NJC 9758 2022 Promo Solutions

National Junior College Mathematics Department 2022 / SH1 H2 Maths / Promos / Suggested Solutions Page 1 of 20 Qn Suggested Solutions 1 Let $A, $B and $C be the selling price of 1 kg of salmon, 1 kg of tuna and 1 kg of swordfish respectively before discount. 2 3 246 (1) 18 18 (2) 0.85 0.9 0.95 123.3 (3) A B C A C B ABC A B C              By G.C., 48, 60, 30A B C   The selling price of 1 kg of salmon, tuna and swordfish before discount is $48, $60 and $30 respectively. Qn Suggested Solutions 2(a) (b)  hyx   1 h y x    1 h y xp     1 h y xp     1 h yq xp   2y 0.5y 0.5y 0.5y 0.5 1 1.5 yq q      1,0 1x 1xp  1xp 12 3 xp p     x y O

National Junior College Mathematics Department 2022 / SH1 H2 Maths / Promos / Suggested Solutions Page 2 of 20 Qn Suggested Solutions 3(i) (ii) Solve for intersection between the curves: 1xb xbxa           11 or 1 1 1 1 0 or 0 x b x b x b x bx a x a x b x b x a x a                     11 or x b x a xa    or 11 or x b x a xa    Therefore, we have or or x b x a x a   From graph in part (i), xa or 0x x y for (ii)

National Junior College Mathematics Department 2022 / SH1 H2 Maths / Promos / Suggested Solutions Page 3 of 20 Qn Suggested Solutions 4(a)     2 22 22 2 21 1 d .43 1 d d4 3 4 3 1 6 1 3 d d6 4 3 3 23 1 1 3ln 4 3 tan62 23 x xx x xxxx x xxx x xxA                    (b)   2 1 1 1 or 1 yx yx y x y x            2211 00 Volume required π 1 d π 1 d 8.3776 (4 d.p.) by GC x x x x       Alternatively (Shell Method) – For Teacher’s Ref only   2 2 0 Volume required 2 π 1 1 d 8.3776 (4 d.p.) by GC y y y     

National Junior College Mathematics Department 2022 / SH1 H2 Maths / Promos / Suggested Solutions Page 4 of 20 Qn Suggested Solutions 5(i) By Ratio Theorem,  1 22 2 2 BA BCBR               a b c b a b c  : 2 , 2 BRl BR          r b b a b c (ii) Given that  : 2 ,AQl      r a a b c To show that lines BR and AQ meet, we let         222 1 1 222                       b a b c a a b c a b c a b c Since the points O, A, B and C do not lie on the same plane, we can compare the coefficients of a, b and c to obtain 1 2 (1)2 1 (2) (3)2         Sub (3) into (2), we obtain 21 23     . Hence    11 33OG         b a 2b c a b c (iii) Length of projection of OS   onto OA   OAOS OA       

National Junior College Mathematics Department 2022 / SH1 H2 Maths / Promos / Suggested Solutions Page 5 of 20   2 2 2 2π2 cos 3 12 2 1                 b a a a b a a a a b a a a ba ba ba ab Geometrical Method The length of projection of OS   onto OA   xy x  aa To find y: Consider the shaded right-angled triangle, πcos 32 y y   b b Therefore, length of projection of OS   onto OA   xy   ab O A B S x y a

National Junior College Mathematics Department 2022 / SH1 H2 Maths / Promos / Suggested Solutions Page 6 of 20 Qn Suggested Solutions 6(i) 2dtan sec d xx m t m t t   22 2 2 2 2 22 2 2 2 2 22 22 1 d 1 sec d tan 11 sec d since 0 1 tan 1 sec d sec πsec d sec sec sec since 0 2 ln sec tan ln OR ln x mx m t t m m t m t t m m m mm t tt t t t t t t t t t C x m x Cm x m x A                               (ii)   1 22 2 122 2 22 1 d 2 mxx xC mx m x C         (iii) 1 22 tan dxx xmmx      1 22 dtan , d x v xu mx mx        1 1 2 22 dtan d 1 m x m x u mu m x m x        From part (ii),      1 22 21 2 2 2 2 2 1 2 d 1 1 2d 2 2 mxv x m x v m xx       

National Junior College Mathematics Department 2022 / SH1 H2 Maths / Promos / Suggested Solutions Page 7 of 20 Qn Suggested Solutions 1 22 1 2 2 2 2 22 1 2 2 22 22 2 2 1 2 2 1 2 2 tan d tan d 1tan d tan ln from (i) OR tan ln from (i) xx xmmx xm m x m x xm m x x m x m xm mx x x m xm x m D mm xm x m x m x K m                                   

National Junior College Mathematics Department 2022 / SH1 H2 Maths / Promos / Suggested Solutions Page 8 of 20 Qn Suggested Solutions 7(i) 223 2 0x xy y k    Differentiating with respect to x:       dd6 2 2 2 0 dd d6 2 2 2 d d3 d yyx x y y xx yx y y x x y x y x y x          (ii) At stationary points, d3 0d y x y x y x   . 3 0 3x y y x     Sub 3yx into equation of C:     22 2 2 2 2 2 3 2 3 3 0 3 6 9 0 12 12 x x x x k x x x k xk kx             For C to have no stationary points, 2 12 kx  has no solution. Since 2 0x  , therefore 0k  . (iii) Tangents parallel to the y-axis  d d y x is undefined (or d 0d x y  ). Thus, we have 0y x y x    . Sub yx into equation of C: 2 2 2 2 2 3 2 0 4 4 or 44 or 22 x x x k xk kx kkxx kkxx           ,22 kk    (Given that 1k  so 1k . Thus k is well-defined.) Sub 2 kx  into C:

National Junior College Mathematics Department 2022 / SH1 H2 Maths / Promos / Suggested Solutions Page 9 of 20 Qn Suggested Solutions 2 2 2 2 2 3 2 022 30 4 04 022 kk y y k k k y y k ky k y kkyy                                 OR Discriminant   2 40 4 kk       , thus only one solution. Since there is only one solution, the tangent meets C at exactly one point. Therefore, the tangent does not intersect C again.

National Junior College Mathematics Department 2022 / SH1 H2 Maths / Promos / Suggested Solutions Page 10 of 20 Qn Suggested Solutions 8(i) 4 for 1,f ( ) cos for 1 π. xxx xx   Since every horizontal line yk , k cuts the graph of  fyx at most once , f is one-one. Therefore f has an inverse. OR Since every horizontal line yk ,    , 3 1,0.540k     cuts the graph of  fyx exactly once , f is one-one. Therefore f has an inverse. (ii) 4 for 1,f ( ) cos for 1 π. xxx xx   When 1x , 44y x x y     . From graph of  fyx , range is ( , 3]  . When 1 π,x 1cos cosy x x y    . From graph of  fyx , range is [ 1,0.540) . 1 1 4 for 3,f ( ) cos for 1 0.540 xxx xx         (iii)   2 π π πf ff f cos f 0 42 2 2                    y x O