A Level Mathematics 9709 Paper 13, May/June 2025

2 + (12)(-2)x^{-3}

2 + (12)(-2)(-2)^{-3} [= 5]

Either (their $5)=\frac{y-(-1)}{x-(-2)}$ or -1 = (their $5)\times (-2)+c\Rightarrow c=\dots$

y = 5x + 9

$r=\frac{8{\sin }^3\theta }{4{\sin }^2\theta }$ $[=2\sin \theta ]$

$\frac{4{\sin }^2\theta }{1-2\sin \theta }=\frac{1}{2}$

Expect $8{\sin }^2\theta +2\sin \theta -1$ [= 0]

$[\theta =]$ ${\sin }^{-1}$ $\frac{1}{4}$, 0.25

$(\frac{1}{-1\times 4})a(4x-3{)}^{-1}+2x$

Apply correct limits, x = 3 & 1, to their integral

$\frac{-a}{36}+6-(\frac{-a}{4}+2)=12\Rightarrow [\frac{8a}{36}+4=12]$

a = 36

32[x^0]

-120x

+180x^2

x = 0.01 or 1/100

(Their 32 - 120x + 180x^2) with x = (their 0.01)

30.8[18]

Replace $\tan \theta$ with $\frac{\sin \theta }{\cos \theta }$

Replace ${\sin }^2\theta$ with $1-{\cos }^2\theta$ leading to a 3-term quadratic

$9{\cos }^2\theta +\cos \theta -4$ [= 0]

Attempt to solve (their quadratic in $\cos \theta )$ using a valid method

Any two of $\pm 52.2$, $\pm 136.4$

All four values

a + 2(N - 1) = 55

$\frac{1}{2}\times 3N(2a+2(3N-1))=5760$

Eliminate either a or N

N^2 + 56N - 1920 or a^2 - 226a + 1953 [= 0]

N = 24

a = 9

3x^2 + 10x - 8 < 0 or 3(x + 5/3)^2 - 49/3 < 0

-4 and 2/3

-4 < x < 2/3

Identify x = -4 as the x-value of the maximum point.

[y = ] 3/3 x^3 + 10/2 x^2 - 8x [+ c]

Use of y = 27, x = their(-4 or 2/3) in their integral.

c = -21 => y = x^3 + 5x^2 - 8x - 21

$\frac{1}{2}\pi -2({\tan }^{-1}\frac{4}{12}$ $\text{ or }$ ${\sin }^{-1}\frac{4}{\sqrt{160}}$ $\text{ or }$ ${\cos }^{-1}\frac{12}{\sqrt{160}})$ Or $2{\sin }^{-1}[\frac{\frac{1}{2}EF}{AE}=\frac{\sqrt{32}}{\sqrt{160}}=\frac{\sqrt{2}}{\sqrt{10}}]$ or other valid method.

$[E\widehat{A}F=0.927295\cdots =]$ 0.9273

[AE or AF =] $\sqrt{160}$ or $\frac{4}{\sin E\widehat{A}B}$

$[r\theta =]$ (their $12.65)\times 0.9273$

[11.729... + 8 + 8 =] 27.7 AWRT

$\frac{1}{2}$(their $12.65{)}^2\times 0.9273$ [= 74.184]

144 - 24 - 24 - {$\frac{1}{2}$(their $12.65{)}^2\times 0.9273$}

[Area =] 21.8

Gradient of PR $=\frac{k-5}{2+13}$ or gradient of RQ $=\frac{k-1}{2-5}$

(their gradient of PR)$ \times $(their gradient of RQ) = -1 and attempt to simplify

k^2 - 6k - 40 [= 0]

k = 10, k = -4

[Centre is] (-4, 3)

Radius gradient $=\frac{(their3)-10}{(their-4)-2}$ $[=\frac{7}{6}]$

Either $\frac{-1}{their-\frac{7}{6}}=\frac{y-10}{x-2}$ Or $10=\frac{-1}{their-\frac{7}{6}}\times 2+c\Rightarrow c=\dots$

$y-10=-\frac{6}{7}(x-2)$ or $y=(-\frac{6}{7})x+\frac{82}{7}$

6x + 7y - 82 = 0 or 82 - 6x - 7y = 0

$-9\times 2(2x-5{)}^{-2}+2$

(their -18(2x - 5)^{-2} + 2) = 0 and rearrange to form a quadratic. [(2x - 5)^2 = 9 or 8x^2 - 40x + 32 = 0]

(1, -6)

(4, 6)

$-18\times -2\times 2(2x-5{)}^{-3}$ [= 72(2x - 5)^{-3} or $\frac{144x-360}{(2x-5{)}^4}]$

Use (their x = 1 and x = 4) in (their 72(2x - 5)^{-3}) To determine the nature of both turning points.

For x = 1, [d^2y/dx^2 =] -72/27 or < 0 => maximum; For x = 4, [d^2y/dx^2 =] 72/27 or > 0 => minimum

(1, -13)

[y = ] $\frac{9}{2(x\pm 3)-5}+2(x\pm 3)-5\pm 7$

$[y=]-(\frac{9}{2(x\pm 3)-5}+2(x\pm 3)-5\pm 7)$

$y=-\frac{9}{2x+1}-2x-8$

[f(x) = ] (x + 2a)^2 - 4a^2 + a

(their(-4a^2 + a)) = -33

4a^2 - a - 33 [= 0]

-11/4, 3

[g(x) = ] 1/2(x^3 + 4)

[g(0) = ] $1\mathrm{/}2((0{)}^3\pm 4)$ [g(0) = 2]

gg(0) = 6

f(their 6) [= 36 + 24a + a]

36 + 24a + a = 96

[a = ] 12/5