Tiêu đề 2024 M2 detailed solution.pdf ✅

2024-DSE-MATH-EP(M2)-1 1. Let O, A and B be three distinct points in a rectangular coordinate plane such that OB OA AB = + , where O is the origin. (a) Describe the geometric relationship between O, A and B. (b) Suppose that OA = + 3 4 i j and OB = 20 . Find AB . (3 marks) Solution (a) O, A and B are collinear and A lies on line segment OB. 1A (b) Let OB kOA k k = = + 3 4 i j , where k > 0. 1M OB = 20 2 2 (3 ) (4 ) 20 k k + = 2 25 400 k = k = 4 OB OA = = + 4 12 16 i j AB OB OA = − = + 9 12 i j 1A
2024-DSE-MATH-EP(M2)-2 2. Let x > 0. (a) Find 0 lim h x h x → h + − . (b) Find d d x e x from first principles. (5 marks) Solution (a) 0 lim h x h x → h + − 0 lim h x h x x h x h x h x → + − + + =  + + 1M 0 lim ( ) h x h x h x h x → + − = + + 1 2 x = 1 (b) 0 d lim d x h x h e e x h + → − = 1M 0 ( 1) lim x x h x h e e x h x x h x h + − → − + −   =     + −   1 2 x e x   =     1M 2 x e x = 1A
2024-DSE-MATH-EP(M2)-3 3. Let m be a positive integer. Consider the expansion of 24 m 2 x x   −     in descending powers of x. (a) Find the first 3 terms of the expansion. (b) If the 19th term of the expansion is a constant, find m and the coefficient of x 60 in the expansion. (5 marks) Solution (a) Let T(r) be the (r + 1)th term in the expansion of 24 m 2 x x   −     . 24 24 24 24 ( 1) 2 T( ) ( ) ( 2) r m r r m m r r r r C x C x x − − +   − = = −     1M T(0) = 24 0 24 0 ( 2) m C x − = 24m x T(1) = 24 1 24 1 1 ( 2) m m C x − − − = 23 1 48 m x − − T(2) = 24 2 24 2( 1) 2 ( 2) m m C x − + − = 22 2 1104 m x − 1A (b) 24 18 24 18( 1) 24 18 6 18 T(18) ( 2) ( 2) 18 18 m m m C x C x − + − = − = − 6m – 18 = 0 m = 3 1A When the power of x = 60, 72 – 4r = 60 r = 3 The coefficient of x 60 = 24 3 3 C ( 2) − 1M = –16192 1A
2024-DSE-MATH-EP(M2)-4 4. (a) Let 0 2 x    . Prove that csc2 cot2 tan x x x − = . (b) Solve the equation (csc3 cot 3 )(csc cot ) 1     − − = , where 6 3      . (5 marks) Solution (a) csc2 cot2 x x − 1 cos2 sin2 sin2 x x x = − 2 2 sin 2 sin cos x x x = 1M sin cos x x = = tanx 1A (b) (csc3 cot 3 )(csc cot ) 1     − − = 3 tan tan 1 2 2   = 1M 3 1 tan 2 tan 2   = 3 tan tan 2 2 2      = −     1M 3 2 2 2    = − 4   = 1A