SULIT
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JABATAN PEPERIKSAAN
PANDAI EDUCATION
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SIJIL PELAJARAN MALAYSIA 2024
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MATEMATIK TAMBAHAN
Kertas 2 (Set 1)
3472/2
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 1. |
| (a) |
\(a=18\) cm, \(r=0.8\) |
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\(18(0.8)^9\) |
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\(2.4159\) cm |
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7m |
| (b) |
\(18(0.8)^{n-1}\lt 2\) |
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\((n-1)\lg{(0.8)}\lt \lg{\left(\dfrac{1}{9}\right)}\) |
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\(n=11\) |
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1m
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| (c) |
\(\dfrac{18}{1-0.8}\) |
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\(90\) cm |
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1m
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 2. |
| (a) |
(i) |
\(m_{UT}=\dfrac{2}{5}\), \(m_{UV}=-\dfrac{5}{2}\) |
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\(\dfrac{q-6}{p-5}=-\dfrac{5}{2}\) |
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\(p=\dfrac{37-2q}{5}\) |
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(ii) |
\(\begin{aligned} \triangle{TUV}&=\dfrac{1}{2}|(5q+4p)-(20+6p)|\\ &=p-\dfrac{5}{2}q+10 \end{aligned}\) |
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6m |
| (b) |
\(p-\dfrac{5}{2}q+10=29\) |
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\(\dfrac{37-2q}{5}-\dfrac{5}{2}q+10=29\) |
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\(V(9,-4)\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 3. |
| (a) |
(i) |
\((x-\alpha)(x-\beta)=0\) DAN |
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\(x^2-(\alpha+\beta)x+\alpha\beta=0\) ATAU \(x^2+\left(\dfrac{b}{a}\right)x+\dfrac{c}{a}=0\) |
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Guna \(\alpha +\beta =-\dfrac{b}{a}\) DAN \(\alpha\beta=\dfrac{c}{a}\) |
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(ii) |
\(SOR=\dfrac{15}{2}\), \(POR=-3\) |
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2m
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7m |
| (b) |
\(4x^2-6x-23=0\) |
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\(\dfrac{-(-6)\pm\sqrt{(-6)^2-4(4)(-23)}}{2(4)}\) |
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\(\dfrac{3\pm\sqrt{101}}{4}\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 4. |
| (a) |
\(\dfrac{dy}{dx}=3x^2-12\), \(m=15\) |
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\(y+9=15(x-3)\) |
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\(y=15x-54\) |
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7m |
| (b) |
\(P=y^2-12y+16\) ATAU setara |
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\(\dfrac{dP}{dy}=2y-12\) |
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\(P=(6)^2-12(6)+16\) |
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\(-20\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 5. |
| (a) |
(i) |
\(\log_2{\left(\dfrac{2\times3^2}{5}\right)}\) |
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\(1+2h-k\) |
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(ii) |
\(\log_2{(3^2\times2\times5)}\) |
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\(1+k+3h\) |
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7m |
| (b) |
\(\dfrac{8+\sqrt{24}}{\sqrt{2}+\sqrt{8}}\times\dfrac{\sqrt{2}-\sqrt{8}}{\sqrt{2}-\sqrt{8}}\) ATAU setara |
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\(\dfrac{8\sqrt{2}-8\sqrt{8}+\sqrt{48}-\sqrt{192}}{2-\sqrt{16}+\sqrt{16}-8}\) ATAU setara |
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\(\dfrac{4\sqrt{2}+2\sqrt{3}}{3}\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 6. |
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(a)
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\(\begin{aligned} \cos{2x}&=\cos^2{x}-\sin^2{x} \\ &=(1-\sin^2{x})-\sin^2{x} \\ &=1-2\sin^2{x} \end{aligned}\) ATAU \(\begin{aligned} &=(\sin^2{x}+\cos^2{x})-2\sin^2{x}\\ &=\cos^2{x}-\sin^2{x} \\ &=\cos{2x} \end{aligned}\) |
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6m |
| (b) |
(i) |
\(y=\cos{2x}\) |
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(ii) |
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Bentuk |
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Graf dipantulkan atas paksi-\(x\) |
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Bilangan penyelesaian: \(4\) |
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Skema Jawapan |
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| 7. |
| (a) |
\(\pi r^2h=81\pi\) |
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\(L=\pi r(2r)+2\pi r\left(\dfrac{81}{r^2}\right)+\pi r^2\) |
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\(L=3\pi \left(r^2+\dfrac{54}{r}\right)\) |
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10m |
| (a) |
Bezakan \(L\) terhadap \(r\), |
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\(\dfrac{dL}{dr}=3\pi \left(2r-\dfrac{54}{r^2}\right)\) |
