Mathematics > QUESTIONS & ANSWERS > MATH 1311 2016 Hailey bury trial 2016 (All)

MATH 1311 2016 Hailey bury trial 2016

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MATHS METHODS 3 & 4 TRIAL EXAMINATION 2 SOLUTIONS 2016 SECTION A – Multiple-choice answers 1. C 9. C 17. E 2. A 10. D 18. B 3. D 11. E 19. D 4. A 12. C 20. B 5. A 13. C 6. C 14. B 7. E 15 ... . C 8. D 16. D SECTION A – Multiple-choice solutions Question 1 period=π n where n=3 π 4 =π÷ 3 π 4 = 43 The answer is C. Question 2 Sketch the graph of y=(x−1)2 . The range of y=(x−1)2 is y∈[0,∞) . For the function f, the range is restricted to [1,∞) . The graph of f could therefore be So possible domains of f are x∈(−∞,0] or x∈[2,∞) or x∈(−∞,0]∪[2,∞) . Only the first domain is offered. _____________________________________________________________________ © THE HEFFERNAN GROUP 2016 Maths Methods 3 & 4 Trial Exam 2 solutions P.O.Box1180 Surrey HillsNorth VIC 3127 Phone0398365021 Fax 0398365025 [email protected] www.theheffernangroup.com.au THE GROUP HEFFERNAN 1 2 1 ( 2 , 1 ) x y y  ( x  1 ) 2 2 2 1 1 ( 2 , 1 ) 1 ( 2 , 1 ) x O R O R x x y y y2 The answer is A. ©THE HEFFERNAN GROUP 2016 Maths Methods 3 & 4 Trial Exam 2 solutions3 Question 3 The graph of g is an upright quartic which touches the origin so we need a positive x2 term. This eliminates options A, B and C. Note that a must be a negative number since the point (a,0) lies to the left of the origin. Intuitively, we think that we need an (x+a) factor but because a is a negative number, for example –1, this factor would become (x−1) which is incorrect. So we need an (x−a) factor which eliminates option E. The answer is D. Question 4 We have f :[0 , a]→R , f (x )=cos(2(x+ π 3)) . The inverse function f −1 exists if the function f is 1:1. Sketch the graph of y=cos(2(x+ π3)), x≥0. Looking at this graph, we see that it is a 1:1 function for x∈[0, π6] . (It’s actually a 1:1 function for smaller intervals such as x∈[0,12 π ] etc but we want the largest interval possible ie. we want the maximum value of a.) So if the function f with rule f (x )=cos(2(x+ π 3 )) is to have an inverse, then the largest possible value of a is π6 . The answer is A. Question 5 y=loge(ax), a>0 dy dx = a ax = 1x When x=1 a , y=loge(a ×1 a )=loge(1)=0 ©THE HEFFERNAN GROUP 2016 Maths Methods 3 & 4 Trial Exam 2 solutions x y 1 - 1 - 0 . 5 2 3 6 3 2  6 5  , 0 3 c o s 2      y         x      x Equation of tangent is y−0=a(x−1a ) y=ax−1 y-intercept occurs when x=0 y=−1 . The answer is A.4 At (1 a ,0), dy dx =11 a =a [Show More]

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