Advertisement
Guest User

Temp

a guest
Jun 25th, 2018
117
0
Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
text 14.52 KB | None | 0 0
  1. \documentclass{article}
  2. \usepackage[utf8]{inputenc}
  3. \usepackage{graphicx}
  4.  
  5.  
  6. \title{MATHEMATICS: GRADE 12 PAPER 1 NOVEMBER 2017 MEMO}
  7. \author{Marks: 150}
  8. \date{Time: 3 hours}
  9.  
  10. \begin{document}
  11.  
  12. \maketitle
  13.  
  14. \section*{Question 1}
  15.  
  16. 1.1.1
  17. {Answer}{}
  18. $x=-7$ or $x=-2$\\
  19. {}{}
  20. {Memo}{}
  21. $x^{2}+9x+14=0$\\
  22. $(x+7)(x+2)=0$\\
  23. $x=-7$ or $x=-2$
  24. {}{}
  25. \\\\
  26.  
  27. 1.1.2
  28. {Answer}{}
  29. $x=0,29$ or $x=-2,54$\\
  30. {}{}
  31. {Memo}{}
  32. $4x^{2}+9x-3=0$\\
  33. $x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$\\
  34. $=\frac{-9\pm\sqrt{9^{2}-4(4)(-3)}}{2(4)}$\\
  35. $=\frac{-9\pm\sqrt{129}}{8}$\\
  36. $x=0,29$ or $x=-2,54$\\
  37. OR \\
  38. $x^{2}+\frac{9}{4}x + \frac{81}{64} = \frac{3}{4} + \frac{81}{64}$\\
  39. $(x+\frac{9}{8})^{2}=\frac{129}{64}$\\
  40. $x+\frac{9}{8}=\pm\frac{\sqrt{129}}{8}$\\
  41. $x=\frac{-9\pm\sqrt{129}}{8}$\\
  42. $x = 0,29$ or $x=-2,54$
  43. {}{}
  44. \\\\
  45.  
  46. 1.1.3
  47. {Answer}{}
  48. $x=5$\\
  49. {}{}
  50. {Memo}{}
  51. $\sqrt{x^{2}-5}=2\sqrt{x}$\\
  52. $x^{2}-5=4x$\\
  53. $x^{2}-4x-5=0$\\
  54. $(x-5)(x+1)=0$\\
  55. $x=5$ or $x=-1$\\
  56. $x=5$
  57. {}{}
  58. \\\\
  59.  
  60. 1.2
  61. {Answer}{}
  62. $y=-1$ or $y=17$\\
  63. {}{}
  64. {Memo}{}
  65. $3x-y=4$\\
  66. $y = 3x-4$\\
  67. $x^{2}+2xy-y^{2}=-2$\\
  68. $x^{2}+2x(3x-4)-(3x-4)^{2}=-2$\\
  69. $x^{2}+6x^{2}-8x-(9x^{2}-24x+16)=-2$\\
  70. $7x^{2}-8x-9x^{2}+24x-16=-2$\\
  71. $-2x^{2}+16x-14=0$\\
  72. $x^{2}-8x+7=0$\\
  73. $(x-7)(x-1)=0$\\
  74. $x=1$ or $x=7$\\
  75. $y = 3(1)-4$ or $y=3(7)-4$\\
  76. $y=-1$ or $y=17$\\
  77. OR\\
  78. $3x-y=4$\\
  79. $x= \frac{y+4}{3}$\\
  80. $x^{2}+2xy-y^{2}=-2$\\
  81. $(\frac{y+4}{3})^{2} + 2(\frac{y+4}{3})y-y^{2}=-2$\\
  82. $y^{2}+8y+16+6y^{2}+24y-9y^{2}-18$\\
  83. $-2y^{2}+32y+34=0$\\
  84. $y^{2}-16y-17=0$\\
  85. $(y-17)(y+1)=0$\\
  86. $y = -1$ or $y = 17$\\
  87. $x= \frac{-1+4}{3}$ or $x= \frac{17+4}{3}$\\
  88. $x=1$ or $x= 7$
  89. {}{}
  90. \\\\
  91.  
