Mock set 4 paper 2 q14 a level maths (4 distinct points)

leoishush

15

I'm stuck on the last two lines of the mark scheme. where did the "6-" bit come from?

(edited 10 months ago)

Reply 1

mqb2766

21

Its (double) the minimum (left) root of the quadratic x^2-6x-k=0. If its >0, then so is the other one.

(edited 10 months ago)

Reply 2

leoishush

OP

15

Original post by mqb2766

Its (double) the minimum (left) root of the quadratic x^2-6x-k=0. If its >0, then so is the other one.

sorry i still dont get it

Reply 3

mqb2766

21

Original post by leoishush

sorry i still dont get it

The roots of the previous quadratic are
6 +/- sqrt(36+4k) / 2
You want them both to be positive. As
6 - sqrt(36+4k) / 2 < 6 + sqrt(36+4k) / 2
You simply need to show (the values of k which satisfiy)
0 < 6 - sqrt(36+4k)

(edited 10 months ago)

Reply 4

leoishush

OP

15

Original post by mqb2766

The roots of the previous quadratic are
6 +/- sqrt(36+4k) / 2
You want them both to be positive. As
6 - sqrt(36+4k) / 2 < 6 + sqrt(36+4k) / 2
You simply need to show (the values of k which satisfiy)
0 < 6 - sqrt(36+4k)

ahhhhhhhhhhhh i get it now, thank you very much. Also for this question: normally when you do y-b = gradient(x-a) you sub in the y coordinate in b, but in line 2 theyve subbed in 0 in y instead of the - e^2x/x^3 +4 Screenshot 2023-06-12 7.09.10 PM.png

(edited 10 months ago)

Screenshot 2023-06-12 7.39.00 PM.png59.3KB

Screenshot 2023-06-12 7.40.09 PM.png24.3KB

Reply 5

mqb2766

21

On line 2 they sub (p,y(p)) into the equation you mention, so a=p, b=y(p)=....
On line 3, they then sub (0,0) into the line 2 equation as it says it passes through the origin, so x=0, y=0.

(edited 10 months ago)

Reply 6

leoishush

OP

15

Original post by mqb2766

On line 2 they sub (p,y(p)) into the equation you mention, so a=p, b=y(p)=....
On line 3, they then sub (0,0) into the line 2 equation as it says it passes through the origin, so x=0, y=0.