PAS MATEMATIKA BLOG: Agustus 2020

Contoh Soal Lanjutan Nilai Mutlak

$\begin{array}{ll}\\ 6.&\textrm{Perhatikanlah gambar berikut}\\ \end{array}$

$\begin{array}{ll}\\ &\textrm{Persamaan yang sesuai dengan grafik di atas adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=\left | x-3 \right |\\ \textrm{b}.&y=-\left | x-3 \right |\\ \color{red}\textrm{c}.&y=\left | x+3 \right |\\ \textrm{d}.&y=\left | x+6 \right |\\ \textrm{e}.&y=-\left | x+6 \right |\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ \end{array}$

$\begin{array}{l}\\ 7.&\textrm{Diketahui suatu fungsi}\: \: y=\left | 2x \right |\\ &\textrm{Nilai saat}\: \: -2\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&-2\\ \textrm{c}.&0\\ \textrm{d}.&2\\ \color{red}\textrm{e}.&4\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}y&=\left | 2x \right |\\ \textrm{saat}&\: \: x=-2,\: \: \textrm{maka}\\ y&=\left | 2(-2) \right |\\ &=\left | -4 \right |\\ &=-(-4)\\ &=4 \end{aligned} \end{array}$

$\begin{array}{l}\\ 8.&\textrm{Diketahui suatu fungsi}\: \: f(x)=\left | x-2 \right |\\ &\textrm{Nilai}\: \: f(-2)\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&2\\ \color{red}\textrm{c}.&4\\ \textrm{d}.&6\\ \textrm{e}.&8\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}f(x)&=\left | x-2 \right |\\ \textrm{Nilai}&\: \: f(-2)\: \: \textrm{berarti nilai}\\ \textrm{saat}&\: \: x=-2,\: \: \textrm{maka}\\ f(x)&=\left | x-2 \right |\\ f(-2)&=\left | -2-2 \right |\\ &=\left | -4 \right |\\ &=-(-4)\\ &=4 \end{aligned} \end{array}$

$\begin{array}{l}\\ 9.&\textrm{Diketahui fungsi}\: \: f(x)=\left | 2-4x\right |\\ &\textrm{Nilai dari}\: \: f(1)\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \textrm{b}.&0\\ \textrm{c}.&1\\ \color{red}\textrm{d}.&2\\ \textrm{e}.&4\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}f(x)&=\left | 2-4x \right |\\ \textrm{Nilai}&\: \: f(1)\: \: \textrm{berarti nilai}\\ \textrm{saat}&\: \: x=1,\: \: \textrm{maka}\\ f(x)&=\left | 2-4x) \right |\\ f(1)&=\left | 2-4(1) \right |\\ &=\left | -2 \right |\\ &=-(-2)\\ &=2 \end{aligned} \end{array}$

$\begin{array}{l}\\ 10.&\textrm{Nilai}\: \: k\: \: \textrm{yang memenuhi}\: \: \left | 4k \right |=16\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&\pm 2\\ \color{red}\textrm{d}.&\pm 4\\ \textrm{e}.&\pm 8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left | 4k \right |&=16\\ (4k)&=\pm 16\\ k&=\pm \displaystyle \frac{16}{4}\\ &=\pm 4 \end{aligned}\end{array}$

Contoh Soal Nilai Mutlak (Kelas X Matematika Wajib)

$\begin{array}{ll}\\ 1.&\textrm{Penyelesaian}\: \: -5\left |x-7 \right |+2=-13\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x=-10\: \: \textrm{dan}\: \: x=-4\\ \textrm{b}.&x=-10\: \: \textrm{dan}\: \: x=-2\\ \textrm{c}.&x=-2\: \: \textrm{dan}\: \: x=4\\ \textrm{d}.&x=4\: \: \textrm{dan}\: \: x=-10\\ \color{red}\textrm{e}.&x=4\: \: \textrm{dan}\: \: x=10\end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}-5\left |x-7 \right |&=-13-2\\ \left |x-7 \right |&=\displaystyle \frac{-15}{-5}\\ \left |x-7 \right |&=3\\ x-7&=\pm 3\\ x&=\mp 3+7\\ x&=\begin{cases} +3 & \text{ + } 7=10 \\\\ -3 & \text{ + } 7=4 \end{cases} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui persamaan}\: \: \left |x-5 \right |+\left | x+9 \right |=22\\ &\textrm{dan diberikan pernyataan-pernyataan berikut}\\ &\textrm{untuk}\\ &(\textrm{i})\quad x<-9,\: \textrm{maka}\: \: -x+5-x+9=22\\ &(\textrm{ii})\: \: \: -9\leq x< 5,\: \textrm{maka}\: \: -x+5-x+9=22\\ &(\textrm{iii})\: \: x<-9,\: \textrm{maka}\: \: -x+5-x-9=22\\ &(\textrm{iv})\: \: \: x\geq 5,\: \textrm{maka}\: \: x-5+x+9=22\\ &\textrm{Pernyataan yang tepa ditunjukkan oleh}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{ii})\\ \textrm{b}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \color{red}\textrm{c}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{iv})\\ \textrm{d}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{e}.&(\textrm{iii})\: \: \textrm{dan}\: \: (\textrm{iv}) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned} \end{aligned}\end{array}$

