SULIT
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JABATAN PEPERIKSAAN
PANDAI EDUCATION
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SIJIL PELAJARAN MALAYSIA 2024
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MATEMATIK TAMBAHAN
Kertas 1 (Set 1)
3472/1
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 1. |
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(a)
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\(\begin{aligned} ck+d&=0 \\ ck&=-d \\ k&=-\dfrac{d}{c} \end{aligned}\)
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Katakan \(g^{-1}(x)=y\)
\(\begin{aligned} g(y)&=x\\ \dfrac{ay+b}{cy+d}&=x \\ ay+b&=xcy+xd \\ y&=\dfrac{dx-b}{-cx+a} \end{aligned}\)
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\(g^{-1}(x)=\dfrac{-dx+b}{cx-a}\), \(x\ne \dfrac{a}{c}\)
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1m
1m
1m
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6m |
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(b)
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(i)
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\(g^{-1}(x)=\dfrac{3x-4}{2x-7}\), \(x\ne \dfrac{7}{2}\)
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(ii)
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\(g^{-1}(x)=\dfrac{5x-3}{x-2}\), \(x\ne 2\)
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1m
1m
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(c)
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\(g(x)=g^{-1}(x)\) jika \(a=-d\).
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1m
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Skema Jawapan |
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Jumlah Markah |
| 2. |
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\(\begin{aligned} 5&\lt 2x^2+x+4 \\ 2x^2+x-1&\gt 0 \\ (2x-1)(x+1)&\gt 0 \\ \end{aligned}\)
\(x\lt -1\), \(x\gt\dfrac{1}{2}\)
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\(\begin{aligned} 2x^2+x-6&\lt 0\\ (2x-3)(x+2)&\lt0 \end{aligned}\)
\(-2\lt x\lt \dfrac{3}{2}\)
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\(x\lt -1\), \(x\gt \dfrac{1}{2}\) ATAU \(-2\lt x\lt \dfrac{3}{2}\)
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\(-2\lt x \lt -1\) ATAU \(\dfrac{1}{2}\lt x \lt \dfrac{3}{2}\)
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1m
1m
1m
1m
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4m |
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Skema Jawapan |
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Jumlah Markah |
| 3. |
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(a)
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Bukan, kerana terdapat persamaan yang mempunyai hasil darab dua pemboleh ubah.
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1m
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5m |
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(b)
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\(\begin{aligned} x-2y&=4 \quad \cdots\boxed{1} \\ 2x-3y+2z&=-2 \quad \cdots\boxed{2} \\ 4x-7y+2z&=6 \quad\cdots\boxed{3} \end{aligned}\)
Dari \(\boxed{1}\),
\(x=4+2y\quad\cdots\boxed{4}\)
Gantikan \(\boxed{4}\) ke dalam \(\boxed{2}\),
\(\begin{aligned} 2(4+2y)-3y+2z&=-2\\ 8+4y-3y+2z&=-2\\ 8+y+2z&=-2\\ y+2z&=-10\quad\cdots\boxed{5} \end{aligned}\)
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Gantikan \(\boxed{4}\) ke dalam \(\boxed{3}\),
\(\begin{aligned} 4(4+2y)-7y+2z&=6\\ 16+8y-7y+2z&=6\\ 16+y+2z&=6\\ y+2z&=-10\quad\cdots\boxed{6} \end{aligned}\)
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\(\boxed{5}-\boxed{6}\),
\(0=0\)
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Maka, sistem persamaan linear ini mempunyai penyelesaian yang tidak terhingga kerana \(0=0\).
