Onlinetuition.com SPM Additional Mathematics
Question 8 (3 marks):It is given that the sum of the first n terms of an arithmetic progression is S n = n 2 [ 13−3n ] Find the nth term.Solution: S n = n 2 [ 13−3n ] S n−1 = n−1 2 [ 13−3( n−1 ) ] = 1 2 ( n−1 )( … Read more
Question 4 (3 marks):Diagram 4 shows a trapezium ABCD.Diagram 4 Given p ˜ =( 3 4 ) and q ˜ =( k−1 2 ), where k is a constant, find value of k. Solution: p ˜ =m q ˜ ( 3 4 )=m( k−1 2 ) ( 3 4 )=( mk−m 2m ) mk−m=3 ………. ( 1 ) 2m=4 ……………. ( 2 ) From( 2 ): 2m=4 m=2 … Read more
Question 3 (3 marks):Diagram 3 shows vectors AB → , AC → and AD → drawn on a square grid with sides of 1 unit.Diagram 3 ( a ) Find | − BA → |. ( b ) Given AB → = b ˜ and AC → = c ˜ , express in terms of b ˜ and c ˜ … Read more
Question 10:(a) It is found that 60% of the students from a certain class obtained grade A in English in O level trial examination.If 10 students from the class are selected at random, find the probability that(i) exactly 7 students obtained grade A.(ii) not more than 7 students obtained grade A.(b) Diagram below shows a … Read more
Question 8:(a) A survey is carried out about red crescents in a school.It is found that the mean of the number of red crescents is 315, the variance is 126 and the probability that a student participate in red crescents is p.(i) Find the value of p.(ii) If 8 students from the school are chosen … Read more
Question 9: Diagram 4 shows part of a curve x = y2 + 2. The gradient of a straight line QR is –1.Find (a) the equation of PQ, [2 marks] (b) the area of shaded region, [4 marks] (c) the volume of revolution, in terms of π, when the shaded region is rotated through 360o about the … Read more
Question 7:Diagram below shows a curve y= 1 4 x 2 +3 which intersects the straight line y = x + 6 at point A.(a) Find the coordinates of A.(b) Calculate(i) the area of the shaded region M,(ii) the volume generated, in terms of π, when the shaded region N is revolved 360o about the … Read more
Question 13: Given that y= x 2 2x−1 , show that dy dx = 2x( x−1 ) ( 2x−1 ) 2 . Hence, evaluate ∫ −2 2 x( x−1 ) 4 ( 2x−1 ) 2 dx . Solution: y= x 2 2x−1 dy dx = ( 2x−1 )( 2x )−x( 2 ) ( 2x−1 ) 2 = 4 x 2 −2x−2 … Read more
Question 9: Given y= 5x x 2 +1 and dy dx =g( x ), find the value of ∫ 0 3 2g( x )dx. Solution: Since dy dx =g( x ), thus y= ∫ g( x ) dx ∫ 0 3 2g( x )dx=2 ∫ 0 3 g( x )dx =2 [ y ] 0 3 =2 [ 5x x 2 +1 … Read more
Question 5: Given ∫ ( 6 x 2 +1 )dx=m x 3 +x +c, where m and c are constants, find (a) the value of m. (b) the value of c if ∫ ( 6 x 2 +1 )dx=13 when x=1. Solution: (a) ∫ ( 6 x 2 +1 )dx=m x 3 +x +c 6 x 3 3 +x+c=m x 3 +x+c 2 x 3 +x+c=m x 3 +x+c Compare the both sides, ∴ m=2 (b) … Read more