$\begin{array}{l}\\ 3.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &3\left |x-1 \right |-2\left | x+1 \right |=0\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x=-\displaystyle \frac{1}{5}\: \: \textrm{dan}\: \: x=5\\ \textrm{b}.&x=-5\: \: \textrm{dan}\: \: x=-\displaystyle \frac{1}{5}\\ \color{red}\textrm{c}.&\displaystyle x=\frac{1}{5}\: \: \textrm{dan}\: \: x=5\\\\ \textrm{d}.&x=-5\: \: \textrm{dan}\: \: x=\displaystyle \frac{1}{5}\\\\ \textrm{e}.&x=-5\: \: \textrm{dan}\: \: x=5\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}3\left |x-1 \right |&-2\left | x+1 \right |=0 \\ 3\left |x-1 \right |&=2\left | x+1 \right |\\ 3^{2}(x-1)^{2}&=2^{2}(x+1)^{2}\\ 9\left ( x^{2}-2x+1 \right )&=4\left ( x^{2}+2x+1 \right )\\ 9x^{2}-4x^{2}&-18x-8x+9-4=0\\ 5x^{2}-26x&+5=0\\ (5x-1)&(x-5)=0\\ x=\displaystyle \frac{1}{5}\quad &\textrm{atau}\quad x=5 \end{aligned} \end{array}$

$\begin{array}{l}\\ 4.&\textrm{Himpunan penyelesaian dari}\: \: \left | 5-\displaystyle \frac{2}{3}x \right |-9=8\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ -33,-18 \right \}\\ \color{red}\textrm{b}.&\left \{ -18,33 \right \}\\ \textrm{c}.&\left \{ 33,8 \right \}\\ \textrm{d}.&\left \{ 8,-33 \right \}\\ \textrm{e}.&\left \{ -8,33 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\left | 5-\displaystyle \frac{2}{3}x \right |&-9=8\\ \left | 5-\displaystyle \frac{2}{3}x \right |&=8+9=17\\ 5-\displaystyle \frac{2}{3}x&=\pm 17\\ -\displaystyle \frac{2}{3}x&=\pm 17-5\\ x&=\displaystyle \frac{\pm 17-5}{\left ( -\displaystyle \frac{2}{3} \right )}\\ x_{1}&=\displaystyle \left (\frac{17-5}{-2} \right )\times 3\\ &=-18\\ x_{2}&=\displaystyle \left (\frac{-17-5}{-2} \right )\times 3\\ &=33 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Harga buku di toko "KITA" adalah}\: \: 28.000,-\\ &\textrm{Jika harga buku tulis di toko "FAMILI" memiliki selisih}\: \: 7.000,-\\ &\textrm{Harga buku tulis di toko "FAMILI" dapat dinyatakan dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left | x-7000 \right |=28000\\ \textrm{b}.&\left | x+7000 \right |=28000\\ \color{red}\textrm{c}.&\left | x-28000 \right |=7000\\ \textrm{d}.&\left | x+28000 \right |=7000\\ \textrm{e}.&\left | x-21000 \right |=7000\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ \end{array}$

Contoh Soal Lanjutan 4 Limit Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 21.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\tan 3x}{1-\cos ^{2}2x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -\frac{3}{2}\\\\ \textrm{b}.&\displaystyle -\frac{3}{4}\\\\ \textrm{c}.&\displaystyle -\frac{1}{2}\\\\ \textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{e}.&\displaystyle \frac{3}{4} \end{array} \end{array}$

Jawab : e

Berikut pembahasannya

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\tan 3x}{1-\cos ^{2}2x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\tan 3x}{\sin ^{2}2x}\\ &=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\times \tan \textcircled{3}x}{\sin \textcircled{2}x\times \sin \textcircled{2}x}\\ &=\displaystyle \frac{1\times 3}{2\times 2}\\ &=\displaystyle \frac{3}{4} \end{aligned}$

$\begin{array}{ll}\\ 22.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-9}{\sin (x-3)}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle 9\\\\ \textrm{b}.&\displaystyle 7\\\\ \textrm{c}.&\displaystyle 6\\\\ \textrm{d}.&\displaystyle 3\\\\ \textrm{e}.&\displaystyle -6 \end{array} \end{array}$

Jawab : c

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-9}{\sin (x-3)}&\quad \textrm{perhatikan bahwa}\\ &\color{black}\textrm{misal}\: \: x=3+h,\: \: \textrm{maka}\: \: h\rightarrow 0\\ &\color{red}\textrm{selanjutnya gunakan}\: \: h=x-3,\: \: \textrm{dengan}\: \: h\rightarrow 0\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{(3+h)^{2}-9}{\sin h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{h^{2}+6h+9-9}{\sin h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \:\frac{h(h+6)}{\sin h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \:(h+6)\\ &=0+6=6 \end{aligned}$

atau boleh juga

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-9}{\sin (x-3)}&=\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{(x+3)(x-3)}{\sin (x-3)}\\ &=\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle (x+3)\\ &=3+3=6 \end{aligned}$

