Please note this is not the official markscheme and could therefore contain mistakes. Use accordingly

1(a)6x+4

(b)(i)-2/3

(ii)x<-2/3

2(a)(i)u2=40

(ii)u3=28,u4=33

(b)(i)u5=35

(ii)851

3(a)3x²-13x-30=0

(b)(i)6, -3/5

(ii)reject -3/5 as log negative is invaid

4(a)A=55

(b)H=55(e^(-0.136x))+30

5(a)Show that k=12

(b)(0,-28)

6(a)-(2y+3x²)/(6y+2x)

(b)13x-11y+81=0

7(a)Show that AD//BC

(b)6.99

8(a)2√17 cos(x-1.326)

(b)9√17, 1.326

9(a)Show thow that y=3ln(x+5)

(b)3x-25y=-150ln5

10(a)(2k+3)/(x+4)+(k-3)/(x-2)

(b)(21/ln5)-6

11(a)1/9-2/27x+1/27x²

(b)0.03304

(c)6ln(17/16)-45/136

12(a)Show that dh/dt=λ/√h

(b)h^(3/2)=0.513t+1.728

(c)18.4

13(a)5000

(b)3 because |x-3|

(c){t:0≤t<3/5∪t>13/5}

14(a)sin2θ(23cos²θ-8cosθ-15)=0

(b)360°,540°,491°

15)Consider sinx>0 and cos x<0 for obtuse or any other valid arguments

1(a)6x+4

(b)(i)-2/3

(ii)x<-2/3

2(a)(i)u2=40

(ii)u3=28,u4=33

(b)(i)u5=35

(ii)851

3(a)3x²-13x-30=0

(b)(i)6, -3/5

(ii)reject -3/5 as log negative is invaid

4(a)A=55

(b)H=55(e^(-0.136x))+30

5(a)Show that k=12

(b)(0,-28)

6(a)-(2y+3x²)/(6y+2x)

(b)13x-11y+81=0

7(a)Show that AD//BC

(b)6.99

8(a)2√17 cos(x-1.326)

(b)9√17, 1.326

9(a)Show thow that y=3ln(x+5)

(b)3x-25y=-150ln5

10(a)(2k+3)/(x+4)+(k-3)/(x-2)

(b)(21/ln5)-6

11(a)1/9-2/27x+1/27x²

(b)0.03304

(c)6ln(17/16)-45/136

12(a)Show that dh/dt=λ/√h

(b)h^(3/2)=0.513t+1.728

(c)18.4

13(a)5000

(b)3 because |x-3|

(c){t:0≤t<3/5∪t>13/5}

14(a)sin2θ(23cos²θ-8cosθ-15)=0

(b)360°,540°,491°

15)Consider sinx>0 and cos x<0 for obtuse or any other valid arguments

(edited 8 months ago)

Scroll to see replies

Thank you. 2h was definitely not enough…literally panicking after seeing this

Original post by Anonymo us

1(a)6x+4

(b)-2/3

(c)x<-2/3

2(a)(i)u2=40

(ii)u3=28,u4=33

(b)(i)u5=35

(ii)851

3(a)3x²-13x-30=0

(b)(i)6, -3/5

(ii)reject -3/5 as log negative is invaid

4(a)A=55

(b)H=55(e^(-0.136x))+30

5(a)Show that k=12

(b)(0,-28)

6(a)-(2y+3x²)/(6y+2x)

(b)13x-11y+81=0

7(a)Show that AD//BC

(b)6.99 8(a)2√17 cos(x-1.326)

(b)9√17, 1.326

9(a)Show that y=6ln(x+5)

(b)3x-25y=-150ln5

10(a)(2k+3)/(x+4)+(k-3)/(x-2)

(b)(21/ln5)-6 11(a)1/9-2/27x+1/27x²

(b)0.03304

(c)6ln(17/16)-45/136

12(a)Show that dh/dt=λ/√h

(b)h^(3/2)=0.513t+1.728

(c)18.4

13(a)5000

(b)3 because |x-3|

(c){t:0≤t<3/5∪t>13/5}

14(a)sin2θ(23cos²θ-8cosθ-15)=0

(b)360°,540°,491°

15)Consider sinx>0 and cos x<0 for obtuse

(b)-2/3

(c)x<-2/3

2(a)(i)u2=40

(ii)u3=28,u4=33

(b)(i)u5=35

(ii)851

3(a)3x²-13x-30=0

(b)(i)6, -3/5

(ii)reject -3/5 as log negative is invaid

4(a)A=55

(b)H=55(e^(-0.136x))+30

5(a)Show that k=12

(b)(0,-28)

6(a)-(2y+3x²)/(6y+2x)

(b)13x-11y+81=0

7(a)Show that AD//BC

(b)6.99 8(a)2√17 cos(x-1.326)

(b)9√17, 1.326

9(a)Show that y=6ln(x+5)

(b)3x-25y=-150ln5

10(a)(2k+3)/(x+4)+(k-3)/(x-2)

(b)(21/ln5)-6 11(a)1/9-2/27x+1/27x²

(b)0.03304

(c)6ln(17/16)-45/136

12(a)Show that dh/dt=λ/√h

(b)h^(3/2)=0.513t+1.728

(c)18.4

13(a)5000

(b)3 because |x-3|

(c){t:0≤t<3/5∪t>13/5}

14(a)sin2θ(23cos²θ-8cosθ-15)=0

(b)360°,540°,491°

15)Consider sinx>0 and cos x<0 for obtuse

For 15, I got to sin2x>0 by expanding the bracket, could be wrong though, probs multiple methods.

Original post by AshRevise

For 15, I got to sin2x>0 by expanding the bracket, could be wrong though, probs multiple methods.

Yes i got this. If you solve for x, there are no solutions that are obtuse, hence the contradiction!

Original post by AshRevise

For 15, I got to sin2x>0 by expanding the bracket, could be wrong though, probs multiple methods.

Yea any valid method will do

same

Original post by AshRevise

For 15, I got to sin2x>0 by expanding the bracket, could be wrong though, probs multiple methods.

For 15, I got to sin2x>0 by expanding the bracket, could be wrong though, probs multiple methods.

Same. I think you could use that and then say for obtuse x u can't pos sin2x

Original post by LEBB2460

How many marks was subscriber question (Q13)

10

12c isn’t 18. You had to 5-1.44 then input that value of h. H wasn’t 5

Original post by emma7393

is there another value for 14b of 450° as it's the sin2x graph for [360,540] ?

In the q it said x≠ 450

Original post by Hiim

10

Original post by anonymous747

In the q it said x≠ 450

360 540 and 490 something

Original post by anonymous747

In the q it said x≠ 450

omg yeah i remember now thats so annoying 😭 thanks

Original post by emma7393

is there another value for 14b of 450° as it's the sin2x graph for [360,540] ?

Original post by LBJLBJ975310

12c isn’t 18. You had to 5-1.44 then input that value of h. H wasn’t 5

No you didn't since the model takes that into account with +c so it's 18 and H=5

Original post by navbh

360 540 and 490 something

Yh 490.7

By the way, the negative log is still valid because x^2 makes any answer positive. So, there are no invalid answers!

Original post by LBJLBJ975310

12c isn’t 18. You had to 5-1.44 then input that value of h. H wasn’t 5

h was 5 as it was a model for the total height not change in height. Otherwise h^3/2 would be 0 when t = 0

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