront of the term involving n, which evaluates to n, we have n - 1 terms, which can be seen as pairs which always have a value of -1.

Therefore n - 1/2(n - 1) >= 100

2n - (n - 1) >= 200

n + 1 >= 200
n >= 199

The smallest value of n is 199 (c)

I: we want x^2 - 1 to be a factor. Therefore f(1) = 0 and f(-1) = 0.

f(1) = 0
n^2 - 25n + 150 = 0
(n - 10)(n - 15) = 0
n = 10, n = 15

f(-1) = 0

We need to consider when n is even or odd, as (-1)^x will be 1 if x is even and -1 if x is odd

When n is even
-n^2 + 25n - 150 = 0
-(n^2 - 25n + 150) = 0
-(n - 10)(n - 15) = 0
n = 10, n = 15
n can only be 10 here as 15 is not even

When n is odd
-n^2 - 25n - 150 = 0
-(n^2 + 25n + 150) = 0
-(n + 10)(n + 15) = 0
n = -10, n = -15
n can only be -15 here as -10 is not odd

The only consistent value of n is 10 and hence the answer is (B)

MAT Prep Thread - 2nd November 2016

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