$\begin{array}{ll}\\ 23.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{2}}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6x}{x-\displaystyle \frac{\pi }{2}}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle 6\\\\ \textrm{b}.&\displaystyle 3\\\\ \textrm{c}.&\displaystyle 2\\\\ \textrm{d}.&\displaystyle -2\\\\ \textrm{e}.&\displaystyle -6 \end{array} \end{array}$

Jawab : e

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle \frac{\pi }{2}}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6x}{x-\displaystyle \frac{\pi }{2}}&\quad \color{black}\textrm{perhatikan juga bahwa}\\ &x=\displaystyle \frac{\pi }{2}+h,\quad \textrm{dengan}\: \: \: h\rightarrow 0\\ &\color{red}\textrm{atau}\: \: \: h=x-\displaystyle \frac{\pi }{2}\quad \textrm{dengan}\: \: \: h\rightarrow 0\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6\left ( \displaystyle \frac{\pi }{2}+h \right )}{h}=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ( \displaystyle 3\pi +6h \right )}{h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ( \displaystyle \pi +6h \right )}{h}=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{-\sin \left ( \displaystyle 6h \right )}{h}\\ &=-6 \end{aligned}$

Daftar Pustaka

1. Astuti, A. N., Miyanto, dan Ngapiningsih. 2020. Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas XII. Yogyakarta: PT. PENERBIT INTAN PARIWARA.

Contoh Soal Lanjutan 3 Limit Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 16.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \cos 4x=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -1\\\\ \textrm{b}.&-\displaystyle \frac{1}{2}\sqrt{3}\\\\ \textrm{c}.&\displaystyle -\frac{1}{2}\\\\ \textrm{d}.&\displaystyle 0\\\\ \textrm{e}.&\displaystyle 1 \end{array} \end{array}$

Jawab : e

Cukup jelas

Dengan substitusi langsung kita akan dapat mentukan hasilnya, yatu:

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \cos 4x=\cos 4(0)=\cos 0=1 \end{aligned}$

$\begin{array}{ll}\\ 17.&\textrm{Diketahui}\: \: f(x)=\begin{cases} \sin 2x, & \text{ saat } x< 0 \\ \tan x, & \text{ saat } x\geq 0 \end{cases}\\\\ &\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0^{+}}{\textrm{Lim}}\: \: f(x)=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -2\\\\ \textrm{b}.&-\displaystyle 1\\\\ \textrm{c}.&\displaystyle 0\\\\ \textrm{d}.&\displaystyle 1\\\\ \textrm{e}.&\displaystyle \textrm{tidak ada} \end{array} \end{array}$

Jawab : c

Berikut penjelasannya

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0^{+}}{\textrm{Lim}}\: \: \displaystyle f(x)&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \tan x=\tan 0=0 \end{aligned}$

$\begin{array}{ll}\\ 17.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{1-\cos 2x}{5\sin ^{2}x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -\frac{2}{5}\\\\ \textrm{b}.&\displaystyle \frac{2}{5}\\\\ \textrm{c}.&\displaystyle 1\\\\ \textrm{d}.&\displaystyle 2\\\\ \textrm{e}.&\displaystyle \infty \end{array} \end{array}$

Jawab : b

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 1-\cos 2x}{5\sin ^{2}x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 2\sin ^{2}x}{5\sin ^{2}x}\\ &=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{2\color{red}\sin \textcircled{1}x\times \sin \textcircled{1}x}{5\sin \color{red}\textcircled{1}x\times \sin \textcircled{1}x}\\ &=\displaystyle \frac{2\times 1\times 1}{5\times 1\times 1}\\ &=\displaystyle \frac{2}{5}\\ &\\ &\\ \color{black}\textrm{Perlu diingat}\: &\textrm{bahwa}\quad 1-\cos 2x=2\sin ^{2}x \end{aligned}$

$\begin{array}{ll}\\ 18.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{2}}{\textrm{Lim}}\: \: \displaystyle \frac{3-\sin x}{\cos 2x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -4\\\\ \textrm{b}.&\displaystyle -2\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&\displaystyle 1\\\\ \textrm{e}.&\displaystyle 2 \end{array} \end{array}$

Jawab : b

$\begin{array}{ll}\\ 19.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \pi }{\textrm{Lim}}\: \: \displaystyle \frac{\sin x-2}{\cos x-3}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle -1\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&\displaystyle 1\\\\ \textrm{e}.&\displaystyle 2 \end{array} \end{array}$

Jawab : a

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle \pi }{\textrm{Lim}}\: \: \displaystyle \frac{ \sin x-2}{\cos x-3}&\qquad\color{black}\textrm{dengan substitusi langsung didapatkan}\\ &=\underset{x\rightarrow \displaystyle \pi }{\textrm{Lim}}\: \: \displaystyle \frac{ \left (\sin \pi \right ) -2}{\left (\cos \pi \right ) -3}\\ &=\underset{x\rightarrow \displaystyle \frac{\pi }{2}}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}0-2}{-1-3}\\ &=\displaystyle \frac{-2}{-4}\\ &=\displaystyle \frac{1}{2}\\ \end{aligned}$

