2023 JC1 Promo Practice Paper C (solutions) VJC

2023 PROMO PRACTICE PAPER C Solutions Qn Suggested solution Comments 1 Method 1 As 2 2 1 11 1130 2 44xx x ++= ≥ > ++  for all real values of x . OR Method 2 As (coefficient of 2 1 ) 0x = > and discriminant 2 4(1)(3) 11 0(1) − = −<= , we have that 2 30x x++> all real values of x . 2 2 2 2 2 26 1 12 4 26 1 0, 4 and 3( 4)( 3) 4 2 6 ( 3) 0( 4)( 3) 3 0( 4)( 3) xx xx x xx xxxx x xx x xx xx xx ++ ≥−− − ++ − ≥ ≠ ≠−−+ − + +− + ≥−+ ++ ≥−+ This implies 1 0( 4)( 3)xx ≥−+ since 2 30x x++> all real values of x . ( 4)( 3) 0xx− +> 3 or 4xx<− > 2 ( ) ( ) ( ) 2 2 2 2 4i 1 i 1 i1 i i 4i 4 1 5 4i 1 z λλ λλ λλ λ λ λλ λ −+= ×−+ + −+ = + +− = + arg( )z π= ⇒ , 0 and Im( ) 0zz z∈< = 2 40 2λλ−= ⇒ = ± When 102, 2 5zλ = = = (Rejected since 0z< ) When 102, 25zλ −= −= = −

2023 PROMO PRACTICE PAPER C Solutions Alternatively, Sketch indicates that λ< 0. 4// 1OA OB λ λ −−⇒= 2 4 2 (since < 0) λ λλ −= − =− Hence, ( )2 1 2i2 4i 21 2i 1 2iz −+−−= = =−++ 3i ( ) ( ) ( ) 222 22 2 cos 5 6 2(5)(6)cos 25 36 60 cos cos sin sin 3461 60 cos sin55 61 36cos 48sin PQ PR QR PR QR PRQ PRS PRS PR S θ θθ θθ θθ =+−⋅ ∠ =+− ∠ + =+− ∠ − ∠ = −−  = −+ ( ) 1 2 61 36cos 48sin (shown)PQ θθ∴=− + A(1, ) B( , –4) RE IM O

2023 PROMO PRACTICE PAPER C Solutions 3ii ( ) ( ) ( )( ) 1 2 1 2 1 2 1 2 2 2 2 211 222 22 2 61 36cos 48sin 61 36 1 48 2 25 48 18 48 1851 25 25 1 48 18 4851 2 25 25 2! 25 24 9 28851 25 25 625 24 6351 25 625 245 5 PQ θθ θ θ θθ θθ θθ θ θθ θ θθ θ = −+  ≈− −+  =++ = ++  −  ≈+ + +       = ++ − = +− = + 263 24 63, , 125 5 125pqθ−= = − 4i d ed xy xyx −+= 2 2 dd edd xyy xyxx −+ += − 32 32 d d dd ed d dd xy yyyxx x xx −+ ++= When 0x= , 1y= , d 1d y x = , 2 2 d 2d y x =− , 3 3 d 1d y x =− . Let y = f(x). The Maclaurin series for y is: 23 f (0) f (0) f (0) f (0) ...2! 3! xxyx ′ ′′ ′′′= ++ + + 23 11 ...6y xx x=+− − + f(0) = 1 ( )1 when 0yx= = 4ii Since 23 11 ...6y xx x=+− − + , differentiating y w.r.t. x, 2d1 1 2 ...d2 y xxx = −− + 5ai 2 1 d25 xxx++ ⌠⌡ ( ) 2 2 1 d 12 x x = ++∫ 111tan22 x C− += +

2023 PROMO PRACTICE PAPER C Solutions 5aii ( ) 3 ln dx xx ⌠ ⌡ ( ) 4 ln 4 x C= + f(x) = ln x, 1f '( )x x= ( ) ( ) 3 4 f ' ()f () d f( ) 4 xxx x C= + ∫ Note that ( ) 3 ln 3lnxx ≠ 5b [ ] [ ] ( ) π 2 ππ 22 π 2 π 2 0 0 0 0 0 cos d (sin ) sin d π sin d2 π cos2 π 012 π 12 x xx x x xx xx x = − =+− = + = +− = − ∫ ∫ ∫ dd d d dd uvv x uv u xxx = −∫∫ Take note the correct choices of function to be d d u x (integrated) and v (to be differentiated) choose d sind u xx = and v = x 6a 22 0x ay bx cy+ ++= Since C passes through ( )5, 3− and 13,22  − , ( ) ( ) ( ) 225 35 3 0 9 5 3 25 - 1 a bc abc + − + +− = ⇒+−= − ( ) 22 1 31 3 02 22 2 19 1 3 044 2 2 9 2 6 1 - 2 a bc abc abc    + − + +− =       ++−= ⇒+−= − At 13,22  − , tangent is parallel to the y-axis. ( ) dd22 0dd d22 d 12d 2 3d 2 2 1 3 yyx ay b c xx yay c x b x by x ac b ac + ++ = + = −− −−= −+  −−= −+ Since tangent is parallel to the y-axis. ( )3 0 - 3ac− += By GC, 1, 5 and 3ab c= −= − = − . Note that 13,22  − is also a point on the curve and needs to be substituted to get another equation Tangent parallel to the y-axis is vertical, so d1 d3 yb x ac −−= −+ is undefined, meaning that denominator is zero and NOT numerator. (See diagram)

