ACJC JC1 H1 Maths Rev A-2 Complete Solution Graphing Tech

2023 JC1 H1 REVISION SET A-2 COMPLETE SOLUTIONS GRAPHING TECHNIQUES 1 DHS 2014 Promo Q3 ( ) 32 32 32 2 2 2 2 ln 0 2 1 ln 11 ln 2 x x x x x x x x x x x x − − + + = − − + =− − − + =− From graph, since number of points of intersection between the graphs of 32 1y x x x= − − + and 1 ln2yx=− is 3, the number of solutions to 322 2 2 2 ln 0x x x x− − + + = is 3. 2 22 6 2 1 0x y x y+ − + + = 22( 3) ( 1) 9xy − + + = 22( 3) ( 1) 9xy − + + = Coordinates of centre are (3, −1) and radius is 3. 3 y x O 2 3 4− 6 (3, 1)− 1−

4 5 (i) 5 (ii) Circle with centre = Axes of symmetry : 5 (iii) Asymptotes : 6 ax c c abya x b x b ++= = +−− Asymptotes are x = b and y = a Comparing with x = 1 and y = 2, b = 1 and a = 2 2 1 xcy x += − At intersection with x-axis, y = 0 and x = −3/2  0 = 2(−3/2) + c so c = 3 (-0.5, 0) (-1, 0) y x 0 (4, 0) (0, 4) x = - ½ y = - ½ x y (0, 4) O y x y = 2 x ( 13 , 0) (2, 9) (3, 4) y = 213 x− y = 2 36 x y (0, 13)

7 ( )ln 3y kx=− Vertical asymptote : 330kx x k− =  = x-intercept: 40 3 1y kx x k=  − =  = 4 ,0k   ( ) ( )e ln 2 3 1 ln 2 3 exx xx −− =  − = Sketch the graph of y = e−x to the same diagram. From the GC, the two graphs intersect at x = 2.07. 8 (i) Asymptotes are y = 3, x = –1 and x = –1 (ii) 9 y = 4 1 5 22 2 2 2 x xx + =+−− Asymptotes x = 1 and y = 2 Intersections with axes (0, −1/2) and (−1/4, 0) 22 2 log (4 1) log (2 2) 41log 22 x x x x x = + − − += − 41 222 xx x += − y = e−x O y = ln(2x−3) x y x= 3/2 2 1 x = 2.07 x y y = 3 1.26 – 1.59 6 x = 1 x = –1 (0.172, 5.91) (5.83, 3.09)

At intersection, x = −0.769 (3 sf) or x = 2.10 (3 sf) But 4x + 1 and 2x − 2 must both be positive, so x > 1 only. Therefore, x cannot be −0.769, giving x = 2.10 only (3 sf). 10 (a) 3234 2Sub into equation of : 327 22 34Hence the point , lies on .27 a aax H y a a a H  −= = =  +   10 (b) ( ) ( ) ( ) 222 2 1 1 ( 1) 1 1x x k x k x k+ + + = + + + − = + + Curve is quadratic with minimum point at ( )1, k− . Hence, from the graph, there will always be two points of intersection, giving two x-coordinates. Hence there will always be 2 real roots for 2k  . y 2 2 xay xa −= + 2y= x 12− 2xa=− 2a f ( )yx= y x 1x=− 2y= O ( ) 2 1y x k= + +