  92. 1.3.1
  93. {Answer}{}
  94. $x \in (-\infty; -4)$ or $x \in(-4; \infty)$\\
  95. {}{}
  96. {Memo}{}
  97. $x^2+8x+16>0$\\
  98. $(x+4)(x+4)>0$\\
  99. $x \in R, x\neq -4$\\
  100. OR\\
  101. $x \in (-\infty; -4)$ or $x \in(-4; \infty)$\\
  102. OR\\
  103. $x<-4 $ or $ x >-4$\\
  104. OR
  105. $x^2+8x+16>0$\\
  106. $(x+4)(x+4)>0$\\
  107. The function values remain positive\\
  108. $x \in R, x\neq -4$
  109. {}{}
  110. \includegraphics{graph8.png}
  111. \\\\
  112.  
  113. 1.3.2
  114. {Answer}{}
  115. $0 < p < 16 $\\
  116. {}{}
  117. {Memo}{}
  118. For two negative unequal roots: $0 < p < 16$\\
  119. OR \\
  120. $x^{2} + 8x+16=p$\\
  121. $x^{2}+8x+16-p=0$\\
  122. $0<16-p<16$\\
  123. $-16<-p<0$\\
  124. $0<p<16$\\
  125. OR\\
  126. $x^{2}+8x+16-p=0$\\
  127. $x = \frac{-8\pm\sqrt{64-4(16-p)}}{2}$\\
  128. $0 < 64-4(16-p)<64$\\
  129. $0<4p<64$\\
  130. $0<p<16$\\
  131. OR\\
  132. $x^{2}+8x+16=p$\\
  133. $x^{2}+8x+16-p=0$\\
  134. Roots are real and unequal:\\
  135. $8^{2}-4(16-p)>0$\\
  136. $4p>0$\\
  137. $p>0$\\
  138. Roots are: $\frac{-8\pm\sqrt{4p}}{2}$\\
  139. For both roots to be negative:\\
  140. $\sqrt{4p}<8$\\
  141. $4p<64$\\
  142. $p<16$\\
  143. $0<p<16$\\
  144. {}{}
  145. \includegraphics{graph1.png}
  146. \\\\
  147.  
  148. \section*{Question 2}
  149.  
  150. 2.1.1
  151. {Answer}{}
  152. $-6$\\
  153. {}{}
  154. {Memo}{}
  155. First differences: $-9;-15;-21$\\
  156. Second difference: $-6$\\
  157. {}{}
  158. \includegraphics{graph2.png}
  159. \\\\
  160.  
  161. 2.1.2
  162. {Answer}{}
  163. $T_{n}=-3n^{2}+8$\\
  164. {}{}
  165. {Memo}{}
  166. $T_{n}=an^{2}+bn+c$\\
  167. $a =\frac{second difference}{2} = -3$\\
  168. $3a + b = -9$\\
  169. $3(-3)+b=-9$\\
  170. $b=0$\\
  171. $a+b+c=5$\\
  172. $-3+0+c=5$\\
  173. $c=8$\\
  174. $T_{n}=-3n^{2}+8$\\
  175. OR\\
  176. $T_{n} = T_{1} + (n-1)d_{1} + \frac{(n-1)(n-2)d_{2}}{2}$\\
  177. $= 5+(n-1)(-9)+\frac{(n-1)(n-2)(-6)}{2}$\\
  178. $= 5-9n+9-3n^{2}+9n-6$\\
  179. $T_{n}=-3n^{2}+8$\\
  180. {}{}
  181. \\\\
  182.  
  183. 2.1.3
  184. {Answer}{}
  185. The $93^{rd}$ term has a value of $-25939$.\\
  186. {}{}
  187. {Memo}{}
  188. $-3n^{2}+8=-25939$\\
  189. $-3n^{2}=-25947$\\
  190. $n^{2}=8649$\\
  191. $n=-93$ or $n=93$\\
  192. The $93^{rd}$ term has a value of $-25939$.
  193. {}{}
  194. \\\\
  195.  