$\begin{array}{ll}\\ 20.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{6}}{\textrm{Lim}}\: \: \displaystyle \left (\cos ^{2}x-\sin ^{2}x \right )=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -\frac{1}{2}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&\displaystyle 1\\\\ \textrm{e}.&\displaystyle 2 \end{array} \end{array}$

Jawab : b

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle \frac{\pi }{6}}{\textrm{Lim}}\: \: \displaystyle \left ( \cos ^{2}x-\sin ^{2}x \right )&\qquad \color{black}\textrm{dengan substitusi langsung}\\ &=\cos ^{2}\left ( \displaystyle \frac{\pi }{6} \right )-\sin ^{2}\left ( \displaystyle \frac{\pi }{6} \right )\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )^{2}-\left ( \displaystyle \frac{1}{2} \right )^{2}\\ &=\displaystyle \frac{3}{4}-\frac{1}{4}=\frac{2}{4}\\ &=\displaystyle \frac{1}{2} \end{aligned}$

Daftar Pustaka

1. Tim. 2020. Modul Matematika (Peminatan Matematika dan Ilmu-Ilmu Alam Kelas XII. Tangerang: RAHMA GEMILANG.

Contoh Soal Lanjutan 2 Limit Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 11.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{15x\tan x}{3x\sin 5x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{8}\\\\ \textrm{b}.&\displaystyle \frac{1}{6}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&\displaystyle \frac{3}{2}\\\\ \textrm{e}.&\displaystyle 1 \end{array} \end{array}$

Jawab : e

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 15x\tan x}{3x\sin 5x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\textcircled{15}x\tan \textcircled{1}x}{\color{red}\textcircled{3}x\tan \color{red}\textcircled{5}x}\\ &=\displaystyle \frac{15\times 1}{3\times 5}\\ &=\displaystyle \frac{15}{15}\\ &=1 \end{aligned}$

$\begin{array}{ll}\\ 12.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{18x\tan x}{3x\sin 4x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{8}\\\\ \textrm{b}.&\displaystyle \frac{1}{6}\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&\displaystyle \frac{3}{2}\\\\ \textrm{e}.&\displaystyle 2 \end{array} \end{array}$

Jawab : d

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 18x\tan x}{3x\sin 4x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\textcircled{18}x\tan \textcircled{1}x}{\color{red}\textcircled{3}x\tan \color{red}\textcircled{4}x}\\ &=\displaystyle \frac{18\times 1}{3\times 4}\\ &=\displaystyle \frac{18}{12}\\ &=\displaystyle \frac{3}{2} \end{aligned}$

$\begin{array}{ll}\\ 13.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{12x\sin x}{2x^{2}}=....\\ &\begin{array}{ll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle 3\\\\ \textrm{d}.&\displaystyle 6\\\\ \textrm{e}.&\displaystyle 12 \end{array} \end{array}$

Jawab : d

Sebagai pembahasannya adalah sebagai berikut

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 12x\sin x}{2x^{2}}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\textcircled{12}x\sin \textcircled{1}x}{\color{red}\textcircled{2}x\times \textcircled{1}x}\\ &=\displaystyle \frac{12\times 1}{2\times 1}\\ &=\displaystyle \frac{12}{2}\\ &=6 \end{aligned}$

$\begin{array}{ll}\\ 14.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin^{2} 3x}{12x^{2}}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle \frac{3}{4}\\\\ \textrm{d}.&\displaystyle 4\\\\ \textrm{e}.&\displaystyle 16 \end{array} \end{array}$

Jawab : c

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ \sin^{2} 3x}{12x^{2}}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\sin \textcircled{3}x\times \sin \textcircled{3}x}{\color{red}\textcircled{12}x\times \textcircled{1}x}\\ &=\displaystyle \frac{3\times 3}{12\times 1}\\ &=\displaystyle \frac{9}{12}\\ &=\displaystyle \frac{3}{4} \end{aligned}$

$\begin{array}{ll}\\ 15.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan^{2} 3x}{21x^{2}}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\\\\ \textrm{b}.&\displaystyle \frac{3}{7}\\\\ \textrm{c}.&\displaystyle \frac{3}{4}\\\\ \textrm{d}.&\displaystyle 3\\\\ \textrm{e}.&\displaystyle 7 \end{array} \end{array}$

Jawab : b

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ \tan^{2} 3x}{21x^{2}}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\tan \textcircled{3}x\times \tan \textcircled{3}x}{\color{red}\textcircled{21}x\times \textcircled{1}x}\\ &=\displaystyle \frac{3\times 3}{21\times 1}\\ &=\displaystyle \frac{9}{21}\\ &=\displaystyle \frac{3}{7} \end{aligned}$

Sampai di sini semoga ini semua dapat menambah khazanah dalam menyelesaikan permasalahan jika suatu ketika pembaca mendapati soal yang semisal

