Leaving Cert Maths Samples

_________________________________________ leavingcertsolutions.com ________________________________________ 2015 Question 7 – Higher Paper 1 Solution (a) (i) Given 0 is the origin (0,0). To show d = 0, we know ( 5 marks ) 3 2 (0) 0 0.0024(0) 0.018(0) (0) 0 0 0 0 f c d d d = ⇒ + + + = ⇒ + = ⇒ = (ii) 3 2 ( ) 0.0024 0.018 f x x x c = + + We are told (–5,0.15) is on the curve ( 5) 0.15 f ⇒ − = ∴ 3 2 0.0024( 5) 0.018( 5) 0.15 c − + − + = 0.3 0.45 0.15 0.15 0.15 0 c c c − + + = ⇒ + = ⇒ = ( 5 marks ) (b) (i) 3 2 ( ) 0.0024 0.018 f x x x = + 2 ( ) 0.0072 0.036 dy f x x x dx ′ = = + 2 ( 4) 0 .0072( 4) 0.036( 4) 0.029 f ′ − = − + − = − ( 10 marks ) (ii) ( 4) f ′ − gives the slope of the tangent at 4 x = − The slope is the tangent of the angle the line makes with + direction of the x axis 1 tan 0.029 tan ( 0.029) 178.34 θ θ θ − = − ⇒ = − ⇒ = ( 5 marks ) (c) To show the point of inflection is at (–2.5,0.075), we show ( 2.5) 0 f ′ − = ( ) 0.0144 0.036 f x x ′′ = + ( 2.5) 0.0144( 2.5) 0.036 0 f ′′ − = − + = ∴ There is a point of inflection at (–2.5,0.075) ( 10 marks ) (d) (i) We know (–5,0.15) is on the curve from part (a)(ii). We now must show ( ) 5, 0.15 x y − − − + is on the curve Replace x by –5 and y by 0.15: ( ) ( ) 5 5, 0.15 0.15 (0,0) − − − − + = which is on the curve. ∴ True ( 5 marks ) (ii) There are a few ways to do this ( ) 2.5,0.075 ( 5, 0.15) Q P x y − − − − + The translation 2.5, 0.075 x y + − maps P Q → The same translation will map Q R → ( , ) x y → ∴ Image is ( , ) x y Or, since (–2.5,0.075) is the mid point of [RS], then 5 2.5 5 5 2 x a x a x a a x − − + = − ⇒ − − + = − ⇒ − = − ⇒ = then 0.15 0.075 2 y b − + + = 0.15 0.15 y b b y ⇒ − + + = ⇒ = Image is ( x , y ) ( 10 marks ) Comment: Long and tedious. Why do they make the quadratics so difficult? R ( a , b ) (–2.5,0.075) (– x – 5,– y + 0.15 S • We need to show ( 5) 0.15 f x y − − = − + Write ( ) f x as 2 0.0024 ( 7.5) x x + Find 2 3 ( 5) 0.0024( 5) ( 2.5) 0.0024 0.018 0.15 ( ) 0.15 0.15 f x x x x x f x y − − = − − − + = − − + = − + = − + ∴ ( 5, 0.15) x y − − − + is on the curve R •