Advanced Higher Mathematics 2014-2015 :: Discussion Thread

I think it'll be similar to last years... Or at least I hope... But I doubt they would change it by more than +/- 3 or 4 marks in one go (as that's never happened before) (attached comparison of 2014:2007)

See for the question about the matrices and you have to combine m1 and m2, I did m2.m1 because thats the order, but DLBmaths did m1.m2 and got a different answer where the values were positive, but mine were negative. Im kinda worrying if i lost such an easy mark?

I did M2.M1 because that was the order stated in the exam, and we know that's important since matrix multiplication is not commutative.
So in this instance, I believe DLBmaths is wrong and we are correct.

Posted from TSR Mobile

I did M2.M1 because that was the order stated in the exam, and we know that's important since matrix multiplication is not commutative.
So in this instance, I believe DLBmaths is wrong and we are correct.

Posted from TSR Mobile

DLBmaths is not wrong. It is M1.M2.

'Find the 2x2 matrix, M3, associated with an anticlockwise rotation through pi/2 radians about the origin followed by a reflection in the y-axis.'

If you multiply M2 by the position vector of a point, you get the anticlockwise rotation needed. It is then that you apply the reflection in the y-axis, i.e. multiply by M1. Since you multiply the matrix by the position vector of the point, i.e. M3 . X (X being the point), then the M2 matrix must be multiplied by X first, and then the result of this multiplied by M1.

Hence, M3 . X = (M1.M2) . X = M1 . (M2.X)

So it would have to be M1.M2.

I did M2.M1 because that was the order stated in the exam, and we know that's important since matrix multiplication is not commutative.
So in this instance, I believe DLBmaths is wrong and we are correct.

Posted from TSR Mobile

You just got #rekt.

Posted from TSR Mobile

I watched his video again and he seemed pretty certain that because it followed it was that bit first.....

DLBmaths is not wrong. It is M1.M2.

'Find the 2x2 matrix, M3, associated with an anticlockwise rotation through pi/2 radians about the origin followed by a reflection in the y-axis.'

If you multiply M2 by the position vector of a point, you get the anticlockwise rotation needed. It is then that you apply the reflection in the y-axis, i.e. multiply by M1. Since you multiply the matrix by the position vector of the point, i.e. M3 . X (X being the point), then the M2 matrix must be multiplied by X first, and then the result of this multiplied by M1.

Hence, M3 . X = (M1.M2) . X = M1 . (M2.X)

So it would have to be M1.M2.

I'm really not understanding this? I would have thought that if it said anticlockwise rotation followed by reflection in y axis that you would first multiply a point by the anticlockwise rotation matrix and then multiply by the y axis matrix.
And that would therefore be M2.M1?

Posted from TSR Mobile

M1 is the matrix associated with a reflection in the y-axis.
M2 is the matrix associated with an anti-clockwise rotation of pi/2 radians.

You said it yourself: 'you first multiply a point by the anticlockwise rotation matrix and then multiply by the y-axis matrix'. Hence, you first multiply a point by matrix M2 and then multiply by M1.

Keep in mind that, when applying a transformation matrix to a point, the transformation matrix comes first in the multiplication i.e. M . X (M being the transformation matrix and X the position vector of the point) and NOT X . M.

Hence, we do M2 . X, in order to apply the rotation, and then we multiply M1 by the result of that.

So M1 . (M2 . X) gives us the desired point. In accordance with the properties of matrices, M1 . (M2 . X) = (M1 . M2) . X, and so M3 = M1 . M2.

For the odd and even functions question. I made (and stated) the assumption of f(x)=ax^2 +bx +c. This would make f(-x)=ax^2-bx+c. So f(x) +f(-x)=2ax^2+2c . Since all the powers of x are even, g(x) is an even function.
h(x)=f(x)-f(-x) = 2bx all the powers of x are odd so h(x) is an odd function.

Would I get marks for doing it this way?