  196. 2.2.1
  197. {Answer}{}
  198. $T_{15} = 59$\\
  199. {}{}
  200. {Memo}{}
  201. $2k-7; k+8$ and $2k-1$\\
  202. $k+8-(2k-7)=2k-1-(k+8)$\\
  203. $-k+15=k-9$\\
  204. $2k=24$\\
  205. $k=12$\\
  206. $2k-7; k+8$ and $2k-1$\\
  207. $17;20;23.......$\\
  208. $d = 3$\\
  209. $T_{15}=17+14(3)$\\
  210. $=59$
  211. {}{}
  212. \\\\
  213.  
  214. 2.2.2
  215. {Answer}{}
  216. $S_{30} = 3210$\\
  217. {}{}
  218. {Memo}{}
  219. Sequence is $17; 20; 23; 26; 29; 32 .....$\\
  220. Every alternate term of the sequence will be even. \\
  221. $20 + 26 + 32 + ....$\\
  222. $S_{30}=\frac{30}{2}[2(20)+(29)(6)]$\\
  223. $=15[40+174]$\\
  224. $= 3210$\\
  225. OR \\
  226. $T_{30}=20+29(6)$\\
  227. $=94$\\
  228. $S_{30}=\frac{30}{2}(20+194)$\\
  229. $=3210$
  230. {}{}
  231. \\\\
  232.  
  233. \section*{Question 3}
  234.  
  235. 3.1
  236. {Answer}{}
  237. $a=\frac{9-9r}{4}$ OR $a= \frac{2}{1+r}$ OR $a= \frac{1-r}{4r^{2}}$\\
  238. {}{}
  239. {Memo}{}
  240. $a+ar=2$\\
  241. $a(1+r)=2$\\
  242. $a=\frac{2}{1+r}$\\
  243. OR\\
  244. $\frac{a}{1-r} - 2 = \frac{1}{4}$\\
  245. $4a-8(1-r)=1-r$\\
  246. $4a-8+8r=1-r$\\
  247. $4a=9-9r$\\
  248. $a=\frac{9-9r}{4}$\\
  249. OR\\
  250. $S_{n}= \frac{a(r^{n}-1}{r-1}$\\
  251. $2= \frac{a(r^{2}-1}{r-1}$\\
  252. $2 = \frac{a(r-1)(r+1)}{r-1}$\\
  253. $2 = a(r+1)$\\
  254. $a = \frac{2}{r+1}$\\
  255. OR\\
  256. $\frac{ar^{2}}{1-r} = \frac{1}{4}$\\
  257. $a=\frac{1-r}{4r^{2}}$
  258. {}{}
  259. \\\\
  260.  
  261. 3.2
  262. {Answer}{}
  263. $a = \frac{3}{2}$ and $r = \frac{1}{3}$\\
  264. {}{}
  265. {Memo}{}
  266. $S_{\infty} = T_{1}+T_{2}+\sum{\infty}{n=3}{T_{n}}$\\
  267. $S_{\infty}=2+\frac{1}{4}$\\
  268. $\frac{a}{1-r}=2+\frac{1}{4}$\\
  269. $\frac{a}{1-r}=\frac{9}{4}$\\
  270. $(\frac{2}{1+r})x(\frac{1}{1-r}) = \frac{9}{4}$\\
  271. $\frac{2}{1-r^{2}} = \frac{9}{4}$\\
  272. $8 = 9 - 9r^{2}$\\
  273. $9r^{2}=1$\\
  274. $r = \frac{1}{3}$\\
  275. $a = \frac{3}{2}$\\
  276. OR\\
  277. $S_{\infty} = T_{1}+T_{2}+\sum{\infty}{n=3}{T_{n}}$\\
  278. $S_{\infty} = 2+\frac{1}{4}$\\
  279. $\frac{a}{1-r} = 2 + \frac{1}{4}$\\
  280. $\frac{a}{1-r} = \frac{9}{4}$\\
  281. $4a=9-9r$\\
  282. $r=\frac{9-4a}{9}$\\
  283. $a+a(\frac{9-4a}{9})=2$\\
  284. $9a+9a-4a^{2} = 18$\\
  285. $2a^{2} - 9a+9 = 0$\\
  286. $(a-3)(2a-3)=0$\\
  287. $a=\frac{3}{2}$ or $a=3$\\
