In an arithmetic progression, the sum of the first five...

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TJC JC1 Promotional Exams 2011_Solutions and Marker’s Comments 1 In an arithmetic progression, the sum of the first five terms is twice the sum of the next five terms. Given that the first term is 60, find the 13 th term. [4]
TJC JC1 Promotional Exams 2011_Solutions and Marker’s Comments 2 In the diagram, P and A are fixed points 500 m apart on horizontal ground. point B is vertically above A , and represents a balloon which is ascending at a steady rate of 3 m s 1 . The balloon is being observed from a moving point Q on the line PA . At time t = 0 s, the balloon is at A and the observer is at P observation point Q moves towards A with steady speed 4 m s 1 . At time t , the angle BQA is radians. Show that d d t 1500 9 t 2 500 4 t 2 . [4] The . The t.
TJC JC1 Promotional Exams 2011_Solutions and Marker’s Comments 3 Using an algebraic method, solve the inequality 2 2 2 1 x x x . [3] Hence solve 2 cos 2 cos2 2 2 cos2 1 where 0 2 . [2] Solution: 2 2 3 2 2 2 2 2 0 1 1 1 x x x x x x x x x x   3 1 0 x x x 0 or 1 3 x x Replacing x by cos2 , cos2 0 or 1 cos2 3 N.A. since 1 cos2 1 Hence, 3 4 2 p p or 5 7 4 4 p p . Marker’s Comments : Some students did cross-multiplication for inequality which was a serious mistake. Most students did not know how to solve inequality involving trigonometrical functions even though GC was allowed.
TJC JC1 Promotional Exams 2011_Solutions and Marker’s Comments 4

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