Soalan 14:
(a) Diberi bahawa g(x) = |1 – 3x| untuk domain -2 ≤ x ≤ 2.
(i) Tentukan sama ada g(x) ialah fungsi diskret atau fungsi selanjar.
Beri justifikasi anda.
(ii) Lakar graf g(x).
[3 markah]
(b) Diberi bahawa f : x → x + 3.
(i) Cari fn(x), dengan keadaan n ialah integer positif.
(ii) Seterusnya, cari h(x) dengan keadaan hf10(x) = (x – 30)(x + 30).
[5 markah]
Jawapan:
(a)(i) g(x) ialah fungsi selanjar kerana semua nilai x ditakrifkan bagi domain -2 ≤ x ≤ 2 dan satu garis lurus yang selanjar tanpa sebarang jurang diperoleh apabila dilukis.
(a)(ii) $$ \begin{aligned} g(2) & =|1-3(2)| \\ & =|-5| \\ & =5 \end{aligned} $$
$$ \begin{aligned} g(-2) & =|1-3(-2)| \\ & =7 \end{aligned} $$
$$ \begin{aligned} |1-3 x| & =0 \\ 1-3 x & =0 \\ x & =\frac{1}{3} \end{aligned} $$
(b)(i) $$ \begin{aligned} &f(x)=x+3\\ &\begin{aligned} f^2(x)=f f(x) & =f(x+3) \\ & =(x+3)+3 \\ & =x+6 \end{aligned}\\ &\begin{aligned} f^3(x)=f f^2(x) & =f(x+6) \\ & =(x+6)+3 \\ & =x+9 \end{aligned} \end{aligned} $$
$$ \begin{aligned} f^n(x)=f f^{n-1}(x) & =f[x+3(n-1)] \\ & =f(x+3 n-3) \\ & =(x+3 n-3)+3 \\ & =x+3 n \\ \therefore f^n(x)=x+3 n & \end{aligned} $$
(b)(ii) $$ \begin{aligned} h f^{10}(x) & =(x-30)(x+30) \\ h[x+3(10)] & =(x-30)(x+30) \end{aligned} $$
$$ \begin{aligned} &\text { Katakan, } \quad k=x+30\\ &k-30=x \end{aligned} $$
$$ \begin{aligned} & h(k)=(k-30-30)(k) \\ & h(k)=(k-60)(k) \\ & h(k)=k^2-60 k \\ & \therefore h(x)=x^2-60 x \end{aligned} $$
(a) Diberi bahawa g(x) = |1 – 3x| untuk domain -2 ≤ x ≤ 2.
(i) Tentukan sama ada g(x) ialah fungsi diskret atau fungsi selanjar.
Beri justifikasi anda.
(ii) Lakar graf g(x).
[3 markah]
(b) Diberi bahawa f : x → x + 3.
(i) Cari fn(x), dengan keadaan n ialah integer positif.
(ii) Seterusnya, cari h(x) dengan keadaan hf10(x) = (x – 30)(x + 30).
[5 markah]
Jawapan:
(a)(i) g(x) ialah fungsi selanjar kerana semua nilai x ditakrifkan bagi domain -2 ≤ x ≤ 2 dan satu garis lurus yang selanjar tanpa sebarang jurang diperoleh apabila dilukis.
(a)(ii) $$ \begin{aligned} g(2) & =|1-3(2)| \\ & =|-5| \\ & =5 \end{aligned} $$
$$ \begin{aligned} g(-2) & =|1-3(-2)| \\ & =7 \end{aligned} $$
$$ \begin{aligned} |1-3 x| & =0 \\ 1-3 x & =0 \\ x & =\frac{1}{3} \end{aligned} $$
(b)(i) $$ \begin{aligned} &f(x)=x+3\\ &\begin{aligned} f^2(x)=f f(x) & =f(x+3) \\ & =(x+3)+3 \\ & =x+6 \end{aligned}\\ &\begin{aligned} f^3(x)=f f^2(x) & =f(x+6) \\ & =(x+6)+3 \\ & =x+9 \end{aligned} \end{aligned} $$
$$ \begin{aligned} f^n(x)=f f^{n-1}(x) & =f[x+3(n-1)] \\ & =f(x+3 n-3) \\ & =(x+3 n-3)+3 \\ & =x+3 n \\ \therefore f^n(x)=x+3 n & \end{aligned} $$
(b)(ii) $$ \begin{aligned} h f^{10}(x) & =(x-30)(x+30) \\ h[x+3(10)] & =(x-30)(x+30) \end{aligned} $$
$$ \begin{aligned} &\text { Katakan, } \quad k=x+30\\ &k-30=x \end{aligned} $$
$$ \begin{aligned} & h(k)=(k-30-30)(k) \\ & h(k)=(k-60)(k) \\ & h(k)=k^2-60 k \\ & \therefore h(x)=x^2-60 x \end{aligned} $$
Soalan 15:
(a)(i) Rajah 15 menunjukkan segi tiga BCA.