  288. $r=\frac{1}{3}$ or $r = -\frac{1}{3}$\\
  289. OR\\
  290. $r = \frac{2-a}{a}$\\
  291. $\frac{ar^{2}}{1-r} = \frac{1}{4}$\\
  292. $4ar^{2} = 1-r$\\
  293. $4a(\frac{2-a}{a})^{2} = 1-\frac{2-a}{a}$\\
  294. $16-16a+4a^{2}=2a+2$\\
  295. $2a^{2}-9a+9=0$\\
  296. $(2a-3)(a-3)=0$\\
  297. $a=\frac{3}{2}$, $a\neq3$\\
  298. $r = \frac{1}{3}$, $r\neq\frac{1}{3}$\\
  299. OR\\
  300. $S_{\infty} = T_{1}+T_{2}+\sum{\infty}{n=3}{T_{n}}$\\
  301. $S_{\infty} = 2+\frac{1}{4}$\\
  302. $\frac{a}{1-r} = 2+\frac{1}{4}$\\
  303. $\frac{a}{1-r}=\frac{9}{4}$\\
  304. $(\frac{1-r}{r^{2}})x(\frac{1}{1-r}) = \frac{9}{4}$\\
  305. $\frac{1}{4r^{2}}= \frac{9}{4}$\\
  306. $4 = 36r^{2}$\\
  307. $9r^{2}=1$\\
  308. $r = \frac{1}{3}$\\
  309. $a = \frac{3}{2}$
  310. {}{}
  311. \\\\
  312.  
  313. \section*{Question 4}
  314.  
  315. 4.1
  316. {Answer}{}
  317. $a = \frac{1}{2}$ and $b = 2$\\
  318. {}{}
  319. {Memo}{}
  320. $f(x) = -ax^{2} + bx + 6$\\
  321. $f'(x) = -2ax+b$\\
  322. $-2ax+b=3$\\
  323. at $x=-1$\\
  324. $2a+b=3$ [1]\\
  325. $f(-1)=\frac{7}{2}$\\
  326. $-a-b+6=\frac{7}{2}$\\
  327. $-2a-2b+12=7$\\
  328. $2a+2b=5$ [2]\\
  329. $[2]-[1]$\\
  330. $b=2$\\
  331. $2a + 2 = 3$\\
  332. $a = \frac{1}{2}$\\
  333. OR\\
  334. $f'(x) = -2ax+b$\\
  335. $3=2a+b$\\
  336. $b=3-2a$\\
  337. $\frac{7}{2}=-a(-1)^{2}+(3-2a)(-1)+6$\\
  338. $a+3=\frac{7}{2}$\\
  339. $a=\frac{1}{2}$\\
  340. $b=2$
  341. {}{}
  342. \\\\
  343.  
  344. 4.2
  345. {Answer}{}
  346. $(-2;0)$ $(6;0)$\\
  347. {}{}
  348. {Memo}{}
  349. $f(x) = -\frac{1}{2}x^{2} + 2x+6$\\
  350. x-intercepts:\\
  351. $-\frac{1}{2}x^{2} + 2x+6 = 0$\\
  352. $-x^{2} + 4x+12=0$\\
  353. $x^{2}-4x-12=0$\\
  354. $(x-6)(x+2)=0$\\
  355. $(-2;0)$ $(6;0)$
  356. {}{}
  357. \\\\
  358.  
  359. 4.3
  360. {Answer}{}
  361. TP$(2;8)$\\
  362. {}{}
  363. {Memo}{}
  364. $f(x)=-\frac{1}{2}x^{2} + 2x+6$\\
  365. $f'(x) = 0$ or $x = -\frac{b}{2a}$ or $x= \frac{-2+6}{2}$\\
  366. $-x+2=0$ and $x=-\frac{2}{2(-\frac{1}{2})}$ and $x = 2$\\
  367. $x = 2$ and $x = 2$\\
  368. $y = -\frac{1}{2}(2)^{2} + 2(2)+6$\\
  369. $= -2+4+6$\\
  370. $= 8$\\
  371. TP$(2;8)$\\
  372. OR\\
  373. $y = -\frac{1}{2}(x^{2}-4x-12)$\\
  374. $= -\frac{1}{2}[(x-2)^{2}-4-12]$\\
  375. $= -\frac{1}{2}(x-2)^{2} + 8$\\
  376. TP$(2;8)$
  377. {}{}
  378. \\\\
  379.  
