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Transformations of functions

Determine relative extrema algebraically.\newlineWilliam Artiaga 0404/1818/2424 1212:3535 AM\newline−2-2−4-4 Determine\newlineThis quiz: 55 point(s) possible\newlineThis question: 11 point(S) possible\newlineResume later\newlinealgebraically.\newlineQuestion 33 of 55\newlineSubmit quiz\newlineFind the xx-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema.\newlinef(x)=5+(4+3x)23f(x)=5+(4+3x)^{\frac{2}{3}}\newlineA. There are no relative minima. The function has a relative maximum of \newlineoxed{} at \newlinex=x= \newlineoxed{} T.\newline(Use a comma to separate answers as needed.)\newlineB. There are no relative maxima. The function has a relative minimum of \newlineoxed{} at \newlinex=x= \newlineoxed{}\newline(Use a comma to separate answers as needed.)\newlineC. The function has a relative maximum of \newlineoxed{} at \newlinex=x= \newlineoxed{} and a relative minimum of \newlineoxed{} at \newlinex=x= \newlineoxed{} \newlineoxed{}\newline(Use a comma to separate answers as needed.)\newlineD. There are no relative extrema.
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11. Diketahui f(x)=2x+5 \mathrm{f}(\mathrm{x})=2 \mathrm{x}+5 dan g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 , maka (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots \newline22. Fungsi f f dan g g adalah pemetaan dari R R ke R R yang dirumuskan oleh f(x)= f(x)= 3x+5 3 \mathrm{x}+5 dan g(x)=2xx+1,x≠−1 \mathrm{g}(\mathrm{x})=\frac{2 x}{x+1}, x \neq-1 . Rumus (gof)(x) adalah g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 00\newline33. Diketahui fungsi g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 11 dan g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 22. Nilai komposisi fungsi g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 33 adalah ...\newline44. Diketahui fungsi g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 44, dan g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 55. Nilai komposisi fungsi g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 66\newlineLATIHAN\newline55. Diketahui fungsi-fungsi g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 77 didefinisikan dengan g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 88 g(x)=x−1x+4,x≠−4 \mathrm{g}(\mathrm{x})=\frac{x-1}{x+4}, x \neq-4 99 didefinisikan dengan (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 00. Hasil dari fungsi (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 11 adalah ...\newline66. Diketahui (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 22 dirumuskan oleh (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 33 dan (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 44 −6-6 . Jika (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 55, nilai (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 66\newline77. Diketahui (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 22 dirumuskan oleh (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 88 dan (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 99 f f 00. Jika f f 11, maka nilai f f 22 yang memenuhi adalah ...\newline88. Jika f f 33 dan f f 44, maka f f 55\newline99. Diketahui f f 66 dan f f 77. Rumus fungsi f f 88 adalah...\newline1010. Suatu pemetaan (fog)(x)=… (\mathrm{fog})(\mathrm{x})=\ldots 22 dengan g g 00 dan g g 11, maka g g 22
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Tugas Mandiri 22\newline11. Obligasi dengan nilai nominal 100100.000000 , tingkat bunga obligasi 8% 8 \% , berjangka waktu 33 tahun, diterbitkan tg⁡ \operatorname{tg} 1/1/2005 1 / 1 / 2005 . Bunga dibayarkan setiap tanggal 1/1,1/5 1 / 1,1 / 5 dan 1/9 1 / 9 . Hitunglah nilai jual obligasi, buatlah tabel amortisasi dan buatlah jurnal pada tanggal 11 Januari 20052005, 11 Mei 20052005, 11 September 20052005, 3131 Desember 20052005, dan 11 Januari 20062006. Jika pada saat diterbitkan tingkat bunga pasar sebesar :\newlinea. 9% 9 \% \newlineb. 6% 6 \% \newline22. Bila obligasi pada point 11 diatas dilunasi pada tanggal 1/1/2007 1 / 1 / 2007 dengan harga pelunasan 110110.000000 .. Buatlah jurnalnya. Jika pada saat diterbitkan tingkat bunga pasar sebesar :\newlinea. 9% 9 \% \newlineb. 6% 6 \%
