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C2 maths help

Could someone please help me with question 21a, I've tried it using the remainder theorem and simultaneous equations but the answers I get do not work with part b. Could someone go through it for me and help me understand?! ImageUploadedByStudent Room1383047308.898817.jpg

ImageUploadedByStudent Room1383047308.898817.jpg

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Reply 1

ChildishHambino

Original post by Mathsphysicspe

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that should work, what were your workings?

EDIT: just tried it, i believe there is a mistake in the question :smile:

(edited 10 years ago)

Reply 2

ExcitinglyMundane

Original post by ChildishHambino

that should work, what were your workings?

EDIT: just tried it, i believe there is a mistake in the question :smile:

I agree, it is possible to find a value for a and b, however this produces a function not divisible by

(x-2)

.

I believe that the question should read: Show that

(x-3)

is a factor of

f(x)

Reply 3

Mathsphysicspe

Original post by ExcitinglyMundane

I agree, it is possible to find a value for a and b, however this produces a function not divisible by

(x-2)

.

I believe that the question should read: Show that

(x-3)

is a factor of

f(x)

I looked at the answer and it's got different values for a and b. I got a=-5 and b=-12 I think and the answer book got a=-12 and b=16 this however seems to work for both part a and b but I don't see how they got the answers for a and b!

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Reply 4

danlocke

Looks like a mistake to me as well:

Reply 5

Mathsphysicspe

Original post by ChildishHambino

If a=-12 and b=16 that would suggest that x-4 is a factor which it says it isn't in part one, the book is wrong :smile:

But if you sub in a=-12 and b=16 using f(4) 64-48+16=32 which shows 4 is not a factor doesn't it?! I'm so confused!!

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Reply 6

ChildishHambino

Original post by Mathsphysicspe

But if you sub in a=-12 and b=16 using f(4) 64-48+16=32 which shows 4 is not a factor doesn't it?! I'm so confused!!

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Sorry can't even add :tongue:

, anyway if you do f(-2) you don't get -10 as a remainder showing it is wrong (hopefully).

Reply 7

Mathsphysicspe

Original post by ChildishHambino

Sorry can't even add :tongue:

, anyway if you do f(-2) you don't get -10 as a remainder showing it is wrong (hopefully).

Yeah I agree!! Stupid question is so frustrating! But at least now I know someone else agrees that it doesn't work!!

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Reply 8

Mathsphysicspe

Could someone help me with this question as well 24b :/ ... ImageUploadedByStudent Room1383051059.027831.jpg

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Reply 9

ExcitinglyMundane

Original post by Mathsphysicspe

Yeah I agree!! Stupid question is so frustrating! But at least now I know someone else agrees that it doesn't work!!

Okay, they made a mistake. We all assumed the two factors and their remainders were correct, and as such we got

a=-5, b=-12

, however that makes part b wrong, which works if the question were to show that

(x-3)

is a factor. The answers in the book show that actually what's wrong is the initial factors. Either way, it is wrong as it is, but making one correction makes it work (but with different answers).

Original post by Mathsphysicspe

Could someone help me with this question as well 24b :/ ... ImageUploadedByStudent Room1383051059.027831.jpg

]

The C2 method for this is as follows:
1) Work out the gradient of the line between the centre of the circle and P.
2) As this is the radius of the circle passing through point P, it therefore must be perpendicular to the tangent to the circle at point P. (Think circle theorems at GCSE level)
3) You should know:

G_1 = -\frac{1}{G_2}

, with

G_1

being the gradient of a line and

G_2

being the gradient of the perpendicular line.
4) You can therefore workout the gradient of the tangent.
5) As you know the gradient of the tangent, and that it must go through P, you can use