  380. 4.4
  381. {Answer}{}
  382. look at memo\\
  383. {}{}
  384. {Memo}{}
  385. look at graph
  386. {}{}
  387. \includegraphics{graph3.png}\\
  388. \\\\\
  389.  
  390. 4.5
  391. {Answer}{}
  392. $ 0<x<4$ or (0 ; 4)\\
  393. {}{}
  394. {Memo}{}
  395. $ 0<x<4$ or (0 ; 4)
  396. {}{}
  397. \\\\
  398.  
  399. 4.6
  400. {Answer}{}
  401. look at memo\\
  402. {}{}
  403. {Memo}{}
  404. look at graph
  405. {}{}
  406. \includegraphics{graph3.png}\\
  407. \\\\
  408.  
  409. 4.7
  410. {Answer}{}
  411. $x \leq -2 $ or $-1\leq x \leq 6$\\
  412. OR\\
  413. $(-\infty; 2] $ OR $ [-1 ; 6]$\\
  414. {}{}
  415. {Memo}{}
  416. $x \leq -2 $ or $-1\leq x \leq 6$\\
  417. OR\\
  418. $(-\infty; 2] OR [-1 ; 6]$
  419. {}{}
  420. \\\\
  421.  
  422. \section*{Question 5}
  423.  
  424. 5.1
  425. {Answer}{}
  426. $y \in R$ ; $y \neq 1$\\
  427. {}{}
  428. {Memo}{}
  429. $y < -1 $ or $ y>-1$\\
  430. OR\\
  431. $y \in (-\infty ; -1)$ or $ y \in (-1 ; \infty)$\\
  432. OR\\
  433. $R-\{-1\}$
  434. {}{}
  435. \\\\
  436.  
  437. 5.2
  438. {Answer}{}
  439. D(2; -1)\\
  440. {}{}
  441. {Memo}{}
  442. $g(x) = \frac{2}{x-2}-1 $
  443. {}{}
  444. \\\\
  445.  
  446. 5.3
  447. {Answer}{}
  448. t = 3\\
  449. {}{}
  450. {Memo}{}
  451. $f(x) = log_{3}x$\\
  452. $log_{3}t = 1$\\
  453. $t = 3$\\
  454. OR\\
  455. $g(x) = \frac{2}{x-2}-1 $\\
  456. $1 = \frac{2}{t-2}-1 $\\
  457. $2 = \frac{2}{t-2}$\\
  458. $2t -4 = 2$\\
  459. $t = 3$
  460. {}{}
  461. \\\\
  462.  
  463. 5.4
  464. {Answer}{}
  465. $y=3^x$\\
  466. {}{}
  467. {Memo}{}
  468. $x = log_{3}y$\\
  469. $y=3^x$
  470. {}{}
  471. \\\\
  472.  
  473. 5.5
  474. {Answer}{}
  475. $x < 1$\\
  476. {}{}
  477. {Memo}{}
  478. $3^x < 3^1$\\
  479. $x < 1$\\
  480. OR\\
  481. $3^x < 3^1$\\
  482. $ x \in (-\infty; 1)$\\
  483. {}{}
  484. \\\\
  485.  
  486. 5.6
  487. {Answer}{}
  488. Point of intersection: B (1 ; 0)\\
  489. {}{}
  490. {Memo}{}
  491. Equation of the axis of symmetry: $y = -x+1 $\\
  492. $x$-intercept of the axis of symmetry is at $x$ = 1\\
  493. has an $x$-intercept at B(1 ; 0) which is the same as the\\
  494. $x$-intercept of the axis of symmetry\\
  495.  
  496. OR\\
  497.  
  498. Since BE = ED = 1 and D lies on the axis of symmetry\\
  499. and the gradient of the axis of symmetry is –1, B will\\
  500. also lie on the axis of symmetry. But B also lies on $f$.\\
  501. Therefore B(1 ; 0) is the point of intersection between $f$\\
  502. and the axis of symmetry with a negative gradient.
  503. {}{}
  504. \\\\
  505.  
  506. \section*{Question 6}
  507.  