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1m
1m
1m
1m
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 4. |
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(a)
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\(\begin{aligned} &3^{-n}(15^n-27^n)\\ &=3^{-n}(3^n5^n-3^n9^n)\\ &=3^{-n}\cdot3^n(5^n-9^n)\\ &=5^n-9^n \end{aligned}\)
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1m
1m
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7m |
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(b)
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\(\begin{aligned} \sqrt{4x}-\sqrt{x}&=\dfrac{3}{\sqrt{a}-\sqrt{b}} \\ 2\sqrt{x}-\sqrt{x}&=\dfrac{3}{\sqrt{a}-\sqrt{b}}\times\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}\\ \sqrt{x}&=\dfrac{3(\sqrt{a}+\sqrt{b})}{a-b}\\ x&=\dfrac{9(\sqrt{a}+\sqrt{b})^2}{(a-b)^2}\\ h&=\dfrac{9}{(a-b)^2} \end{aligned}\)
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1m
1m
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(c)
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\(\begin{aligned} 5e^{n-3}&=e^{-2}\\ \dfrac{e^{n-3}}{e^{-2}}&=\dfrac{1}{5}\\ e^{n-1}&=\dfrac{1}{5}\\ \ln{e^{n-1}}&=\ln{\left(\dfrac{1}{5}\right)}\\ n-1&=\ln{\left(\dfrac{1}{5}\right)} \\ n&=\ln{\left(\dfrac{1}{5}\right)}+1 \\ m&=\dfrac{1}{5} \end{aligned}\)
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1m
1m
1m
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 5. |
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(a)
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Pak Abu, \(A_n:100,104,108,_\cdots\)
Pak Kasim, \(K_n:400,402,404,_\cdots\)
\(A_n=100+(n-1)(4)\\ K_n=400+(n-1)(2)\)
\(100+(n-1)(4)\gt 400+(n-1)(2)\)
\(\begin{aligned} 2n&\gt302\\ n&\gt151\\ n&=152 \end{aligned}\)
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1m
1m
1m
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6m |
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(b)
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Jarak ayunan lengkap : \(400,340,289,_\cdots\)
\(a=400\), \(r=0.85\)
\(\begin{aligned} S_n&\lt 2434\\ \dfrac{400[1-0.85^n]}{1-0.85}&\lt2434\\ 0.85^n&\gt0.08725\\ \lg{0.85^n}&\gt\lg{0.08725}\\ n\lg{0.85}&\gt\lg{0.08725}\\ n&\lt\dfrac{\lg{0.08725}}{\lg{0.85}}\\ n&\lt15.007\\ n&=15 \end{aligned}\)
\(\therefore\) Bilangan ayunan lengkap ialah \(15\) ayunan.
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2m
1m
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 6. |
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\(\dfrac{y}{x^3}=m-\dfrac{n}{x}\) ATAU \(\dfrac{y}{x^2}=mx-n\)
\(\begin{aligned} h&=m-3 \\ n&=6h \\ n&=6m-18 \end{aligned}\)
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1m
1m
1m
1m
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4m |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 7. |
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\(\dfrac{1}{2}(2x-10)(5x-20)\sin{30^\circ}=1700\)
\(x=\dfrac{-5\pm\sqrt{5^2-4(5)(-3500)}}{2(5)}\)
\(x=26\)
\(62\) m
\(110\) m
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1m
1m
1m
1m
1m
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5m |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 8. |
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\(16a+12=64a\)
\(a=\dfrac{1}{4}\) DAN \(p=4\)
\(\pi\int_4^{16}4y\ dy=\pi[2y^2]_4^{16}\)
\(\pi[2(16)^2-2(4)^2]\)
\(480\pi\)
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1m
1m
1m
1m
1m
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5m |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 9. |
| (a) |
\(|-\overrightarrow{OQ}|=\sqrt{3^2+(-4)^2}\) |
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\(5\) unit |
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1m
1m
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4m |
| (b) |
\(\overrightarrow{RQ}=\overrightarrow{RO}+\overrightarrow{OQ}\) |
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\(-r+q\) |
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1m
1m
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 10. |
| (a) |
\(\angle{BOC}=\pi-2a\) |
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\(10\pi-20a\) |
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1m
1m
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6m |
| (b) |
\(\theta=1\) rad |
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\(\dfrac{1}{2}\times10^2\times2.142\) |
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\(\dfrac{1}{2}\times10^2\times2.142-\dfrac{1}{2}\times10\times10\times\sin{122.71^\circ}\) |
