JPJC 2021 J1 H2 Maths promo Exam solutions

Jurong Pioneer Junior College H2 Mathematics JC1-2021 End-of-Year Exam P1 solution Question Solution Remarks 1(i) Set min t= 2 in window setting Need to label and exclude the end-point 1(ii) d d 1 2e , 2 edt dt t tx y     d 2 e d 1 2e t t y x   At P, d 1 d 3 y x  Gradient of normal at P = 3 At P, x = 2, y = 1 The equation of normal is y – 1 = 3(x – 2) y = 3x + 7 Note that d (e ) edt t t  Note that e0 = 1 2(i) (1, 3): 3 (1)4 a b c    (5,21): 5 21 (2)2 a b c    2 d d ( 3) y a bx x   At x = 5: d 0 0 (3)d 4 y a bx      Using GC: a = 8, b = 2, c = 7 So the equation of C is 8 2 73y x x   Since there are 3 unknowns, we need 3 equations. Note that C has a stationary point (5, 21) means that C (i) passes through (5, 21); (ii) d 0d y x  at x = 5 (ii) The two equations of asymptotes are: x = 3 y = 2x – 7 Vertical asymptote: x = 3 Oblique asymptote: y = bx + c (1.73,3.39) x y O

2 Question Solution Remarks 3 2 1 3tan 4x y y   Differentiate wrt x,   2 2 d 1 d 2 0d 1 d y yx y xx y x         2 2 d 1 2d 1 y x xyx y       2 2 d 2 d 1 1 y xy x xy        2 2 2 2 2 1 1 xy y x x y     when y = 1 2 1 3(1) tan 1 4x   2 3 4 4x    2x  x  Since x > 0, x    2 2 2 1 1d d 1 (1) y x        = 4 1 2   (exact) We use degree only for vectors (or otherwise stated). Hence 1tan 1 4   and not 45. 4(i) The semi-circle must intersect with x-axis You need to indicate the vertical asymptote of the ln graph (GC does not show) (ii) The point of intersection is (2.70, 1.31) So the solution is 1< x < 2.70 (iii) Replace x by x2: 1< x2 < 2.6997 Method 1 Using Graph, 1.64 < x < 1.64 Proper method (in getting final answer) must be shown in order to gain full credit.              x y x=-1 (2.70,1.31) O (3,0) (3,0) (0,3) y=2.6997 1y 1.64 1.64

3 Method 2 Since x2  0, 1< x2 < 2.6997 can be simplified to x2 < 2.6997 (x – 1.64)(x + 1.64) < 0 1.64 < x < 1.64 It is incorrect to reject the part on 1< x2. Use of identity a2 – b 2=(a + b)(a – b) Question Solution Remarks 5(i) Let y =   21 e 13 x  2e 1 3 2 ln(3 1) ln(3 1) 2 x y x y x y          f1(x) = ln(3x + 1) + 2, x > 0 1 ffD R  (ii) As the graph of f 1 intersects the graph of f in the line y = x, the solution of f(x) = f1(x) is the same as the solution of f(x) = x. Hence   2 2 1 e 13 e 3 1 x x x x       Use GC, the solution is x = 4.72 Clear explanation must be given as well as how 2e 3 1x x   is obtained. Intersection point satisfies x > 2 (iii) Rf = (0,  and Dg = (, ) So Rf  Dg, gf exists gf(x) = g(   21 e 13 x  ) =   221 e 1 19 x  