SL P1 Mock Answers 2015/16

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Nickolas Lawson
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1 SL P Mock Answers 0/6. (a) y-intercept is 6, (0, 6), y 6 N [ mark] (b) valid attempt to solve (M) (c) ( x )( x ) 0, 4 x, one correct answer x, x N N Note: The shape must be an approximately correct concave up parabola. Only if the shape is correct, award the following: for the y-intercept in circle and the vertex approximately on x, below y 6, for both the x-intercepts in circles, for both end points in ovals. Total [7 marks]

2 SL P Mock Answers 0/6. (a) correct approach () d u u, d N [ marks] (b) correct approach () u8 7, listing terms u8 N [ marks] (c) correct approach () S 8 8 ( ), listing terms, 8 () 7() S8 00 N [ marks]. (a) substituting for f (x) x ( ), x 4 ( ) (may be seen in intral) correct intration, x 4 dx = x () substituting limits into their intrated function and subtracting (in any order)(m), ( 4) ( f (x)) dx = (= 6.) N (b) attempt to substitute limits or function into formula involving f (M) ( f (x)) dx, π x 4 dx π (= 6.π) N [ marks]

3 [ marks] SL P Mock Answers 0/6 4. (a) (i) log 7 = N (ii) log8 = 8 N N (iii) log6 4 = (b) correct equation with their three values = log 4 x, + ( ) = log 4 x () involving powers () log x x=4, 4 =4 4 x =8 0. (a) (i) f ( ) = (ii) f () = 0 (accept y = 0 ) (b) N M4//MATME/SP/ENG/TZ/XX/M domain of f is range of f Rf = Df N N [ marks] (R) correct answer x, x [, ] (accept < x <, y ) N [ marks] (c) N Note: Graph must be approximately correct reflection in y = x. Only if the shape is approximately correct, award the following: for x-intercept at, and for endpoints within circles. [ marks]

4 SL P Mock Answers 0/6 6) (a) METHOD approach involving Pythagoras theorem + x =, labelling correct sides on triangle finding third side is (may be seen on diagram) (M) cos A = METHOD approach involving sin θ + cos θ = + cos θ =, x + 69 = cos θ = cos A= AG (M) AG N0 N0 [ marks] (b) correct substitution into cosθ (),, 0 69, cosa = 9 69, () N Total [ marks] 7) 6 (a) summing probabilities to (M) =, x = 0 () p =, p = p = 0 N (b) correct substitution into formula for E( X ) ( p) E( X ) = (.) 0 N () ()

5 SL P Mock Answers 0/6 8) (a) () 6 p 6 N [ marks] (b) multiplying along correct branches () 6 P( C L) N [ marks] (c) multiplying along the other branch (M) adding probabilities of their mutually exclusive paths 6 6 P( L) 4 (M) () N continued

6 SL P Mock Answers 0/6 (d) recognizing conditional probability (seen anywhere) (M) P( C L ) correct substitution of their values into formula P( C L) N () (e) valid approach (M) X B, 4, 4 4,, three ways it could happen correct substitution , , () () N Total [ marks]

7 9. SL P Mock Answers 0/6 8. (a) (i) correct value 0, or 6 p A N (ii) correct equation which clearly leads to p = 6 p = 0, 6 = p (b) METHOD p = AG N0 valid approach x = b a ( 6) (), x = 6 6 (M) correct answers N x =, y = 0 ; (, 0) METHOD valid approach f (x) = 0, factorisation, completing the square x x + = 0, (x )(x ), f (x) = (x ) (M) correct answers N x =, y = 0 ; (, 0) METHOD valid approach using derivative f (x) = 0, 6x 6 (M) correct equation 6x 6 = 0 correct answers N x =, y = 0 ; (, 0) (c) x = N [ mark] (d) (i) a = N (ii) h = N (iii) k = 0 N

8 SL P Mock Answers 0/ (a) derivative of x is (must be seen in quotient rule) () derivative of x + is x (must be seen in quotient rule) correct substitution into quotient rule (x + )() (x)(x), (x + ) 4x (x + ) (x + ) which clearly leads to given answer x +0 4x, x +0 4x (x + ) x 4 +0x + () 0 x f (x) = (x + ) AG N0 (b) valid approach using substitution or inspection (M) u = x +, du = xdx, ln (x + ) x dx = x + du u () du = ln u + c u () ln (x + ) + c N4

9 SL P Mock Answers 0/6 (c) correct expression for area () q ln x q ( + ) x, x + dx substituting limits into their intrated function and subtracting (in either order) ln (q + ) ln + ln (q + ) ln0, ln q + 0 equating their expression to ln7 (seen anywhere) ln (q + ) ln0 = ln7, ln q + 0 = ln7, ln (q + ) = ln7 + ln0 correct equation without logs q + 0 = 7, q + = 70 q = 6 (M) () (M) () () q = 6 N Note: Award A0 for q = ± 6. [7 marks] Total [ marks]

Function Practice. 1. (a) attempt to form composite (M1) (c) METHOD 1 valid approach. e.g. g 1 (5), 2, f (5) f (2) = 3 A1 N2 2

Function Practice. 1. (a) attempt to form composite (M1) (c) METHOD 1 valid approach. e.g. g 1 (5), 2, f (5) f (2) = 3 A1 N2 2 1. (a) attempt to form composite e.g. ( ) 3 g 7 x, 7 x + (g f)(x) = 10 x N (b) g 1 (x) = x 3 N1 1 (c) METHOD 1 valid approach e.g. g 1 (5),, f (5) f () = 3 N METHOD attempt to form composite of f and g