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\(65.03\) |
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1m
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 11. |
| (a) |
\(\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}\) |
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\(\overrightarrow{PR}=4\utilde{u}+18\utilde{v}\) |
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1m
1m
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7m |
| (b) |
(i) |
\(\overrightarrow{TX}=4m\utilde{u}\) |
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(ii) |
\(\overrightarrow{PX}=\overrightarrow{PT}+\overrightarrow{TX}\) ATAU \(\overrightarrow{PX}=\lambda\overrightarrow{PR}\) ATAU \(\overrightarrow{PR}=\lambda\overrightarrow{PX}\) |
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\(4m\utilde{u}+9\utilde{v}=\lambda(4\utilde{u}+18\utilde{v})\) ATAU \(4\utilde{u}+18\utilde{v}=\lambda(4m\utilde{u}+9\utilde{v})\) |
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\(4m=4\lambda\) ATAU \(18\lambda=9\) |
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\(200\) meter |
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1m
1m
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1m
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 12. |
| (a) |
\(P(\text{Merah})=\dfrac{9}{4}\) DAN \(P(\text{Biru})=\dfrac{5}{14}\) |
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\({}^8C_0\left( \dfrac{5}{14}\right)^0\left(\dfrac{9}{14}\right)^8\) ATAU \({}^8C_1\left(\dfrac{5}{14}\right)^1\left(\dfrac{9}{14}\right)^7\) ATAU \({}^8C_2\left(\dfrac{5}{14}\right)^2\left(\dfrac{9}{14}\right)^6\) ATAU \({}^8C_3\left(\dfrac{5}{14}\right)^3\left(\dfrac{9}{14}\right)^5\) ATAU \({}^8C_4\left(\dfrac{5}{14}\right)^4\left(\dfrac{9}{14}\right)^4\) |
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\(0.1945\) |
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1m
1m
1m
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5m |
| (b) |
\(1-{}^8C_0\left( \dfrac{5}{14}\right)^0\left(\dfrac{9}{14}\right)^8-{}^8C_1\left(\dfrac{5}{14}\right)^1\left(\dfrac{9}{14}\right)^7-{}^8C_2\left(\dfrac{5}{14}\right)^2\left(\dfrac{9}{14}\right)^6-{}^8C_3\left(\dfrac{5}{14}\right)^3\left(\dfrac{9}{14}\right)^5\) |
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\(0.3090\) |
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1m
1
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 13. |
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1m
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8m |
| (b) |
\(|2x-3|-1=0\) |
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\(x=1\), \(x=2\) |
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1m
2m
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| (c) |
\(2x-3\le2\) DAN \(2x-3\ge-2\) (atau setara) |
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\(\dfrac{1}{2}\le x\le \dfrac{5}{2}\) |
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1m
1m
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| (d) |
\(|2x-3|-1=x\) |
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\(x=4\) DAN \(x=\dfrac{2}{3}\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 14. |
| (a) |
\(\dfrac{r}{\sin{\theta}}\) |
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\(R=r\left(1+\dfrac{1}{\sin{\theta}}\right)\) |
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1m
1m
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8m |
| (b) |
Luas sektor \(=\dfrac{1}{2}(3r)^2\left(\dfrac{\pi}{3}\right)=\dfrac{3}{2}\pi r^2\) |
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\(\dfrac{\pi r^2}{\dfrac{3}{2}\pi r^2}\) |
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\(\dfrac{2}{3}\) |
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1m
1m
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| (c) |
Panjang lengkok \(=5\left(\dfrac{2\pi}{3}\right)\) @ \(\dfrac{5}{\tan{30^\circ}}\) |
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\(\dfrac{10\pi}{3}+2(5\sqrt{3})\) |
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\(27.79\) m |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 15. |
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\(0.1x+0.4y+0.6z=225\)
\(x+y+z=500\)
\(x=2y\)
Hapuskan pemboleh ubah \(z\),
\(5x+2y=750\)
Hapuskan pemboleh ubah \(x\) atau \(y\),
\(5(2y)+2y=750\) ATAU \(5x+2\left(\dfrac{x}{2}\right)=750\)
\(x=125\)
\(y=62.5\)
\(z=312.5\)
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8m |
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