S4 AM WA2 Solution

2026 Sec 4 Additional Math WA2 Mark Scheme S/No. Answer Success Criteria 1(a) 23 22 2 d sin( ) cos tand 2 cos( ) 3cos sin sec x xxx x x xx x  ++ = −+ I can use the product rule to find the derivative. I can find the derivative of trigonometric functions. Content L Complexity M Context L Response Strategy Routine Assessment Objective AO1 S/No. Answer Success Criteria 1(b) 5 2 3 3 2 1 4 2 4 23 d 2 3 d 23 14 2 6 2 xx xx xxx xx c x c x − − − = − = −+ − = ++ ∫ ∫ Or 14 262 x xc − ++ I can integrate power functions. Content L Complexity M Context L Response Strategy Routine Assessment Objective AO1

S/No. Answer Success Criteria 2(a) 2 2 2ln 34 ln( 2) ln(3 4) xy x xx −= + = −− + 2 d1 6 d 23 4 yx xx x= −−+ When 4x= , 2 d 1 6(4) d 4 2 3(4) 4 1 26 y x = −−+ = I can simplify the expression using the quotient law of logarithm. I can find the derivative of ln functions. Content L Complexity M Context L Response Strategy Routine Assessment Objective AO1 S/No. Answer Success Criteria 2(b) d dd ddd 1d5 26 d d 5 26d 130 units/s y yx t xt x t x t = × −= × = −× =− Rate of change of x is -130 units/s. I can simplify the expression using the quotient law of logarithm. I can find the derivative of ln functions. Content L Complexity M Context L Response Strategy Routine Assessment Objective AO1

S/No. Answer Success Criteria 3 2 22 2 22 2 22 d ( 1) (2 ) d ( 1) ( 1 2) ( 1) ( 1) ( 1) xx x x y ex e x xx ex x x ex x −− − − − +−= + − ++= + −+= + Since 22( 1) 0x +> , 0 (or e 0)xxe−−−< > and 2( 1) 0x+> , 1x≠− . 2 22 ( 1) 0( 1) xex x −−+ <+ , hence y is a decreasing function for all values of x. I can find the derivate of exponential functions. I can use quotient rule to find d d y x . I can conclude that 22( 1) 0x +≥ , 0xe−−< and 2( 1) 0, 1xx− > ≠− . Content L Complexity M Context M Response Strategy Routine Assessment Objective AO3

S/No. Answer Success Criteria 4(a) 1by ax x= +− 2 d d yb axx= − 2 10 1 52 xy yx += = −+ Gradient of normal 1 2=− Gradient of tangent 2= When x = 1, 2 d 2d 21 2...........(1) y x ba ab = −= −= When x = 1, y = 3 1 31 4.................. (2) by ax x ab ab = +− =+− += (2) - (1), 22 1 b b = = Sub in (2) 14 3 a a += = I can find d d y x . I can form an equation with the value of x and d d y x . I can form an equation with the coordinates of the point the curve passes through. I can solve the equations simultaneously to find the values of the unknown. Content L Complexity M Context M Response Strategy Routine Assessment Objective AO2

4(b) Equation of normal: 13 ( 1)2 17 22 yx yx −= − − = −+ 22 2 1 1731 22 26 2 7 7 9 20 (7 2)( 1) 0 2 or 1 (given)7 xxx x xx x xx xx xx + −= − + + − = −+ − += − −= = = Sub in equation of normal: 12 7()27 2 47 14 y y = −+ = The coordinates are 2 47,7 14   I can find d d y x . I can form an equation with the value of x and d d y x . I can form an equation with the coordinates of the point the curve passes through. I can solve the equations simultaneously to find the values of the unknown. Content L Complexity M Context M Response Strategy Routine Assessment Objective AO2

S/No. Answer Success Criteria 5(a) 164 2 8 2 [2 (4 )] 2 64 4 8 4 16 2 xr x r xr r x rr π π π =+++ =++ =−− I can form an equation by equating the perimeter of the figure to the length of the wire. I can find the area of the rectangle and the semi-circle. Content L Complexity L Context M Response Strategy Routine Assessment Objective AO2 S/No. Answer Success Criteria 5(b) ( ) ( ) 21 4 282A r xrπ= + ( ) ( ) 2 2 2 22 22 1 16 162 8 16 16 2 8 256 32 16 256 32 8 r xr r r rr rr rr rr r π ππ ππ π = + = + −− =+−− =−− I can form an equation by equating the perimeter of the figure to the length of the wire. I can find the area of the rectangle and the semi-circle. Content L Complexity L Context M Response Strategy Routine Assessment Objective AO2

S/No. Answer Success Criteria 5(c) d 256 64 16d A rrr π=−− For stationary value, d 0d A r = ( ) 256 64 16 0 64 16 256 64 16 256 256 64 16 16 4 2.24 (2 d.p.) rr rr r r r r π π π π π −− = += += = + = + = 2 2 d 64 16d A r π= −− ( )0< This value of x gives a maximum value for the area of the diagram. I can find d d A r . I can find the value of r for which d 0d A r = . I can find the stationary value of A. I can show this value is a maximum value using the first or second derivative test. Content L Complexity L Context M Response Strategy Routine Assessment Objective AO2