Semoga bermanfaat

Contoh Soal Lanjutan Limit Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 6.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{12x}{\tan 8x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle \frac{3}{2}\\\\ \textrm{d}.&\displaystyle 4\\\\ \textrm{e}.&\displaystyle 8 \end{array} \end{array}$

Jawab : c

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{12x}{\tan 8x}&=12\times \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \left (\frac{x}{\tan 8x}\times \frac{8}{8} \right )\\ &=12\times \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{8x}{\tan 8x}\times \frac{1}{8}\\ &=\displaystyle \frac{12}{8}\times \left (\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{8x}{\tan 8x} \right )\\ &=\displaystyle \frac{12}{8}\times 1\\ &=\frac{3}{2}\\ \end{aligned}$

Selanjutnya jika ada soal menentukan nilai limit dan fungsinya berbentuk rasional dengan pembilang atau penyebut berupa sinus atau tangen (kadang keduanya muncul bersamaan yang satu diposisi pembilang dan yang satu diposisi penyebut) semisal bentuk di atas dan masing-masing berpangkat sama, maka kita kemungkinan besar tinggal mengambil koefisien dari masing-masing unsur pembilang dan penyebutnya. Demikian untuk langkah mudahnya dalam menyelesaikan soal yang ada.

Sehingga soal di atas dapat juga diselesaikan dengan cara hematnya:

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{12x}{\tan 8x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\textcircled{12}x}{\tan \color{red}\textcircled{8}x}\\ &=\displaystyle \frac{12}{8}\\ &=\displaystyle \frac{3}{2} \end{aligned}$

Coba perhatikan soal berikut

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6x}{\tan 10x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{5}\\\\ \textrm{b}.&\displaystyle \frac{3}{5}\\\\ \textrm{c}.&\displaystyle \frac{5}{9}\\\\ \textrm{d}.&\displaystyle \frac{6}{5}\\\\ \textrm{e}.&\displaystyle 2 \end{array} \end{array}$

Jawab : b

Walau cukup jelas dan sebagai ulasannya adalah sebagai berikut

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6x}{\tan 10x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{magenta}\textcircled{6}x}{\tan \color{magenta}\textcircled{10}x}\\ &=\displaystyle \frac{6}{10}\\ &=\displaystyle \frac{3}{5} \end{aligned}$

Sebagai tambahan kita berikan contoh pula soal berikut

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{16x}{\tan 4x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle 2\\\\ \textrm{d}.&\displaystyle 4\\\\ \textrm{e}.&\displaystyle 8 \end{array} \end{array}$

Jawab : d

Cukup jelas bahwa

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 16x}{\tan 4x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{green}\textcircled{16}x}{\tan \color{green}\textcircled{4}x}\\ &=\displaystyle \frac{16}{4}\\ &=\displaystyle 4 \end{aligned}$

$\begin{array}{ll}\\ 9.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{12x}{\sin 10x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{9}\\\\ \textrm{b}.&\displaystyle \frac{1}{5}\\\\ \textrm{c}.&\displaystyle \frac{5}{9}\\\\ \textrm{d}.&\displaystyle \frac{6}{5}\\\\ \textrm{e}.&\displaystyle \frac{9}{5} \end{array} \end{array}$

Jawab : d

Cukup jelas

$\begin{array}{ll}\\ 10.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan 12x}{\sin 4x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle \frac{3}{2}\\\\ \textrm{d}.&\displaystyle 3\\\\ \textrm{e}.&\displaystyle 4 \end{array} \end{array}$

Jawab : d

Jawaban juga sudah cukup jelas

Contoh Soal Limit Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 1.&\textrm{Perhatikanlah pernyataan-pernyataan berikut}\\ &(\textrm{i})\quad\textrm{Jika}\: \: \underset{x\rightarrow 0^{-} }{\textrm{Lim}}\: \: f(x)=4\: \: \textrm{dan}\: \: \textrm{Jika}\: \: \underset{x\rightarrow 0^{+} }{\textrm{Lim}}\: \: f(x)=2,\\ &\qquad\textrm{maka nilai}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: f(x)=8\\ &(\textrm{ii})\: \: \: \textrm{Jika}\: \: \underset{x\rightarrow 0^{-} }{\textrm{Lim}}\: \: f(x)=4\: \: \textrm{dan}\: \: \textrm{Jika}\: \: \underset{x\rightarrow 0^{+} }{\textrm{Lim}}\: \: f(x)=4,\\ &\qquad\textrm{maka nilai}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: f(x)=4\\ &(\textrm{iii})\: \: \textrm{Jika}\: \: \underset{x\rightarrow 0^{-} }{\textrm{Lim}}\: \: f(x)=4\: \: \textrm{dan}\: \: \textrm{Jika}\: \: \underset{x\rightarrow 0^{+} }{\textrm{Lim}}\: \: f(x)=2,\\ &\qquad\textrm{maka nilai}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: f(x)\: \: \textrm{tidak ada}\\ &(\textrm{iv})\: \: \textrm{Jika}\: \: \underset{x\rightarrow 0^{-} }{\textrm{Lim}}\: \: f(x)=3\: \: \textrm{dan}\: \: \textrm{Jika}\: \: \underset{x\rightarrow 0^{+} }{\textrm{Lim}}\: \: f(x)=2,\\ &\qquad\textrm{maka nilai}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: f(x)=1\\ \end{array}$

Pernyataan di atas yang tepat adalah ...