$$ \text { Berdasarkan segi tiga } B C A \text {, terbitkan } \sin ^2 \theta+\mathrm{kos}^2 \theta=1 \text {. } $$
(ii) $$ \text { Seterusnya, buktikan bahawa } \frac{\sin ^2 \theta+\operatorname{kos} \theta+1}{2-\operatorname{kos} \theta}=\frac{\operatorname{sek} \theta+1}{\operatorname{sek} \theta} \text {. } $$
[5 markah]
(b) $$ \text { Selesaikan persamaan } \sin 2 \theta=-\mathrm{kos} \theta \text { untuk }-180^{\circ} \leqslant \theta \leqslant 540^{\circ} \text {. } $$
[3 markah]
Jawapan:
(a)(i) $$ \begin{aligned} & \sin ^2 \theta+\mathrm{kos}^2 \theta=\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2 \\ & \sin ^2 \theta+\mathrm{kos}^2 \theta=\frac{a^2}{c^2}+\frac{b^2}{c^2} \\ & \sin ^2 \theta+\mathrm{kos}^2 \theta=\frac{a^2+b^2}{c^2} \\ & \sin ^2 \theta+\mathrm{kos}^2 \theta=\frac{c^2}{c^2} \\ & \sin ^2 \theta+\mathrm{kos}^2 2 \theta=1 \end{aligned} $$
(a)(ii) $$ \begin{gathered} \frac{\sin ^2 \theta+\mathrm{kos} \theta+1}{2-\mathrm{kos} \theta}=\frac{\operatorname{sek} \theta+1}{\mathrm{sek} \theta} \\ \frac{1-\mathrm{kos}^2 \theta+\mathrm{kos} \theta+1}{2-\mathrm{kos} \theta}=\frac{\frac{1}{\mathrm{kos} \theta}+1}{\frac{1}{\mathrm{kos} \theta}} \\ \frac{-\mathrm{kos}^2 \theta+\mathrm{kos} \theta+2}{2-\mathrm{kos} \theta}=\frac{\frac{1}{\mathrm{kos} \theta}+\frac{\mathrm{kos} \theta}{\mathrm{kos} \theta}}{\frac{1}{\mathrm{kos} \theta}} \end{gathered} $$
$$ \begin{aligned} &\begin{gathered} \frac{-\mathrm{kos}^2 \theta+\mathrm{kos} \theta+2}{2-\mathrm{kos} \theta}=\frac{\frac{1}{\mathrm{kos} \theta}+\frac{\mathrm{kos} \theta}{\mathrm{kos} \theta}}{\frac{1}{\mathrm{kos} \theta}} \\ \frac{\mathrm{kos}^2 \theta-\mathrm{kos} \theta-2}{\mathrm{kos} \theta-2}=\frac{\frac{1+\mathrm{kos} \theta}{\mathrm{kos} \theta}}{\frac{1}{\mathrm{kos} \theta}} \\ \frac{(\mathrm{kos} \theta-2)(\mathrm{kos} \theta+1)}{\mathrm{kos} \theta-2}=\left(\frac{1+\mathrm{kos} \theta}{\mathrm{kos} \theta}\right)\left(\frac{\mathrm{kos} \theta}{1}\right) \\ \mathrm{kos} \theta+1=1+\mathrm{kos} \theta \end{gathered}\\ &\begin{aligned} & \text { Sebelah kiri }=\text { Sebelah kanan }=1+\operatorname{kos} \theta \end{aligned}\\ &\text { ∴ Terbukti } \end{aligned} $$
(b)
$$ \begin{aligned} \sin 2 \theta & =-\mathrm{kos} \theta \\ 2 \sin \theta \mathrm{kos} \theta+\mathrm{kos} \theta & =0 \\ \mathrm{kos} \theta(2 \sin \theta+1) & =0 \end{aligned} $$
$$ \begin{aligned} \operatorname{kos} \theta & =0 \\ \operatorname{kos}^{-1} (0) & =90^{\circ}(\text { Sudut asas } ) \\ \theta & =-90^{\circ}, 90^{\circ}, 270^{\circ}, 360^{\circ}+90^{\circ} \\ \theta & =-90^{\circ}, 90^{\circ}, 270^{\circ}, 450^{\circ} \end{aligned} $$
$$ \begin{aligned} 2 \sin \theta+1 & =0 \\ \sin \theta & =-\frac{1}{2}(\text { Sukuan III, IV }) \\ \sin ^{-1}\left(\frac{1}{2}\right) & =30^{\circ}(\text { Sudut asas } ) \\ \theta & =-30^{\circ},-180^{\circ}+30^{\circ}, 180^{\circ}+30^{\circ}, 360^{\circ}-30^{\circ} \\ \theta & =-30^{\circ},-150^{\circ}, 210^{\circ}, 330^{\circ} \end{aligned} $$
$$ \therefore \theta=-150^{\circ},-90^{\circ},-30^{\circ}, 90^{\circ}, 210^{\circ}, 270^{\circ}, 330^{\circ}, 450^{\circ} $$
(a)(i) Rajah 15 menunjukkan segi tiga BCA.
