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   , x > 2 Rgf = (1,  Justification of Rf  Dg should be given (i.e. R f = (0,  & Dg = (, ) Dgf = Df

4 Question Solution Remarks 6(a) u v  0 u v   or u is a zero vector or v is a zero vector Question does not indicate the vector is non- zero vector. (b)(i) Since N lies on l,  1 3ON   a 2b a   for some  Since AN is perpendicular to l,  2 0AN  b a    2 2 2 03              a b b a 2 2 2 4 03 3                        a b a a b b 2 a b Since a and b are perpendicular, 0  a b 2 4 03          a a b b 2 22 4 03 a b       2 (4) 4 (1) 03        1 3   1 1 2 2 3 3 3 3ON     a 2b a a b   Use the concept that a point lies on a line, its position vector satisfies the equation of the line. (ii) Area of triangle OAN = 1 2 OA ON     = 1 2 2 2 3 3      a a b =  1 3  a a b = 1 3   a a a b = 1 3  0 a b = 1 sin 903 a b  = 1 (2)(1)(1)3 = 2 3 Apply area of triangle using cross product.

5 Question Solution Remarks 7a(i) y = f 2x + 1 Note that y = f  2x is a scaling parallel to the x-axis by a factor of 1 2 . Label all intercepts and asymptotes clearly (ii) y = )f( 1 x Label all intercepts and asymptotes clearly. Note that the x- intercept at (1,0) should not be drawn as a ‘sharp’ point, it should be drawn like a min. point. (b) A B 1 1 1 1 1 1 ( 5) 1 4 = ( ) y x x y xy x           Working must be shown clearly, step by step. Credit is not given even if your answer is correct, but you skipped step(s). x = 0 x (– , 0) (0, 1) y y = 3 y y = 1/ 2 0 x –2 1 (–1, –1)

6 Question Solution Remarks 8(a) 2 2 2 (10) 100 100 10 2 r rh r r rh r         Use info given by the question.   2 2 2 2 3 2 1 4 100 10 3 3 2 2 100 103 r rV r h r h r r V r r r                      2 2 d 2 100 3 20 0d 3 3 20 100 0 3 10 10 0 or 20 400 4(3)(100) 20 40 10 10(NA),2( 3) 6 3 V r rr r r r r r                    Either 2nd derivative test:   2 2 2 2 d 2 6 20 0 since 0d 3 or d 2 10 80 10 6 20 0 when d 3 3 3 3 V r rr V rr                       Or 1st derivative test r 3 10 3 4 d d V r 27.2 0 58.6 tangent / - \ 10 is max when 3 2 10 100 100 10000 37100 or 123 3 3 9 3 81 81 V r V              Simplify the expression for V will make the differentiation much simpler. No need to apply product or quotient rule. Qn needs “exact” answer. Required to show working for finding r either by the quadratic formula for factorisation. and working for max V. If using 2 nd derivative test, either substitute in 10/3 to show that 2 2 d 0d V r  or explain that when r > 0, 2 2 d 0d V r  When using the 1st derivative test, numerical values of d d V r should be given. Not just the sign + /-

7 Question Solution Remarks 8(b) Volume of Cone 2 2 31 1 2 (2 )3 3 3V r h r r r     2d 2d V rr   Base area 2A r  d 2d A rr         2 2 2 1 d d d d 2d d d d d15 2 (3) d d 5 d 6 d d d d 5 2 2 (3) 5 cm mind d d d 6 V V r r rt r t t r t r t A A r r rt r t t                    Alternatively,   2 1 2 d d d 1 2 (15) 5cm mind d d 2 A A r dV rt r V dt r          (b) is not related to (a), we only consider a cone. Note where the “ ” is placed. 5 6 , not 5 6  Note units of rate of change of area is cm2 / min 9(i) PQ   = 4 2 1        − 1 1 2         = 3 1 3        The length of projection = 3 1 1 1 3 1 3 4 1 6 3 2 56 14 22 423 3 3                              Not scalar product formula Modulus is to be seen

8 Question Solution Remarks 9(ii) Let M be the point of intersection. 1 3 1 1 2 3 OM                       for some  1 1 3 1 OM            1 3 1 1 1 3 2 3 1                        + 4 = 3   = 1 1 3 2 1 ( 1) 1 0 2 3 5 OM                                 Note notation for vector (iii) Use normal vector 1 0 0         1 2 1 0 1 0 2 0 4 0                           