  508. 6.1
  509. {Answer}{}
  510. $r$ = 6.5\% \\
  511. {}{}
  512. {Memo}{}
  513. $A = P(1 + i)^n $\\
  514. $12 146,72 = 10 000(1+\frac{r}{12})^{36}$\\
  515. $(1+\frac{r}{12})^{36} = 1,214672$\\
  516. $1+\frac{r}{12} = \sqrt[36]{1,214672}$\\
  517. = 1,005416\\
  518. $\frac{r}{12} = 0.005416$\\
  519. $r$ = 0,06500\\
  520. $r$ = 6.5\%
  521. {}{}
  522. \\\\
  523.  
  524.  
  525. 6.2.1
  526. {Answer}{}
  527. His monthly instalment is R 5 536,95\\
  528. {}{}
  529. {Memo}{}
  530. P = $\frac{x[1-(1+i)^{-n}]}{i}$\\
  531. 235000 = $\frac{x[1-(1+\frac{0,11}{12})^{-54}]}{\frac{0,11}{12}}$\\
  532. $x = \frac{235000\cdot\frac{0,11}{0,12}}{[1-(1+\frac{0,11}{12})^{-54}]}$\\
  533. = R 5 536,95
  534. {}{}
  535. \\\\
  536.  
  537.  
  538. 6.2.2
  539. {Answer}{}
  540. Amount of interest = R23 739.60\\
  541. {}{}
  542. {Memo}{}
  543. Amount paid for the year : $(5 536,95 \cdot 12) = $R66 443,40\\
  544. Balance=$235000(1+\frac{0,11}{12})^{12}-\frac{5536,95[(1+\frac{0,11}{12})^{12}-1]}{\frac{0,11}{0,12}}$\\
  545. = 192296,17\\
  546. Interest = $(5536,95 \cdot 12)-(235000-192296,17$\\)
  547. =66443,40 - 42703,83\\
  548. = 23 739,57\\
  549.  
  550. OR\\
  551.  
  552. Total amount paid in first year = $R 5 536.95 \cdot 12$ = R66 443,40\\
  553. Balance on loan after 1 year = P of remaining installments\\
  554. P = $\frac{x[1-(1+i)^{-n}]}{i}$\\
  555. $= \frac{5536,95[1-(1+\frac{0,11}{12})^{-42}]}{\frac{0,11}{12}}$\\
  556. = R192296,20\\
  557. Amount paid off in the first year:\\
  558. R235 000 – R192 296,20 = R42 703,80 \\
  559. Amount of interest = R66 443,40 – R42 703,80\\
  560. = R23 739,60\\
  561.  
  562. OR\\
  563.  
  564. P=$\frac{5536,95[1-(1+\frac{0.11}{12})^{-12}]}{\frac{0.11}{12}}$\\
  565. = R62648.18\\
  566. 235 000 – 62 648,18 = R172 351,82\\
  567. After 12 months, money owed on house is\\
  568. $172351.82(1+\frac{0.11}{12})^{12}$\\
  569. = 192296.17\\
  570. Amount paid after 12 months is\\
  571. $5 536,95 \cdot 12 =$ R 66 443, 40\\
  572. Amount of interest paid:\\
  573. R 66 443, 40 – (235 000 – 192 296,17)\\
  574. = R 23 739, 57\\
  575. {}{}
  576. \\\\
  577.  
  578.  
  579. \section*{Question 7}
  580.  
  581. 7.1
  582. {Answer}{}
  583. $4x-1$\\
  584. {}{}
  585. {Memo}{}
  586. $f(x+h) = 2(x+h)^2 -(x+h)$\\
  587. $= 2(x^2+2xh+h^2)-x-h$\\
  588. $= 2x^2 +4xh+2h^2-x-h$\\
  589. $f(x+h)-f(x) = 2x^2 +4xh+2h^2-x-h-2x^2+x$\\
  590. $= 4xh+2h^2-h$\\
  591. $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
  592. $$= \lim_{h\to0}\frac{4xh+2h^2-h}{h}$$
  593. $$= \lim_{h\to0}\frac{h(4x+2h-1)}{h}$$
  594. $$= \lim_{h\to0} (4x+2h-1)$$
  595. $$= 4x-1$$
  596.  
  597. $$\mathbf{OR}$$\\
  598.  