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Samakan \(\dfrac{dL}{dr}\) dengan \(0\) & selesaikan untuk \(r\), |
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\(3\pi \left(2r-\dfrac{54}{r^2}\right)=0\) |
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\(81\pi\) |
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| (c) |
\(\dfrac{dr}{dt}=\dfrac{1}{3\pi\left(2(6)-\dfrac{54}{(6)^2}\right)}\times63\pi\) |
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\(2\) |
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| (d) |
\(\dfrac{dL}{dr}=(6.002-6)\times\dfrac{63}{2}\pi\) |
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\(0.063\pi\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 8. |
| (a) |
(i) |
\(0=27-(x-2)^3\) |
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\(-27=-(x-2)^3\) |
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\(3^3=(x-2)^3\) |
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\(x=5\) koordinat \((5,0)\) |
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(ii) |
\(\int_0^5 27-(x-2)^3\ dx\) |
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\(\left[ 27x-\dfrac{(x-2)^4}{4} \right]_0^5\) |
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\(\left[ 27(5)-\dfrac{((5)-2)^4}{4} \right]_0^5-0\) |
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\(115.25\) unit\(^2\) |
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| (b) |
\(\pi\int_k^4 (\sqrt{3x+4})^2 \ dx=\pi\int_k^43x+4\ dx=26\pi\) |
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\(\pi\left[ \dfrac{3x^2}{2}+4x\right]_k^4=26\pi\) |
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\(\left[\dfrac{3(4)^2}{2}+4(4)\right]-\left[\dfrac{3(k)^2}{2}+4(k)\right]=26\) |
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\(-3k^2-8k+28=0\) |
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\(k=2\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 9. |
| (a) |
\(\sin{x}\cos{\left(\dfrac{\pi}{6}\right)}+\cos{x}\sin{\left(\dfrac{\pi}{6}\right)}-\left( \sin{x}\cos{\left(\dfrac{\pi}{6}\right)}-\cos{x}\sin{\left(\dfrac{\pi}{6}\right)} \right)\) |
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\(2\cos{x}\sin{\left( \dfrac{\pi}{6}\right)}\) |
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\(2\cos{x}\left(\dfrac{1}{2}\right)\) |
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\(\cos{x}\) |
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1m
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| (b) |
| \(x\) |
\(0\) |
\(\dfrac{\pi}{4}\) |
\(\dfrac{\pi}{2}\) |
\(\dfrac{3\pi}{4}\) |
\(\pi\) |
| \(y\) |
\(5\) |
\(2\) |
\(-1\) |
\(2\) |
\(5\) |
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\(y=\dfrac{8x}{\pi}+1\) |
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- Bentuk kos
- Titik maksimum / minimum
- Kitaran atau pusingan / anjakan
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\(1\) penyelesaian |
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| (c) |
\(\cos{x}(2\sin{x}+1)=0\) |
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\(\cos{x}=0\), \(2\sin{x}+1=0\) |
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\(x=90^\circ\), \(x=270^\circ\) |
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\(x=180^\circ+30^\circ\), \(x=360^\circ-30^\circ\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 10. |
| (a) |
| \(x\) |
\(1\) |
\(2\) |
\(3\) |
\(4\) |
\(5\) |
\(6\) |
\(7\) |
| \(\log_{10}{y}\) |
\(1.75\) |
\(1.48\) |
\(1.40\) |
\(0.95\) |
\(0.80\) |
\(0.53\) |
\(0.25\) |
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Kesemua titik bagi \(\log_{10}{y}\) betul
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Satu titik betul |
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Semua titik betul |
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Garis penyuaian terbaik |
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| (b) |
(i) |
\(y_\text{salah}=25.1\) ; \(y_\text{betul}=17.78\) |
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(ii) |
\(\log_{10}{y}=-(\log_{10}n)x+\log_{10}m\) |
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\(-\log_{10}n=\) kecerunan |
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\(n=1.778\) |
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\(\log_{10}m=2\) |
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\(m=100\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 11. |
| (a) |
(i) |
\(n(0.2)=50\) |
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\(n=250\) |
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(ii) |
\({}^{10}C_4(0.2)^4(0.8)^6\) |
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\(=0.08808\) |
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| (b) |
(i) |
\(\mu=3.1\), \(\sigma^2=0.09\), \(\sigma=0.3\) |
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\(\begin{aligned} P(2.9\lt Z\lt 3.1)&=P\left(\dfrac{2.9-3.1}{0.3}\lt Z\lt \dfrac{3.3-.31}{0.3}\right) \\ &=P\left(-\dfrac{2}{3}\lt Z\lt \dfrac{2}{3}\right)\\ &=0.4950 \end{aligned}\) |
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(ii) |
\(P(X\lt m)=0.25\) |