$\begin{array}{ll}\\ \textrm{a}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{ii})\\ \textrm{b}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{c}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{d}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iv})\\ \textrm{e}.&(\textrm{iii})\: \: \textrm{dan}\: \: (\textrm{iv}) \end{array}$

Jawab : c

Perhatikan definisi limit di sini

$\begin{array}{ll}\\ 2.&\textrm{Perhatikan gambar dan pernyataan-pernyataan berikut}\\ \end{array}$

$\begin{array}{ll}\\ &(\textrm{i})\quad\textrm{Nilai}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{2} }{\textrm{Lim}}\: \: f(x)=4\\ &(\textrm{ii})\: \: \: \textrm{Nilai}\: \: \underset{x\rightarrow \displaystyle \frac{3\pi }{4} }{\textrm{Lim}}\: \: f(x)=\sqrt{2}\\ &(\textrm{iii})\: \: \textrm{Nilai}\: \: \underset{x\rightarrow \displaystyle \pi }{\textrm{Lim}}\: \: f(x)=1\\\\ &(\textrm{iv})\: \: \textrm{Nilai}\: \: \underset{x\rightarrow \displaystyle \frac{5\pi }{4} }{\textrm{Lim}}\: \: f(x)=-1\\ &\textrm{Pernyataan-pernyataan yang tepat ditunjukkan oleh}....\\ \end{array}$

Jawab : a

$\begin{array}{ll}\\ 3.&\textrm{Perhatikan tabel nilai-nilai}\: \: f(x)\: \: \textrm{berikut}\\ \end{array}$

$\begin{array}{|c|c|}\hline x&f(x)\\\hline -1&2,37679\\\hline -0,2&2,01339\\\hline -0,1&2,00334\\\hline -0,01&2,00003\\\hline \cdots &\cdots \\\hline 0&?\\\hline \cdots &\cdots \\\hline 0,01&2,00003\\\hline 0,1&2,00334\\\hline 0,2&2,01339\\\hline 1&2,37679\\\hline \end{array}$

dan

(i) $\underset{x\rightarrow 0 }{\textrm{lim}}\: f(x)=0$

(ii) $\underset{x\rightarrow 0 }{\textrm{lim}}\: f(x)=1$

(iii) $\underset{x\rightarrow 0^{-} }{\textrm{lim}}\: f(x)=2$

(iv) $\underset{x\rightarrow 0^{+} }{\textrm{lim}}\: f(x)=2$

Dari pernyataan yang sesuai tabel di atas

yang tepat adalah ...

Jawab : e

Cukup jelas baik dengan limit kiri dan limit kanan diperoleh nilai limitnya = 2 saat x mendekati nol.

$\begin{array}{ll}\\ 4.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{2} }{\textrm{Lim}}\: \: \cos x=....\\ \end{array}$

$\begin{array}{ll}\\ \textrm{a}.&-1\\\\ \textrm{b}.&-\displaystyle \frac{1}{2}\\\\ \textrm{c}.&0\\\\ \textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{e}.&1 \end{array}$

Jawab : c

$\color{blue}\begin{aligned}\textrm{Kita Selesai}&\textrm{kan dengan substitusi langsung, yaitu:}\\ \underset{x\rightarrow \displaystyle \frac{\pi }{2} }{\textrm{Lim}}\: \: \cos x&=\cos \displaystyle \frac{\pi }{2}=\cos 90^{\circ}=0 \end{aligned}$

$\begin{array}{ll}\\ 5.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{4} }{\textrm{Lim}}\: \: \sin \left ( 2x-\displaystyle \frac{\pi }{6} \right )=....\\ \end{array}$

$\begin{array}{ll}\\ \textrm{a}.&-1\\\\ \textrm{b}.&0\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&\displaystyle \frac{1}{2}\sqrt{2}\\\\ \textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}$

Jawab : e

Berikut pembahasannya

$\color{blue}\begin{aligned}\textrm{Kita Selesaikan deng}&\textrm{an substitusi langsung, yaitu:}\\ \underset{x\rightarrow \displaystyle \frac{\pi }{4} }{\textrm{Lim}}\: \: \sin \left ( 2x-\displaystyle \frac{\pi }{6} \right )&=\sin \left ( 2.\displaystyle \frac{\pi }{4}-\frac{\pi }{6} \right )\\ &=\sin \left ( \displaystyle \frac{\pi }{2}-\frac{\pi }{6} \right )\\ &=\sin \left ( 90^{\circ}-30^{\circ} \right )\\ &=\sin 60^{\circ}\\ &=\displaystyle \frac{1}{2}\sqrt{3} \end{aligned}$

Lanjutan 2 Limit Fungsi Trigonometri

C. Menyelesaikan Limit Fungsi Trigonometri

Rumus dasar limit fungsi trigonometri adalah:

$\begin{array}{|ccc|}\hline \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin x}{x}=1&\textrm{dan}&\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\tan x}{x}=1\\ &&\\ \color{blue}\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{x}{\sin x}=1&\textrm{dan}&\color{blue}\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{x}{\tan x}=1\\ &&\\ \textrm{demikian pula}&&\\ &&\\ \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{ax}{\sin ax}=1&\textrm{dan}&\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{ax}{\tan ax}=1\\ &&\\ \color{magenta}\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin ax}{ax}=1&\textrm{dan}&\color{magenta}\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\tan ax}{ax}=1\\\hline \end{arry}$

Contoh Soal Lanjutan 4 Bangun Ruang Dimensi Tiga (Matematika Wajib Kelas XII)

$\begin{array}{ll}\\ 21.&\textrm{Perhatikanlah kubus ABCD.EFGH berikut} \end{array}$

Jika jarak titik A ke C adalah 8 cm,

maka jarak titik A ke D adalah ... cm

$\begin{array}{ll}\\ \textrm{a}.&4\\ \textrm{b}.&4\sqrt{2}\\ \textrm{c}.&4\sqrt{3}\\ \textrm{d}.&4\sqrt{5}\\ \textrm{e}.&4\sqrt{6} \end{array}$

Jawab : b

Pembahasan diserahkan kepada pembaca yang budiman

$\begin{array}{ll}\\ 22.&\textrm{Perhatikanlah kubus KLMN.OPQR berikut} \end{array}$

Jika jarak K ke Q adalah 9 cm,

maka jarak titik K ke L adalah ... cm

$\begin{array}{ll}\\ \textrm{a}.&3\sqrt{6}\\ \textrm{b}.&3\sqrt{3}\\ \textrm{c}.&9\sqrt{2}\\ \textrm{d}.&6\sqrt{3}\\ \textrm{e}.&6\sqrt{2} \end{array}$

Jawab : b

Pembahasan juga diserahkan kepada pembaca yang budiman

$\begin{array}{ll}\\ 23.&\textrm{Perhatikanlah limas T.ABCD sebagai berikut} \end{array}$

Jika panjang rusuk tegaknya adalah 13 cm,

serta AB = 8 cm, dan BC = 6 cm,

maka jarak T ke bidang ABCD (Titik tengah bidang ABCD)

$\begin{array}{ll}\\ \textrm{a}.&7\\ \textrm{b}.&8\\ \textrm{c}.&9\\ \textrm{d}.&10\\ \textrm{e}.&12 \end{array}$

Jawab : e

Pembahasan juga diserahkan kepada pembaca yang budiman

$\begin{array}{ll}\\ 24.&\textrm{Perhatikanlah balok ABCD.EFGH berikut} \end{array}$

Jika panjang AB = 4 BC = 2 CG = 8 cm,

maka panjang AG adalah ... . cm

$\begin{array}{ll}\\ \textrm{a}.&2\sqrt{21}\\ \textrm{b}.&2\sqrt{19}\\ \textrm{c}.&2\sqrt{17}\\ \textrm{d}.&2\sqrt{15}\\ \textrm{e}.&2\sqrt{13} \end{array}$

Jawab : a

$\begin{aligned}\textrm{AG}&=\sqrt{AC^{2}+CG^{2}}\\ &=\sqrt{AB^{2}+BC^{2}+CG^{2}}\\ &=\sqrt{8^{2}+2^{2}+4^{2}}\\ &=\sqrt{64+4+16}\\ &=\sqrt{84}\\ &=\sqrt{4.21}\\ &=\sqrt{4}.\sqrt{21}\\ &=2\sqrt{21}\: \: \textrm{cm} \end{aligned}$

$\begin{array}{ll}\\ 25.&\textrm{Sebuah kubus panjang diagonal sisinya adalah 4 cm}\\ &{\textrm{Maka panjang rusuk kubus tersebut adalah ... cm}}\\ &\begin{array}{ll}\\ \textrm{a}.&\sqrt{2}\\ \textrm{b}.&2\\ \textrm{c}.&2\sqrt{2}\\ \textrm{d}.&4\sqrt{2}\\ \textrm{e}.&4\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\: \: c \end{array}$

Contoh Soal Lanjutan 3 Bangun Ruang Dimensi Tiga (Matematika Wajib Kelas XII)

$\begin{array}{ll}\\ 16.&\textrm{Perhatikanlah ilustrasi kubus KLMN.OPQR berikut } \end{array}$

pernyataan berikut yang benar adalah ....