  599. $$f'(x)= \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$
  600. $$f'(x)= \lim_{h\to0} \frac{2(x+h)^2 -(x+h)-(2x^2-x)}{h}$$
  601. $$ = \lim_{h\to0}\frac{2x^2 +4xh+2h^2-x-h -2x^2+x}{h}$$
  602.  
  603. $$ = \lim_{h\to0}\frac{4xh+2h^2-h}{h}$$
  604. $$ = \lim_{h\to0}\frac{h(4x+2h-1)}{h}$$
  605. $$ = \lim_{h\to0}(4x+2h-1)$$
  606. $$ = 4x-1$$
  607. {}{}
  608. \\\\
  609.  
  610. 7.2.1
  611. {Answer}{}
  612. $6x-4$\\
  613. {}{}
  614. {Memo}{}
  615. $D_{x}[(x+1)(3x-7)]$\\
  616. $=D_{x}(3x^2-4x-7)$\\
  617. $= 6x-4$
  618. {}{}
  619. \\\\
  620.  
  621. 7.2.2
  622. {Answer}{}
  623. $\frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}}+5x^{-2}$\\
  624. {}{}
  625. {Memo}{}
  626. $y = \sqrt{x^3}-\frac{5}{x}+\frac{1}{2}\pi$\\
  627. $y = x^{\frac{3}{2}}-5x^{-1}+\frac{1}{2}\pi$\\
  628. $\frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}}+5x^{-2}$
  629. {}{}
  630. \\\\
  631.  
  632. \section*{Question 8}
  633.  
  634. 8.1
  635. {Answer}{}
  636. Point of inflection at $x$ = 2\\
  637. {}{}
  638. {Memo}{}
  639. $f(x) = x^3-6x^2+9x$\\
  640. $f'(x) = 3x^2-12x+9$\\
  641. $f''(x)= 6x-12=0$\\
  642. $x=2$
  643. $f''(0) = 6(0)-12$\\
  644. =-12\\
  645. $f''(3) = 6(3)-12$\\
  646. $=6$\\
  647. Point of inflection at $x$=2
  648. {}{}
  649. \includegraphics{graph7.png}
  650. \\\\
  651.  
  652. 8.2
  653. {Answer}{}
  654. look at memo\\
  655. {}{}
  656. {Memo}{}
  657. look at graph
  658. {}{}
  659. \includegraphics{graph4.png}
  660. \\\\
  661.  
  662. 8.3
  663. {Answer}{}
  664. $y = -f(x) $ will be concave down for $ x > 2$\\
  665. {}{}
  666. {Memo}{}
  667. $f$ concave up for $x > 2$\\
  668. $y -f(x) $ will be concave down for $x > 2$\\
  669. {}{}
  670. \\\\
  671.  
  672. 8.4.1
  673. {Answer}{}
  674. (3;7)\\
  675. {}{}
  676. {Memo}{}
  677. (3;7)
  678. {}{}
  679. \\\\
  680.  
  681. 8.4.2
  682. {Answer}{}
  683. Do not agree with Claire as her statement is incorrect.\\
  684. {}{}
  685. {Memo}{}
  686. Between $x$ = 1 and $x$ = 3 the graph of$f$ is decreasing.
  687. Therefore at $x$ = 2 the gradient will have a negative value.
  688.  
  689. OR\\
  690.  
  691. $f'(2)=3(2)^2-12(2)+9$\\
  692. =-3\\
  693. $\neq$1\\
  694. {}{}
  695. \\\\
  696.  
  697. \section*{Question 9}
  698.  
  699. 9.1
  700. {Answer}{}
  701. PB = $\frac{\sqrt{3}}{2} = 0,87$\\
  702. {}{}
  703. {Memo}{}
  704. $y=x^2+2$\\
  705. $P(x;x^2+2)$\\
  706. B(0 ; 3)\\
  707. $PB^2 = (x-0)^2+(x^2+2-3)^2\\$
  708. $= x^2+x^4-2x^2+1$\\
  709. $= x^4-x^2+1$\\
  710.  