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Probabilities and Sample Spaces\newline- Imagne you have two dice. You roll both of them together, and add up the scores.\newline- For example, it you got a 22 on one die and a 66 on the other die, then your total score is 88 .\newline- The table below is a sample space diagram for the outcomes of rolling the tho dice.\newline- Complete the sample space diagram, filling in the total scores for each possibilty.\newline\begin{tabular}{|c|c|c|c|c|c|c|}\newline\hline 1∗ 1^{*} die & 11 & 22 & 33 & 44 & 55 & 66 \\\newline\hline 2∗ 2 * die & & & & & & \\\newline\hline 11 & 22 & 33 & 44 & 00 & 66 & 77 \\\newline\hline 22 & 33 & 44 & 55 & 66 & 77 & 00 \\\newline\hline 33 & 44 & 00 & 66 & 77 & 88 & 99 \\\newline\hline 44 & 55 & 66 & 22 & 88 & 99 & 00 \\\newline\hline 55 & 00 & 22 & 88 & 99 & 1010 & 11 \\\newline\hline 66 & 77 & 88 & 99 & 1010 & 11 & 00 \\\newline\hline\newline\end{tabular}\newlineQuestions\newline11) How many total possibities are there?\newline22) How you would have workod this out without the table?\newline33) Find the probabilhies of these events. 'S' means 'total Score'\newlinea) P(S P(S is 22) ) :\newline11) P P ( S S is a muttiple of 55 ):\newlineb) P P ( S S is 44 ):\newlineg) P P ( S S is tactor of 1010):\newlinec) P P (S is greater than 55 ):\newlineh) P P ( S S is a prime number):\newlined) P P ( S S is loss than 77 ):\newlinei) P P ( S S is an odd number):\newlinee) P P ( S S is 77 or less):\newlinei) 2∗ 2 * 99 ( P(S P(S 00 is a square number):
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11. A number is such that 11 i\newline22. Divide 3434 into two parts in such a way that (47)th  \left(\frac{4}{7}\right)^{\text {th }} of one part is equal to (25)(x2−12)×30 \left(\frac{2}{5}\right)\left(\frac{x}{2}-\frac{1}{2}\right) \times 30 both sides\newline33. Find a number such that when 55 is subtracted from 55 times the number, (x2−15)×30=(x3+ \left(\frac{x}{2}-\frac{1}{5}\right) \times 30=\left(\frac{x}{3}+\right. more than twice the number.\newline44. The digits of a two digit number are differ by 33 . If the digits are inter chan ⇒x2×30−15x \Rightarrow \frac{x}{2} \times 30-\frac{1}{5} x resulting number is added to the original number, we get 143143. Find the original ⇒15x−6=10 \Rightarrow 15 x-6=10 \newline22. Sum of digits of a two digit number is 99 . When we interchange the digits, it isf Transposing −6-6
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11 Relative to the origin O O , the points A A and B B have position vectors a \mathbf{a} and b \mathbf{b} respectively, where a \mathbf{a} and b \mathbf{b} are non-parallel. It is given that ∣a∣=5,∣b∣=10 |\mathbf{a}|=5,|\mathbf{b}|=10 and ∣2a+b∣=18 |2 \mathbf{a}+\mathbf{b}|=18 .\newlinea^=a∣a∣∼b^=b∣b∣=a5 \begin{array}{l} \hat{a}=\frac{a}{|a|} \stackrel{\hat{b}}{\sim}=\frac{b}{|b|} \\ =\frac{a}{5} \end{array} \newline(a) By using the result v⋅v=∣v∣2 \mathbf{v} \cdot \mathbf{v}=|\mathbf{v}|^{2} , or otherwise, show that A A 00.\newline(b) State the geometrical meaning of A A 11 and find its numerical value.\newlineA A 22\newline(c) The points A A 33 and A A 44 have position vectors A A 55 and A A 66 respectively. Given that these three points are collinear, find the value of A A 77.
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11. Una ruletia tiene cinco, giones de igual tamano de color azul, rojo, naranja, amarillo y verde. La flecha en ta ja ha girado 5050 veces durante un experimento. Los resultados se muestran en la tabta a cinuación.\newlineRESULTADOS DEL EXPERIMENTO\newline\begin{tabular}{|l|c|}\newline\hline \multicolumn{11}{|c|}{ Color } & Frecuencia \\\newline\hline Azul & 1212 \\\newline\hline Rojo & 1515 \\\newline\hline Naranja & 66 \\\newline\hline Amarillo & 1010 \\\newline\hline Verde & 77 \\\newline\hline\newline\end{tabular}\newlineSegún los resultados, ¿cuál es la probabilidad experimental de que, en cualquiera de los giros, la flecha se detenga en la sección de color rojo?\newlineA 115 \frac{1}{15} \newlineB 15 \frac{1}{5} \newlineC 37 \quad \frac{3}{7} \newlineD 310 \quad \frac{3}{10} \newline55 ¿Qué expresión es equivalente a −3(2x−8)+4x -3(2 x-8)+4 x ?\newlineA −2x−8 -2 x-8 \newlineB −2x+24 -2 x+24 \newlineC −10x−8 -10 x-8 \newlineD −10x+24 -10 x+24
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Name: Makayla Otunuku\newlineAssessment\newlineAS9157391573: Apply the geometry of conic sections in solving problems (33 Credits)\newlineA star has a planet orbiting at a constant distance of 12AU 12 \mathrm{AU} . Around the star there is also a scientific satellite orbiting in an elliptical orbit with the star as one of the focal points. When the satellite is closest to the star it is 66 AU from the star. When the satellite is furthest from the star it is 88 AU from the planet's orbit.\newline- The satellite has trouble transmitting data whenever it intersects the orbital path of the planet. Find the coordinates where this occurs.\newlineWhen the planet is at the location (−12,0) (-12,0) relative to the star and the satellite is at its greates distance from the star (see diagram below), an asteroid is predicted to follow a parabolic pat passing the planet. The planet is the focal point of the asteroids path and the closest distanc to the asteroid is 7AU 7 \mathrm{AU} from the planet.
