Mathematics Higher Block 3 Practice Assessment A
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1 Mthemtics Higher Block 3 Prctice Assessment A Red crefully 1. Clcultors my be used. 2. Full credit will be given only where the solution contins pproprite working. 3. Answers obtined from reding from scle drwings will not receive ny credit. Questions Assessment Stndrd Subskill detils 1 RC1.3 Differentiting k sin x, k cos x 2 RC1. Integrting functions of the form f ( x) ( x q) n, n 1 3 RC1. Integrting functions of the form f ( x) pcos x nd f ( x) psin x RC1. Clculting definite integrls of polynomil functions with integer limits 5 EF1. Determining the resultnt of vector pthwys in three dimensions 6 EF1. Working with collinerity 7 EF1. Determining the coordintes of n internl division point of line 8 EF1. Evlute sclr product given suitble informtion nd determine the ngle between two vectors 9 EF1.1 Simplifying numericl expression, using the lws of logrithms nd exponents 10 EF1.1 Solving logrithmic nd exponentil equtions, using the lws of logrithms nd exponents 11 EF1.2 Converting cos x bsin x to kcos( x) or ksin( x ), α in 1 st qud k 0 12 RC1.2 Solve trigonometric equtions in degrees, including those involving trigonometric formule or identities, in given intervl
2 1 Differentite the function f ( x) 7sin x with respect to x. (1) RC1.3 2 (#R 2.1, 2) RC1. g ( x) ( x 7), find g( x), x 7. 3 Find 8cos d. (1) RC1. Find 1 3 ( x 5) dx. () 3 RC1. 5 ABCDE is pyrmid with rectngulr bse BCDE. A C B D E The vectors EB, ED, EA re given by: EB 16 ; ED ; EA Express BA in component form. (3) EF1.
3 6 An ir trffic controller is lnding irplnes, he needs to ensure tht: the next 3 irplnes pproching the runwy re in stright line the distnce between irplne 2 nd irplne 3 is three times the distnce between irplne 1 nd irplne 2 Reltive to suitble xes, the position of ech irplne cn be represented by the points A (-2, 5, 7), B (2, 3, 10), nd C (1, 3, 19) respectively. A (-2, 5, 7) B (2, 3, 10) C (1, -3, 19) irplne 1 irplne 2 irplne 3 Hs the ir trffic controller ligned the plnes correctly? You must justify your nswer. (#E 2.1,, #E 2.2 ) EF1. 7 The points P, Q nd R lie in stright line, s shown. Q divides PR in the rtio 2:3. Find the coordintes of Q. R(12, 11, -12) P(2, 1, 3) Q (3) EF1. 8 The digrm shows vectors DE nd DF. D, E nd F hve coordintes D(-3,, 5), E(2, 6, -3) nd F(2, 0, 3). F D E Find the size of the cute ngle EDF. (5) EF1.
4 9 () Simplifylog52 p log56q. (1) EF1.1 (b) Express 7 log x log x in the form k log x. (2) EF Solve log 2( x 1) (2) EF Express 5sin x cos x in the form k sin( x ) where k 0 nd (5) EF Given 5sin x cos x 1cos( x 51.3), solve 5 sin x + cos x = 3.2, 0 < x < 360 (3) RC1.2
5 Question Points of expected response Illustrtive scheme 1 1 differentites correctly 2 #2.1 recognises s differentil eqution nd hence knows to integrte 1 d (7sin x) 7cos x dx #2.1 evidence of setting up integrl form g( x) x 7 dx 1 strts integrtion correctly 1 ( x 7) 3 2 completes integrtion correctly c ** 3 ** To chieve block 2 question 10 the constnt of integrtion must pper t lest once ssocited with correct integrl in either question 10 (block2), 2 (block 3) or 3 (block 3). 3 1 integrtes correctly 1 8sin c ** 1 strts integrtion 1 ( x 5) 2 completes integrtion substitutes limits 1 1 ( 1 5) ( 3 5) 3 evlute definite integrl 60 Notes: 3 nd re only vilble s result of n ttempt t integrtion. Simply substituting these into the integrnd gins no mrks. Cndidtes who differentite the originl expression cnnot gin 3 nd. 5 1 recognise pthwy for PT 1 BA EB EA 2 identify EB vector 3 complete clcultion for BA Note: Do not wrd 3 for (17, -3, 9) BE EB BA
6 6 #2.1 select pproprite strtegy to show collinerity 1 interpret vector 2 interpret multiple of vector 3 complete proof interpret rtio #2.2 Explin solution in context #2.1 show they re colliner by ttempting to demonstrte tht for two pproprite vectors, one is the sclr multiple of the other 1 eg AB eg 12 BC 6 3AB 9 3 BC 3AB hence vectors re prllel, but B is common point so A, B nd C re colliner. interpret rtio BC: AB 3:1 #2.2 Yes, the ir trffic controller hs ligned the irplnes correctly (with sttement relted to previous working). Note: Any pproprite combintion of vectors is cceptble. 7 1 find vector components 1 10 PR uses correct rtio 3 processes vectors nd finds coordintes of S PQ Q (6, 5, 3) NB If cndidtes use the Section formul: 1 begins substitution 2 completes substitution 3 s 3 bove
7 Note: 3 is only vilble if expressed s coordinte. 8 1 find vector components DE 2, DF use sclr product 2 cos EDF DE. DF DE. DF 3 process sclr product 3 DE. DF (25 816) 33 process PQ nd PR ie clculte mgnitudes DE 93 nd PR 5 5 find ngle 5 ngle = 1.0 rdins or 59.3 Note: If the evidence for 2 does not pper explicitly, then 2 is only wrded if working for 5 is ttempted. 9() 1 use log x log y log xy log (2p6 q) log 12 pq (b) 2 log m x mlog x 7log x log x OR log x 2 3 m n OR log x log x log x x m n 3 simplify to klog x 3 3log x Note: For 1 the finl nswer must be simplified to log 15pq strts to solve 1 ( x 1) 2 2 solves correctly 2 x 15 Note: If cndidte simply obtins the solution by inspection, wrd both points.
8 11 1 expnd k sin( x ) 2 compre coefficients 3 process k process 5 stte finl form 1 k sin xcos k cos xsin stted explicitly 2 kcos 5 nd ksin stted explicitly 3 k sin x 38.7 Note: For 1 k(sin xcos cos xsin ) is cceptble sets up cos (..) correctly 2 obtins first ngles 3 obtins finl ngles 1 cos( x 513) x , x 111.3, 351.3
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