2024 Y5 June Holiday Assignment 1 - Solutions

1 2024 Y5 H2 Math Holiday Assignment 1 Solutions Qn Suggested Solution Comments 1 Curve B : 2 2 8 17y y xy= + − Differentiate with respect to x, d d d2 2 8 8d d d d4 d 1 4 y y yy x yx x x yy x y x = + + = −− 2, 1 d 4(1) 1 d 1 1 4(2) 2xy y x == = =−−− d 4 1To find ( , ), let d 1 4 2 1 4 8 91 subst into eqn of curve4 ybab x b a b a b ba = =−−−  − − =− − =  ( ) 2 2 2 912 8 17 4 2 2 9 1 17 17 17 1 ( 1 is given earlier) 9( 1) 1 2.54 bb b b b b b b b by a −= + −  = + − − = =− = −−= =− As there’s only one “a” term in the curve eqn, it’s easier to subst 91 4 ba −= instead of 41 9 ab += . As (2,1) is already provided in the question, students should reject 2, 1ab== & accept 2.5, 1ab=− =− as the final answer. Qn Suggested Solution Comments 2(a) Since f ( 2) 0−= , f (1 3 ) 0 1 3 2 1 x x x −= − =− = Since f(0) = 0, f (1 3 ) 0 1 3 0 1 3 x x x −= −= =  The roots of the eqn are 1 3 and 1. Problem solving: Do not rush in to solve the question. The question did not ask for the sketch of f (1 3 ).yx=− Thus, you may want to think of a simpler approach to solve the question. Solving this question can be done by observing the values that gives you the roots on the f ( )yx= . Concept used: If f( ) 0,a = then a is a root.

2 (b) 3f (| |) 5 0 5f (| |) 3 x x −= = Sketch the graph of f (| |)yx= and 5.3y= From the graph, there are 2 distinct real roots. Concept used: Sketch of f (| |)yx= (c) Concepts used: 1. Sketch of 1 f ( )y x= from f ( )yx= . [see lecture notes] 2. 1f ( ) 0 0 f ( )x x   and vice versa. E.g. 3f ( ) :xx= 3 3 12 0 0 2   . [You can use this to check your sketch when you are done] x = ‒2 x = 0 (−3,0) x y

3 Qn Suggested Solution Comments 3(a) Since 2 2 11 24x t t t = + = + − and 22 t−   , the smallest value of x is 1 4− . Thus there will be no curve when 1 4x− . - End points should be clearly labelled and graph should pass through the origin - Explain clearly why there will be no curve when 1 4x− , by explaining why there are no t-values that give 1 4x− (using discriminant or completing the square, etc) or why smallest value of x is 1 4− (by completing the square or drawing a graph, etc) (b) 2 2 2 dd 1 2 , 2 3dd d 2 3 d 1 2 d 2 3 d 1 2tp xy t t ttt y t t xt y p p xp= = + = + += + += + - We should not sub t = p from the start and differentiate w.r.t p because p is taken to be an unknown constant (c) At A, x = 2 2 2 2 20 ( 2)( 1) 0 2 or 1 tt tt tt t += + − = + − = =− At 322, ( 2) ( 2) 2ty=− = − + −  At 321, (1) (1) 2ty= = + = (shown) 1t= at A (shown) Alternatively, At A, x = 2 and y = 2 2 2 2 20 ( 2)( 1) 0 2 or 1 tt tt tt t += + − = + − =  =− ( ) ( ) 23 32 2 2 2 20 ( 1) 2 2 0 1 no real roots fo