# Binomial expansion a01 ii 1 2 f 1 x x x 2 3 2 2 2 1 2

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Binomial expansion [A01] (ii) 1 2 f( ) 1 x x x    2 3 2 2 2 1 2 1 2 3 1 ( 1)( ) .......... 2! 3! x x x x x x   2 2 3 3 3 1 2 .......... 1 x x x x x x x Common mistake: Did not expand till 3 x . Complete the square 2 1 3 2 4 x instead. Some applied differentiation to get the expansion, leading to much longer working. Able to link Binomial and maclaurin series and apply small angle approximation [A02] (iii) 2 2 2 2 (1 2 ) sin(3 ) 1 f '( ) sin(3 ) 1 3 3 ........ 1 3 3 ...... x x x x x x x x x x       Majority can’t see the link to part (i); hence either solved it by using applying binomial expansion formula or differentiation. A number of students have no knowledge of small angle approximation by writting sin(3 ) x x Validity of binomial expansion and understand the use of approximation. [A03] (iv) For part (ii) , the validity of the expansion: 2 1 x x From G.C., 1.618 0.618 x . Hence, it is not valid to use the answer in part (ii) to find the approximation value of 1 0 f( ) d x x Badly done. However some were able to score 1 mark by writting 2 1 x x ; but unable to solve it .
Page 13 of 20 8 Applications of Differentiation No. Assessment Objectives Solution Feedback Able to use differentiation to find minimum area. (i) 2 3 10 4 x y 2 40 3 y x 2 3 2 2 A x xy 2 2 2 3 40 3 80 3 2 2 2 3 3 A x x x x x 2 d 80 3 3 d 3 A x x x   2 2 80 3 3 0 3 80 3 x x x x 1/3 3 80 80 3 3 x x OR 2 2 3 d 160 3 d 3 A x x When 1/3 80 3 x , 2 2 d 0 d A x least amount of material used when 1/3 80 3 x x 1/3 80 3 1/3 80 3 1/3 80 3 d d A x -ve 0 +ve Slightly more than half of them were able to solve part (i). Common mistakes: 1 3 10 3 2 x x y 2 3 3 2 A x xy (failed to interpret “open tank”) Weak algebraic manipulation; hence failed to solve for x . 1/3 3 80 80 3 3 x x   Some of them forgot to verify that A is min.
Able to form the equation of volume for prism and apply connected rate of change/ implicit differentiation. (ii) cos 6 h k 2 3 k h    2 1 2 5 5 2 3 3 h V h h d 10 d d d 3 V h h t t d 3 d 30 h t h or d d d d d d h V h t t V d 1 1 d 3 10 3 h t h d 3 d 30 h t h When 1, h d 3 m / d 30 h s t or 0.0577 m/s Average. Very few students were able to score full marks. Common mistake:
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