$$ \text { Berdasarkan segi tiga } B C A \text {, terbitkan } \sin ^2 \theta+\mathrm{kos}^2 \theta=1 \text {. } $$
(ii) $$ \text { Seterusnya, buktikan bahawa } \frac{\sin ^2 \theta+\operatorname{kos} \theta+1}{2-\operatorname{kos} \theta}=\frac{\operatorname{sek} \theta+1}{\operatorname{sek} \theta} \text {. } $$
[5 markah]
(b) $$ \text { Selesaikan persamaan } \sin 2 \theta=-\mathrm{kos} \theta \text { untuk }-180^{\circ} \leqslant \theta \leqslant 540^{\circ} \text {. } $$
[3 markah]
Jawapan:
(a)(i) $$ \begin{aligned} & \sin ^2 \theta+\mathrm{kos}^2 \theta=\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2 \\ & \sin ^2 \theta+\mathrm{kos}^2 \theta=\frac{a^2}{c^2}+\frac{b^2}{c^2} \\ & \sin ^2 \theta+\mathrm{kos}^2 \theta=\frac{a^2+b^2}{c^2} \\ & \sin ^2 \theta+\mathrm{kos}^2 \theta=\frac{c^2}{c^2} \\ & \sin ^2 \theta+\mathrm{kos}^2 2 \theta=1 \end{aligned} $$
(a)(ii) $$ \begin{gathered} \frac{\sin ^2 \theta+\mathrm{kos} \theta+1}{2-\mathrm{kos} \theta}=\frac{\operatorname{sek} \theta+1}{\mathrm{sek} \theta} \\ \frac{1-\mathrm{kos}^2 \theta+\mathrm{kos} \theta+1}{2-\mathrm{kos} \theta}=\frac{\frac{1}{\mathrm{kos} \theta}+1}{\frac{1}{\mathrm{kos} \theta}} \\ \frac{-\mathrm{kos}^2 \theta+\mathrm{kos} \theta+2}{2-\mathrm{kos} \theta}=\frac{\frac{1}{\mathrm{kos} \theta}+\frac{\mathrm{kos} \theta}{\mathrm{kos} \theta}}{\frac{1}{\mathrm{kos} \theta}} \end{gathered} $$
$$ \begin{aligned} &\begin{gathered} \frac{-\mathrm{kos}^2 \theta+\mathrm{kos} \theta+2}{2-\mathrm{kos} \theta}=\frac{\frac{1}{\mathrm{kos} \theta}+\frac{\mathrm{kos} \theta}{\mathrm{kos} \theta}}{\frac{1}{\mathrm{kos} \theta}} \\ \frac{\mathrm{kos}^2 \theta-\mathrm{kos} \theta-2}{\mathrm{kos} \theta-2}=\frac{\frac{1+\mathrm{kos} \theta}{\mathrm{kos} \theta}}{\frac{1}{\mathrm{kos} \theta}} \\ \frac{(\mathrm{kos} \theta-2)(\mathrm{kos} \theta+1)}{\mathrm{kos} \theta-2}=\left(\frac{1+\mathrm{kos} \theta}{\mathrm{kos} \theta}\right)\left(\frac{\mathrm{kos} \theta}{1}\right) \\ \mathrm{kos} \theta+1=1+\mathrm{kos} \theta \end{gathered}\\ &\begin{aligned} & \text { Sebelah kiri }=\text { Sebelah kanan }=1+\operatorname{kos} \theta \end{aligned}\\ &\text { ∴ Terbukti } \end{aligned} $$
(b)
$$ \begin{aligned} \sin 2 \theta & =-\mathrm{kos} \theta \\ 2 \sin \theta \mathrm{kos} \theta+\mathrm{kos} \theta & =0 \\ \mathrm{kos} \theta(2 \sin \theta+1) & =0 \end{aligned} $$
$$ \begin{aligned} \operatorname{kos} \theta & =0 \\ \operatorname{kos}^{-1} (0) & =90^{\circ}(\text { Sudut asas } ) \\ \theta & =-90^{\circ}, 90^{\circ}, 270^{\circ}, 360^{\circ}+90^{\circ} \\ \theta & =-90^{\circ}, 90^{\circ}, 270^{\circ}, 450^{\circ} \end{aligned} $$
$$ \begin{aligned} 2 \sin \theta+1 & =0 \\ \sin \theta & =-\frac{1}{2}(\text { Sukuan III, IV }) \\ \sin ^{-1}\left(\frac{1}{2}\right) & =30^{\circ}(\text { Sudut asas } ) \\ \theta & =-30^{\circ},-180^{\circ}+30^{\circ}, 180^{\circ}+30^{\circ}, 360^{\circ}-30^{\circ} \\ \theta & =-30^{\circ},-150^{\circ}, 210^{\circ}, 330^{\circ} \end{aligned} $$
$$ \therefore \theta=-150^{\circ},-90^{\circ},-30^{\circ}, 90^{\circ}, 210^{\circ}, 270^{\circ}, 330^{\circ}, 450^{\circ} $$