Higher Mathematics (2014 on) Expressions and Functions. Practice Unit Assessment B

Transcription

1 Pegass Educational Publishing Higher Mathematics (014 on) Epressions and Functions Practice Unit Assessment B otes: 1. Read the question full before answering it.. Alwas show our working.. Check our paper at the end if ou have time.

2 FRMULAE SHEET Scalar Product: a.b = a b cos θ, where θ is the angle between a and b. a or a.b = a 1b1 + ab + ab where a = a a 1 and b = b b b 1. Trigonometric formulae: sin ( A ± B) = sin Acos B ± cos Asin B cos ( A ± B) = cos Acos Bm sin Asin B sin A = sin Acos A cos A = cos = cos A sin = 1 sin A 1 A A

3 Answer all the questions E&F Assessment Standard (a) Simplif log5 log5. (b) Epress log 8 5 n a + log n a in the form k n a log. (1) (). Solve log 4 ( + 5) =. () E&F Assessment Standard 1.. Epress 7 sin + cos in the form k sin ( + a) where k > 0 and 0 a < 60. (4) 4. The diagram below shows two right-angled triangles. cos. Find the eact value of ( ) (4) 5. Show that ( cos )( 9 7cos ) = 49sin +. () (#.1 ) E&F Assessment Standard π Sketch the graph of = a sin + for 0 π and a > 0, clearl showing the maimum and minimum values and where it cuts the -ais. () 7. The diagram shows the graph of f ( ) = with a maimum turning point at (4, 1) and a minimum turning point at (, 4). 1 (4, 1) 4 Sketch the graph of = f ( +1 ) + 4 (, 4). ()

4 8. The diagram shows the graph of = acos( b) + c. π π -5 Write down the values of a, b and c. () 9. The diagram shows the graph of b ( a) = log. 1 (, 0) (6, 1) 6 Determine the values of a and b. () 10. The functions f and g, defined on suitable domains, are given b f ( ) 7 4 g ( ) = + 1. A third function h ( ) is defined as h ( ) = g( f ( ) ). = and (a) Find an epression for h ( ). (b) Show that the largest domain for ( ) this is. h is given b and eplain wh () (1) (#. ) A function is given b f ( ) =. Find the inverse function f ( ). ()

5 E&F Assessment Standard A PE teacher is laing out cones for a fitness test. He has three cones and needs to set up the test to meet the following conditions: the distance from the start (S) to cone A is three fifths of the distance from S to cone B. The three cones should be in a straight line. Relative to a suitable aes, the centre of each cone can be represented b the points S(, 10, 1), A(11,, 4) and B(17, 10, 6) respectivel. S(, 10, 1) A(11,, 4) B(17, 10, 6) Has the teacher laid the cones correctl? You must justif our answer. (4) (#.1 ) (#. ) 1. The points M, N and P lie in a straight line, as shown. P divides MN in the ratio :7. M( 8, 5, ) P N(19, 1, 7) Find the coordinates of P. () 14. DEFGH is a pramid with a rectangular base DEFG. H G D F E The vectors ED, EF and EH are given b: ED = 8i + j + k EF = i + 10j k EH = i + 7j + 7k Epress HD in component form. () Turn over for question 15

6 15. Points P, Q and R have coordinates A(10, 5, ), B(, 4, 1) and C(4, 8, ). C B A Find the size of the acute angle ABC. (5) End of question paper

7 Epressions and Functions, Practice Assessment B Qu. Points of Epected Response Illustrative Scheme E&F utcome 1.1: Logs and Eponentials 1(a) Use log log =log 1(b) Use log +log =log R log =log Simplif to log log log R 8log +5log 1log Start to solve Solve correctl E&F utcome 1.: Trigonometr Epand sin+ Compare coefficients Process k 4 Process a 4 +5=4 =59 sincos+cossin stated eplicitl cos=7 and sin= stated eplicitl = 58 = 5 4 Process missing sides 4 Epand cos and begin subs Complete substitution Process and 19 cos+sinsin + or equivalent 5 Epand LHS Substitute for sin Simplif and complete #.1 Valid strateg 81 49cos sin 49sin + #.1 Know to use identit 11 + #.1

8 Qu. Points of Epected Response Illustrative Scheme E&F utcome 1.: Trigonometr 6 Ma/min correct n graph ma value a and min value a -intercepts Correct shape and Start and finish at a ote: a The -intercept is not required. 7 Correct horizontal translation Correct vertical translation Correctl annotated diagram otes: For correctl drawn and annotated graphs of =+1 or = 1+ award points of process. For =+1+ ( ) where - coordinates are consistent award 1 point. For =++ ( 1) where - coordinates are consistent award 1 point. -a -coordinates correct -coordinates correct Correct shape and annotation (, 4) (-, -1) 8 Find a Find b Find c 9 Find a Find b a = 4 b = c = -1 a = 1 b = 5 10a Interpret composite process Complete function h=7 4 h= b Interpret domain of h #. Eplain a solution leading to #. In real numbers the square root of a negative number cannot be found. 11 Start inverse process State inverse function += = 16 + #.

9 Qu. Points of Epected Response Illustrative Scheme E&F utcome 1.4: Vectors 1 #.1 Strateg #.1 Show points are not collinear 9 Find vector between two points e.g. SA = 1= 4 1 Find second vector and interpret 15 multiple e.g.sb = 0= Complete proof of collinearit 5SA =SB hence the vectors are parallel. S is a common point so the points S, A and B are collinear. 4 Interpret ratio 4 SA :SB are in the ratio :5 #. Eplain solution in contet #. The orienteering course has been laid out correctl as both conditions have been met. 1 Find component vector MN Use correct ratio Process vectors and find point P 7 MN = MP = 18 9 P = (, 1, 0) Alternative Method: Section formula Begin substitution Complete substitution Process and state point P 14 Find pathwa for HD Identif EH Complete calculation of HD 15 Find component vectors 4 5 Use scalar product Process scalar product Process magnitudes Find angle Alternative Method: Section formula = = P = (, 1, 0) HD = EH +ED R 5 5 Do not award for ( 9, 5, 5) 1 6 = AB 1 and AC = cos= and 06 θ = 61 or 1 07 (radians) #.1 + #.