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77 Gegeben sind die Gerade g g durch die Punkte P(0∣0∣3) P(0|0| 3) und Q(−5∣3∣3) Q(-5|3| 3) und die Gerade h h durch die Punkte R(0∣−1,5∣4,5) R(0|-1,5| 4,5) und S(−5∣4,5∣1,5) S(-5|4,5| 1,5) .\newlinea) Zeigen Sie, dass die Geraden g g und h h sich schneiden und so eine Ebene E E aufspannen.\newlineb) Bestimmen Sie für die Ebene E eine Parametergleichung.\newlinec) Berechnen Sie die Schnittpunkte der Ebene E mit den Koordinatenachsen.\newline88 Untersuchen Sie die gegenseitige Lage der Geraden g und der Ebene E. Bestimmen Sie gegebenenfalls den Durchstoßpunkt.\newlinea) g:xundefined=(408)+t⋅(130) \mathrm{g}: \overrightarrow{\mathrm{x}}=\left(\begin{array}{l}4 \\ 0 \\ 8\end{array}\right)+\mathrm{t} \cdot\left(\begin{array}{l}1 \\ 3 \\ 0\end{array}\right) und P(0∣0∣3) P(0|0| 3) 00\newlineb) P(0∣0∣3) P(0|0| 3) 11 und P(0∣0∣3) P(0|0| 3) 22
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MA111111 Assignment 11...\newline11|Pag e\newlineQuestion 1(4+4+(3+3+4+4)+(3+4+3)+4=32 1(4+4+(3+3+4+4)+(3+4+3)+4=32 marks)\newline(a) Fine the domain and range of f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} .\newline(b) Evaluate the following limits:\newline(i) lim⁡x→25x−25x−5 \lim _{x \rightarrow 25} \frac{x-25}{\sqrt{x}-5} \newline(iii) lim⁡x→−∞x2−34x+5 \lim _{x \rightarrow-\infty} \frac{\sqrt{x^{2}-3}}{4 x+5} \newline(ii) lim⁡θ→0sin⁡(5θ)θ \lim _{\theta \rightarrow 0} \frac{\sin (5 \theta)}{\theta} \newline(iv) lim⁡x→3x−3x2+7−4 \lim _{x \rightarrow 3} \frac{x-3}{\sqrt{x^{2}+7}-4} \newline(c) Consider the piecewise function\newlinef(x)={−x+1, if x<1x−1, if 1<x<25−x2, if x≥2 f(x)=\left\{\begin{array}{ll} -x+1, & \text { if } x<1 \\ x-1, & \text { if } 1<x<2 \\ 5-x^{2}, & \text { if } x \geq 2 \end{array}\right. \newline(i) Find lim⁡x→1f(x) \lim _{x \rightarrow 1} f(x) if it exists.\newline(ii) Show that f f is continuous at x=2 x=2 .\newline(iii) Sketch the graph of f(x) f(x) . using Mathematica\newline(d) Use the definition of derivative to differentiate f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} 00.\newlineQuestion f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} 11 marks)\newline(a) Use the technique of differentiation to find the derivatives of the following functions.\newline(i) f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} 22.\newline(ii) f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} 33.\newline22|Page\newline(iii) f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} 44.\newline(b) Use implicit differentiation to find f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} 55, if f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} 66.\newline(c) Use logarithmic differentiation to solve f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} 55, if f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} 88.\newline(d) For the function f(x)=−2+x+2 f(x)=-2+\sqrt{x+2} 99, find all critical values and determine whether each represents a local maximum, local minimum or neither. Then find the absolute extrema on the interval lim⁡x→25x−25x−5 \lim _{x \rightarrow 25} \frac{x-25}{\sqrt{x}-5} 00. Sketch the graph of