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\(P\left(Z\lt \dfrac{m-3.1}{0.3}\right)=0.25\) |
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\(\dfrac{m-3.1}{0.3}=-0.674\) |
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\(m=3.1-0.674(0.3)\) |
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\(m=2.898\) kg |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 12. |
| (a) |
\(q=\dfrac{100}{80}\times 100\) |
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\(q=125\) |
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| (b) |
(i) |
\(\left(\dfrac{110}{100}\times125\right)\) ATAU \(\left(\dfrac{110}{100}\times140\right)\) |
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\(137.5\) DAN \(154\) |
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(ii) |
\(\dfrac{137.5(2p)+121(6)+112(3)+154(p)}{2p+6+3+p}=128\) |
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\(p=2\) |
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| (c) |
(i) |
Komponen Membeli-belah kerana mempunyai nilai pemberat paling besar. |
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(ii) |
\(\dfrac{151.25}{121}\times100=125\%\) |
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Peratus peningkatan \(=25\%\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 13. |
| (a) |
Titik \(C\) |
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\(AC\) adalah garis terpanjang kerana sudut bertentangannya adalah sudut terbesar bagi kedua-dua segi tiga \(ABC\) dan \(ADC\). |
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(i) |
\(\dfrac{AC}{\sin{93.20^\circ}}=\dfrac{64}{\sin{(180^\circ-93.20^\circ-58^\circ)}}\) |
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\((^*AC)^2=52.4^2+109^2-2(52.4)(109)\cos{\angle{ADC}}\) |
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\(105.05^\circ\) ATAU setara |
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(ii) |
\(\dfrac{1}{2}(52.4)(109)\sin(^*105.05^\circ)\) |
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ATAU |
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\(\dfrac{1}{2}(64)(^*132.64)\sin(58^\circ)\) |
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\(\dfrac{1}{2}(52.4)(109)\sin(^*105.05^\circ)+\dfrac{1}{2}(64)(^*132.64)\sin(58^\circ)\) |
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\(6357.37\) |
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| (c) |
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(ii) |
\(\dfrac{1}{2}\times AE\times 109=^*2757.84\) |
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\(50.60\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 14. |
| (a) |
\(\text{I}:45x+30y\le 330\) |
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\(\text{II}:50x+70y\ge 350\) |
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\(\text{III}:y\ge\dfrac{5}{4}x\) |
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| (b) |
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Lukis sekurang-kurangnya satu garis lurus dengan betul |
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Lukis semua garis lurus dengan betul |
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Lorek rantau dengan betul |
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| (c) |
(i) |
\(4\le y\le 6\) |
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(ii) |
\((4,5)\) |
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\(16(4)+10(5)\) |
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RM \(114\) |
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Skema Jawapan |
Sub Markah |
Jumlah Markah |
| 15. |
| (a) |
(i) |
Ganti dan samakan dengan \(4\), |
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\(2(6)-p=4\) |
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\(p=8\) |
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(ii) |
\((t-3)(t-5)=0\) |
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\(t=3\), \(t=5\) |
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| (b) |
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Bentuk |
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Titik persilangan \((0,15)\), \((3,0)\) dan \((5,0)\) |
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Kamir dengan had yang betul, |
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\(\int_0^3t^2-8t+15\ dt\) ATAU \(\int_3^5t^2-8t+15\ dt\) |
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\(\left[ \dfrac{t^3}{3}-\dfrac{8t^2}{2}+15t \right]_0^3\) ATAU \(\left[ \dfrac{t^3}{3}-\dfrac{8t^2}{2}+15t \right]_3^5\) |
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Gunakan nilai had yang betul, |
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\(\left[ \dfrac{(3)^3}{3}-\dfrac{8(3)^2}{2}+15(3) \right]-\left[ \dfrac{(0)^3}{3}-\dfrac{8(0)^2}{2}+15(0) \right]\) |
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ATAU |
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\(\left[ \dfrac{(5)^3}{3}-\dfrac{8(5)^2}{2}+15(5) \right]-\left[ \dfrac{(3)^3}{3}-\dfrac{8(3)^2}{2}+15(3) \right]\) |
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\(18+\left| -\dfrac{4}{3} \right|\) |
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\(\dfrac{58}{3}\) |
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