$\begin{array}{ll}\\ \textrm{a}.&\textrm{titik R pada rusuk KL}\\ \textrm{b}.&\textrm{titik M pada rusuk LN}\\ \textrm{c}.&\textrm{titik K pada rusuk MN}\\ \textrm{d}.&\textrm{titik N pada rusuk LM}\\ \textrm{e}.&\textrm{titik Q pada rusuk QP} \end{array}$

Jawab : e

Jawaban cukup jelas

$\begin{array}{ll}\\ 17.&\textrm{Perhatikanlah ilustrasi kubus ABCD.EFGH berikut} \end{array}$

Jika panjang rusuk kubus adalah $6\sqrt{2}$ cm,

maka panjang HB adalah .... cm

$\begin{array}{ll}\\ \textrm{a}.&6\sqrt{6}\\ \textrm{b}.&6\sqrt{3}\\ \textrm{c}.&9\\ \textrm{d}.&2\sqrt{6}\\ \textrm{e}.&3\sqrt{6} \end{array}$

Jawab : a

Karena HB adalah diagonal ruang, maka panjang HB adalah

$\begin{aligned}\textrm{HB}&=a\sqrt{3}=sisi.\sqrt{3}=\left ( 6\sqrt{2} \right ).\sqrt{3}=6\sqrt{6}\: \: \textrm{cm} \end{aligned}$

$\begin{array}{ll}\\ 18.&\textrm{Perhatikanlah ilustrasi kubus ABCD.EFGH berikut} \end{array}$

Jika jarak titik A ke titik G adalah 9 cm,

maka jarak titik A ke titik B adalah .... cm

$\begin{array}{ll}\\ \textrm{a}.&9\sqrt{2}\\ \textrm{b}.&3\sqrt{6}\\ \textrm{c}.&3\sqrt{3}\\ \textrm{d}.&6\sqrt{3}\\ \textrm{e}.&6\sqrt{2} \end{array}$

Jawab : c

$\begin{aligned}\textrm{AG}=a\sqrt{3}&=9\\ a\sqrt{3}&=9\\ a&=\displaystyle \frac{9}{\sqrt{3}}\\ &=\displaystyle \frac{9}{\sqrt{3}}\times \displaystyle \frac{\sqrt{3}}{\sqrt{3}}\\ &=\displaystyle \frac{9}{3}\sqrt{3}\\ &=3\sqrt{3}\: \: \textrm{cm}\\ \textrm{Jadi}&\: \: \textrm{panjang rusuk AB adalah}\: \: 3\sqrt{3}\: \: \textrm{cm} \end{aligned}$

$\begin{array}{ll}\\ 19.&\textrm{Perhatikanlah ilustrasi kubus KLMN.OPQR berikut} \end{array}$

Jika panjang rusuknya adalah 12 cm,

maka panjang KT adalah ... cm

$\begin{array}{ll}\\ \textrm{a}.&5\sqrt{6}\\ \textrm{b}.&6\sqrt{6}\\ \textrm{c}.&6\sqrt{10}\\ \textrm{d}.&12\sqrt{2}\\ \textrm{e}.&12\sqrt{3} \end{array}$

Jawab : b

Sebelumnya silahkan perhatikan panjang beberapa unsur kubus berikut

Sehingga panjang KT adalah :

$\begin{aligned}\textrm{KT}&=\displaystyle \frac{1}{2}a\sqrt{6}\\ \textrm{KT}&=\displaystyle \frac{1}{2}.(12).\sqrt{6}\\ &=6\sqrt{6}\: \: \textrm{cm} \end{aligned}$

Jika masih mengalami kesulitan,

pembaca dapat menuliskan pertanyaan di kolom komentar

$\begin{array}{ll}\\ 20.&\textrm{Perhatikanlah ilustrasi kubus KLMN.OPQR berikut} \end{array}$

Jika panjang rusuknya adalah 12 cm,

maka panjang OU adalah ... cm

$\begin{array}{ll}\\ \textrm{a}.&12\sqrt{6}\\ \textrm{b}.&12\sqrt{3}\\ \textrm{c}.&8\sqrt{3}\\ \textrm{d}.&4\sqrt{3}\\ \textrm{e}.&2\sqrt{3} \end{array}$

Jawab : d

Perhatikanlah segitiga siku-siku TOK dengan titik siku-sikunya di titik O.

untuk menentukan panjang OU, maka

$\begin{aligned}\textrm{Luas}\Delta \textbf{TOK}&=\textrm{Luas}\Delta \textbf{TOK}\\ \displaystyle \frac{1}{2}\times \textrm{alas}\times \color{blue}\textrm{tinggi}&=\displaystyle \frac{1}{2}\times \textrm{alas}\times \textrm{tinggi}\\ \textrm{KT}\times \color{blue}\textrm{t}&=\textrm{OT}\times \textrm{OK}\\ \color{blue}\textrm{t}&=\displaystyle \frac{\textrm{OT}\times \textrm{OK}}{\textrm{KT}}\\ &=\displaystyle \frac{\left ( \displaystyle \frac{1}{2}a\sqrt{2} \right )\times a}{\displaystyle \frac{1}{2}a\sqrt{6}}\\ &=\displaystyle \frac{a\sqrt{2}}{\sqrt{6}}\\ &=\displaystyle \frac{a\sqrt{2}}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}\\ &=\displaystyle \frac{a\sqrt{12}}{6},\qquad \textrm{ingat bahwa}\: \: \sqrt{12}=2\sqrt{3}\\ &=\displaystyle \frac{(12).2\sqrt{3}}{6}\\ &=4\sqrt{3} \end{aligned}$

Dan jika Anda mengalami kesulitan dalam memahami solusi di atas,

Anda dapat menyampaikan pertanyaan di kolom komentar

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