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LATIHAN SOAL PELUANG\newlinePWhlah satu pilhaw jawaban yang paling tepat\newline11. 1010 orang finalis suatu lomba kecantikan akan dipilih secara acak, 33 yang terbak. Banyak cara pemilihan tersebut ada cara.\newlineA. 7070\newlineC. 120120\newlineE. 720720\newlineB. 8080\newlineD. 360360\newline22. Banyakmya bilangan antara 20002000 dan 60006000 yang dapat disusun dari angka 0,1,2,3,4,5,6,7 0,1,2,3,4,5,6,7 , dan tidak: ada angka yang sama adalah\newlineA. 16801680\newlineC. 12601260\newlineE. 840840\newlineB. 14701470\newlineD. 10501050\newline33. Dari kota A ke kota B dilayani oleh 44 bus dan dari B ke C oleh 33 bus. Seseorang berangkat dari kota A ke kota C melalui 1818 kenudian kembali lagi ke A juga melalui B. Jika saat kembali dari C ke A. ia tidak mau mengguakan bus yaag sama. maka banyak cara perjalanan orang tersebut adalah\newlineA. 1212\newlineC. 7272\newlineE. 144144\newlineB. 3636\newlineD. 9696\newline44. Banyak garis yang dapon dibuat dari 88 titik yang tersedia, dengan tidak ada 33 titik yang segaris adalah...\newlineA. 336336\newlineC. 5656\newlineE. 1616\newlineB. 168168\newlineD. 2828\newline55. Dalam kantong I terdapat 55 kelereng merah dan 33 kelereng putih, dalam kantong II terdapat 44 kelereng merah dan 66 kelereng hitam. Dari setiap kantong diambil satu kelereng secara acak. Peluang terambilnya kelereng putih dari kantong 11 dan kelereng hitan dari kantong II adalah\newlineA. 39/40 39 / 40 \newlineC. 1/2 1 / 2 \newlineE. 9/40 9 / 40 \newlineB. 9/13 9 / 13 \newlineD. 9/20 9 / 20 \newline66. A,B,C, dan D akan berfoto secara berdampingan. Peluang A dan B selalu berdampingan adalah ...\newlineA. 1/12 1 / 12 \newlineC. 1/3 1 / 3 \newlineE. 2/3 2 / 3 \newlineB. 1/6 1 / 6 \newlineD. 1/2 1 / 2 \newline77. Sebuah kotak berisi 55 bola merah, 44 bola biru, dan 33 bola kuning. Dari dalam kotak diambil 33 bola sekaligus sceara acak, peluang terambil 22 bola merah dan I bola biru adalah\newlineA. 39/40 39 / 40 11\newlineC. 1/6 1 / 6 \newlineE. 39/40 39 / 40 33\newlineB. 39/40 39 / 40 44\newlineD. 39/40 39 / 40 55\newline88. Dalam suatu propulasi keluarga dengan tiga orang anak, peluang kelaarga tersebut mempunyai paling sedikit dua anak laki - laki adalah\newlineA. 39/40 39 / 40 66\newlineC. 39/40 39 / 40 77\newlineE. 33/44\newlineB. 1/3 1 / 3 \newlineD. 1/2 1 / 2 \newline99. Dua buah dadu dilempar bersama - sama, Peluang aunculnya jumbh mata dadu 99 atau 1010 adalah\newlineA. 39/40 39 / 40 44\newlineC. 1/2 1 / 2 11\newlineE. 1/2 1 / 2 22\newlineB. 1/2 1 / 2 33\newlineD. 1/2 1 / 2 44\newline1010. Sebuah dompet berisi uang logam, 55 keping lima ratusan dan 22 keping ratusan rupiah. Dompet yang lain berisi uang logam 33 keping lima ratusan dan I keping ratusan rupiah. Jika sobuah uang logam diambil secara acak dari salah satu dompet, peluang untuk mendapatkan uang logam ratusan rupiah adalah\newlineA. 1/2 1 / 2 55\newlineC. 1/2 1 / 2 66\newlineE. 1/2 1 / 2 77\newlineB. 1/2 1 / 2 88\newlineD. 1/2 1 / 2 99\newlinebimbingar