2023 PROMO PRACTICE PAPER C Solutions 6b ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 22 22 22 2 22 22 10 6 0 55 33 0 5 34 53 144 xy xy xy xy xy −− −= −− −+− =  − −+ = −+ −= Asymptote: 3 5 , 3 5 8, 2 yx y x yx y x +=− += −+ ⇒ =− = −+ Use brackets when completing the square. This is hyperbola. Refer to conics App on GC to see the various forms for various conic sections. Always draw asymptotes first before the curve in order for your curve to approach the asymptotes. Watch for “tail-end behaviour”. Curve should not “steer away” from asymptote. We want a relatively proportional curve here with correct labels – centre, vertices, asymptotes, passing through the origin. A rather common mistake is that many students drew a circle instead. Question: What is the key difference between a hyperbola and ellipse (and circles)? 7i Let x be the length as denoted in the diagram: By similar triangles, 1.5 2 3 4 x h xh = = ( ) ( ) 2 13 824 3 V hh h =  = Be clear about the usage of similar triangles as this is a “show” question – provide clear justification. 7ii d 6d V hh = When volume of water in the drain = 37.2 m , the water level is ( ) 2 2 7.2 3 2.4 2.4 0 h h hh = = = >  x y x 8yx= −2yx= −+ ( )5, 3− ( )10,0 ( ) ( ) 22 22 53 144 xy−+ −= ( )0,0 ( )0, 6− ( )9, 3−( )1, 3−

2023 PROMO PRACTICE PAPER C Solutions ( ) ( ) d dd dd d 1 0.03 0.02 6 2.4 0.0010759 0.00108 h hV tVt= × = ×− = = The rate at which the water level is rising is 0.00108 1m s .− No need to make h the subject to find d d h V . Recall that d 11 dd6 d h VVh h = = . 8i Let nu be the thn term of the geometric series. 1n nu ar −= 3 2 36 36 (1) u ar = = −−−−−−−−−− ( ) 243 2431 243 1 (2) S a r ar ∞ = =− = − −−−−−−− Substitute (2) into (1), ( ) 2 23 243 1 36 243 243 36 0 rr rr −= − −= Using GC to solve for the roots of the cubic equation, 12 (reject 0) or 33rr=−>  2Common ratio, . 3r∴= Substitute 2 3r = into (2), 2243 1 3 81 a = −  = First term, 81.a∴= The deadly mistake in question 7 was the incorrect usage of formulas. Correct formula for nth term of geometric series is 1n nu ar −= Correct formula for sum to infinity is 1 aS r ∞ = − . No need to solve this equation by algebraic means – just use GC! Always read question to see if GC can/should be used. 8ii ( ) ( ) ( ) 3 3 3 281 1 36 21 6 1 22 1 3 23 2 5 243 1 3 26 15 243 243 3 6 15 171 15 165 11 d d d d d d  −  + − = −  += −   +=−  += = = Correct formula for sum of n terms of a geometric series is ( )1 1 nar r − − . Correct formula for sum of n terms of a arithmetic series is [ ]2 ( 1)2 n an d+− . Make sure you know what the “letters” in the formula mean.

2023 PROMO PRACTICE PAPER C Solutions 8iii ( )( ) 2 2 2 281 1 3 1 1 11 21 3 21 11 11 243 1 3 211 10 243 1 3 n n n n n n  −  +− > −  + − > −   −> −   Using GC, 23n≥ . n 11 10n− 2 2243 1 3 n −  22 232 242.999996 23 243 242.999998 24 254 242.999999 The least value of n is 23. Do not be tricked by GC. Always do some verification if in doubt. 2 11 0 2231 3 1 4 x y yx    = = −    −  By evaluating / checking the answer when it seems too unnatural that 2 23 2243 1 2433 × −=  A verification (see above) shows that this isn’t the case. If you are using tables, place your selection over the value to see its unrounded value. 9a There are 2 ways to do this Method 1 ( ) ( ) ( )fffyx yx y x= →− = ⇒ =− Reflect graph in x-axis ( ) ( ) ( )f 2 f 2fy xy xy x= − →−= − ⇒=− Translate resultant graph 2 units in positive y-direction. Method 2 ( ) ( ) ( )f 2f 2fy xy xy x= → += ⇒ = −+ Translate graph 2 units in negative y-direction ( ) ( ) ( )2f 2f 2fyx yx y x= −+ → −= −+ ⇒ =− Reflect resultant graph in x-axis Read question carefully, label all the required features in the graph. The y-intercept and stationary point should be at the same level. Generally, when we reflect a graph in the x-axis, the y- coordinates will change. Vice versa for a reflection in the y- axis. Do note that this applies to the asymptotes as well.