  711. PB will be a minimum if $PB^2$ is a minimum\\
  712. $\frac{d(PB^2)}{dx} = 4x^3-2x$\\
  713. $4x^3-2x=0$\\
  714. $x(2x^2-1)=0$\\
  715. $x=0 $ or $ x^2 = \frac{1}{2}$\\
  716. $x = \frac{1}{\sqrt{2}}$\\
  717.  
  718. $PB^2 = (\frac{1}{\sqrt{2}})^4-(\frac{1}{\sqrt{2}})^2+1$\\
  719. $= \frac{1}{4}-\frac{1}{2}+1$\\
  720. =$\frac{3}{4}$\\
  721. PB = $\frac{\sqrt{3}}{2} = 0.87$\\
  722.  
  723. OR\\
  724.  
  725. Gradient of tangent to curve = $2x$\\
  726. Gradient of line joining B and the curve = $\frac{x^2+2-3}{x-0}$\\
  727. $= \frac{x^2-1}{x}$\\
  728. Shortest distance will be where tangent to curve is perpendicular to the line joining P and the curve\\
  729. $ \frac{x^2-1}{x} = \frac{1}{2x}$\\
  730. $2x(x^2-1)=-x$\\
  731. $2x^3-2x=0$\\
  732. $x(2x^2-1)=0$\\
  733. $x=0$ or $x^2 = \frac{1}{2}$\\
  734. $x = \frac{1}{\sqrt{2}}$\\
  735. $PB^2 = (\frac{1}{\sqrt{2}})^4-(\frac{1}{\sqrt{2}})^2+1$\\
  736. $= \frac{1}{4}-\frac{1}{2}+1$\\
  737. =$\frac{3}{4}$
  738. PB = $\frac{\sqrt{3}}{2} = 0.87$\\
  739.  
  740. OR\\
  741.  
  742. $P(k;k^2+2)$ and B(0;3)\\
  743. BP $\perp$ tangent passing through $y = x^2 + 2$ at P.\\
  744. $m_{tangent at P} = 2k$\\
  745. $M_{BP} = -\frac{1}{2k}$\\
  746. Equation of BP: $y = (-\frac{1}{2k})x+3$\\
  747. $y_p = (-\frac{1}{2k})(k)+3 = 2,5$\\
  748. $\Rightarrow k^2+2=2,5$ and so $ k=\sqrt{0.5}$ and P$(\sqrt{0,5};2,5)$\\
  749. BP = $\sqrt{(\sqrt{0,5}-0)^2+(2,5-3)^2} = \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}=0,87$
  750. {}{}
  751. \\\\
  752.  
  753. \section*{Question 10}
  754.  
  755. 10.1
  756. {Answer}{}
  757. look at memo\\
  758. {}{}
  759. {Memo}{}
  760. memo
  761. {}{}
  762. \includegraphics{graph5.png}
  763. \\\\
  764.  
  765. 10.2
  766. {Answer}{}
  767. $x=42$\\
  768. {}{}
  769. {Memo}{}
  770. $ (49-x) +x+ 8 + 4 + 5+2 + (60-x) + 14 = 100$\\
  771. $ -x + 142 = 100$\\
  772. $x=42$
  773. {}{}
  774. \\\\
  775.  
  776. 10.3
  777. {Answer}{}
  778. P(use only one application) = 27\% \\
  779. {}{}
  780. {Memo}{}
  781. P(use only one application) $= \frac{7+2+18}{100}$\\
  782. $= \frac{27}{100}$ or 27\%
  783. {}{}
  784. \\\\
  785.  
  786. \section*{Question 11}
  787.  
  788. 11.1
  789. {Answer}{}
  790. 2250\\
  791. {}{}
  792. {Memo}{}
  793. $5\cdot 5\cdot 10\cdot 9$\\
  794. =2250
  795. {}{}
  796. \\\\
  797.  
  798. 11.2
  799. {Answer}{}
  800. Codes of two letters and five digits will ensure unique numbers for 700 000 clients.\\
  801. {}{}
  802. {Memo}{}
  803. Codes of two letters and five digits will ensure unique
  804. numbers for 700 000 clients.\\
  805. {}{}
  806. \includegraphics{graph6.png}
  807. \\\\
  808.  
  809. \end{document}
Advertisement
Add Comment
Please, Sign In to add comment
Advertisement