AQA A Level Maths Paper 3 Unofficial Markscheme 2019
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I checked it on the calculator and it is the right value.
Original post by Aadz7
i got that too
Y=-7/30x +1.7
Just put into desmos and is correct
Just put into desmos and is correct
Original post by Aadz7
what did everyone get for the equation of the tangent?
Original post by flying299
I checked it on the calculator and it is the right value.
yeah same, I checked too
Omg I leave it as In(3/2)^4 +7/6 would this affect my marks?
Original post by Aadz7
i got that too
U will drop 1
Original post by Wjoeyxiao
Omg I leave it as In(3/2)^4 +7/6 would this affect my marks?
They wanted it in the form 81/16 so probably only lose 1 mark
Original post by Wjoeyxiao
Omg I leave it as In(3/2)^4 +7/6 would this affect my marks?
Original post by Edexcel Are mugs
Y=-7/30x +1.7
Just put into desmos and is correct
Just put into desmos and is correct
I agree with this.
cool i think i got that too
Original post by Edexcel Are mugs
Y=-7/30x +1.7
Just put into desmos and is correct
Just put into desmos and is correct
Original post by RavenclawOwl
What did people get for the question about Jamal the farmer?
There was a positive correlation so we reject H0 and accept H1
Original post by Edexcel Are mugs
U will drop 1
That’s sad😖😖
What did u do for the prove no stationary point one
Original post by Aadz7
cool i think i got that too
Yes
Original post by unagdi
partial fractions did anyone else get 4 and 7
dy/dx = 0, got x = -0.5y^2 or something like that, subbed back in then got an equation with no real roots
Original post by Edexcel Are mugs
What did u do for the prove no stationary point one
Original post by Edexcel Are mugs
I did contradiction but idk
As long as you’ve categorically proved what they’ve asked you to you’re fine. I also did contradiction and used Pythag with algebra to show that if all were odd a^2+b^2=c^2 came out to be even=odd. This is an impossibility and so there is a contradiction, therefore a b and c cannot all be odd.
Yes I did that, and got like y^6=-48 which is obviously impossible
Original post by Aadz7
dy/dx = 0, got x = -0.5y^2 or something like that, subbed back in then got an equation with no real roots
Original post by Aadz7
dy/dx = 0, got x = -0.5y^2 or something like that, subbed back in then got an equation with no real roots
I personally just showed that for there to be a stationary point, y=root(-2x), separated this out into root(-2) * root(x). As root(-2) is undefined, there cannot be a stationary point.
yeah but if x was negative, say -2 then y would be defined?
Original post by tbofig
I personally just showed that for there to be a stationary point, y=root(-2x), separated this out into root(-2) * root(x). As root(-2) is undefined, there cannot be a stationary point.
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