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  