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Find the mean absolute deviation of values $2,$3,$5,$7 \$ 2, \$ 3, \$ 5, \$ 7 and $10 \$ 10 .\newline\begin{tabular}{|c|l|}\newline\hline Value (x) (\mathrm{x}) & Absolute deviation ∣x−x∣ |\mathrm{x}-\boldsymbol{x}| \\\newline\hline 22 & \\\newline\hline 33 & \\\newline\hline 55 & \\\newline\hline 77 & \\\newline\hline 1010 & \\\newline\hline\newline\end{tabular}\newlineCalculate your M.A.D.\newlineM.A.D. =Σf∣x−xˉ∣n =\frac{\Sigma f|x-\bar{x}|}{n} \newlineKM. travelled mid-pt (x) fx x−xˉ \mathbf{x}-\bar{x} f∣x−xˉ∣ f|x-\bar{x}| \newline00 and under 22\newline22 and under 44\newline44 and under 66\newline66 and under 88\newline88 and under 1010\newlineYou need to find your xˉ \bar{x} value\newlinex‾=Σfxn \overline{\mathrm{x}}=\frac{\Sigma \mathrm{fx}}{\mathrm{n}} \newlineYou need to find your M.A.D. value\newlineM.A.D. =Σf∣x−xˉ∣n =\frac{\Sigma f|x-\bar{x}|}{n} \newlineFind the standard deviation of values $2,$3,$5,$7 \$ 2, \$ 3, \$ 5, \$ 7 and $10 \$ 10 \newline\begin{tabular}{|c|c|c|}\newline\hline Value $10 \$ 10 11 & $10 \$ 10 22 & $10 \$ 10 33 \\\newline\hline 22 & & \\\newline\hline 33 & & \\\newline\hline 55 & & \\\newline\hline 77 & & \\\newline\hline 1010 & & \\\newline\hline\newline\end{tabular}\newlineIf, sic Method\newlines=Σf(x−xˉ)2n−1 s=\sqrt{\frac{\Sigma f(x-\bar{x})^{2}}{n-1}} \newlinewhere $10 \$ 10 44 standard deviation,\newline$10 \$ 10 55 frequency of each of the classes,\newline$10 \$ 10 66 mid-point of each class,\newline$10 \$ 10 77 mean,\newline$10 \$ 10 88 population/sample size.\newline\begin{tabular}{|c|c|}\newline\hline 22 & \\\newline\hline 55 & \\\newline\hline 44 & \\\newline\hline 88 & \\\newline\hline 11 & \\\newline\hline$10 \$ 10 99 & (x) (\mathrm{x}) 00 \\\newline\hline\newline\end{tabular}
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Width of diffraction maximum. We suppose that in a linear crystal there are identical point scattering centers at every lattice point ρm=ma \boldsymbol{\rho}_{m}=m \mathbf{a} , where m m is an integer. By analogy with (2020), the total scattered radiation amplitude will be proportional to F=Σexp⁡[−ima⋅Δk] F=\Sigma \exp [-i m \mathbf{a} \cdot \Delta \mathbf{k}] . The sum over M M lattice points is\newlineF=1−exp⁡[−iM(a⋅Δk]1−exp⁡[−i(a⋅Δk)] F=\frac{1-\exp [-i M(\mathbf{a} \cdot \Delta \mathbf{k}]}{1-\exp [-i(\mathbf{a} \cdot \Delta \mathbf{k})]} \newlineby the use of the series\newline∑m=0M−1xm=1−xM1−x. \sum_{m=0}^{M-1} x^{m}=\frac{1-x^{M}}{1-x} . \newline(a) The scattered intensity is proportional to ∣F∣2 |F|^{2} . Show that\newline∣F∣2≡F∗F=sin⁡212M(a⋅Δk)sin⁡212(a⋅Δk) |F|^{2} \equiv F^{*} F=\frac{\sin ^{2} \frac{1}{2} M(\mathbf{a} \cdot \Delta \mathbf{k})}{\sin ^{2} \frac{1}{2}(\mathbf{a} \cdot \Delta \mathbf{k})} \newline(b) We know that a diffraction maximum appears when a⋅Δk=2πh \mathbf{a} \cdot \Delta \mathbf{k}=2 \pi h , where h h is an integer. We change Δk \Delta \mathbf{k} slightly and define ϵ \boldsymbol{\epsilon} in a⋅Δk=2πh+ϵ \mathbf{a} \cdot \Delta \mathbf{k}=2 \pi h+\epsilon such that ϵ \boldsymbol{\epsilon} gives the position of the first zero in m m 11. Show that m m 22, so that the width of the diffraction maximum is proportional to m m 33 and can be extremely narrow for macroscopic values of M M . The same result holds true for a three-dimensional crystal.
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