2021 SASS AMath Prelims P1 w Ans

/ 1 2 3 Show that p > 0 for the inequality px2 + 1 > 2px - p for all real values of x. [3] px2 + l > 2px - p px2 - 2px + ( l + p) > 0 Since the quadratic function is positive, then D < 0 and coeff of x2 > 0 (-2p) 2 -4p(l + p) < 0 andp > 0 4p2 -4p-4p 2 < 0 ---~ (i) Sketch the graph of y = In (x + 2) for x > -2, indicating clearly the point where the graph cuts the axes and the asymptotes. [2] ( -tr (ii) In order to solve the equation (x + 2)= ex- 3 , a graph with a suitable straight line is drawn on the same set of axes as the graph of y = In (x + 2). Find the equation of this straight line. [2] ( ) x-3 x+2 = e In (x + 2) = x- 3 :. The line to be drawn is y = x - 3. x 2 -4x+2 . . . Express 2 m partial fractions . [4] X -3x+2 x2 -4x+2 = (x2 -3x+2)-x x2 -3x+2 x2 -3x+2 = l+ -x (x-2)(x-1) Let - x = ___A_ + _lL (x-2)(x-1) x-2 x-1 4 2021 Prelims Exam - A Maths 4049/1 Multiplying by (x - 2)(x - 1 ), -x=A(x- l)+B(x-2) When x = 2, A = -2 When x = 1, B = 1 2 :. x -4x+2 = l+-1 ___ 2_ x2-3x+2 x-1 x-2 It is given that 3 2x-2 (2 x+3 ) = 92x-l . (i) Find the exact value of (%) x . 3 2x-2 (2 x+3) = 92x-l 32x -x2x x23 - 92x 32 -9 32xx2x 32 92x - 23 x9 (32x2 r =l 92 8 .. r = ½ (shown) [3] (ii) Hence find the value of x. [2] x= log2 l 9 8 lgl =-8 lgl 9 = 1.3.£ (3 s.f.) 5 (i) Given that sin 0- cos 0= ¾, show that sin 0 cos 0 = 3 7 2 . [3] sin 0- cos 0= l 4 (sin 0- cos 0)2 = -2... 16 sin2 0- 2sin 0cos 0+ cos2 0= f6 6 (iii) 1 - 2 sin 0 cos 0 = (6 2 sin 0cos 0= 1: sin0 cos 0 = l2 (shown) Hence find the value of 3(cot 0 + tan 0). [2] 3(cot 0 +tan 8)=3fc?s0+sin0) \sm0 cos0 = 3 (co~ 2 0+sin 2 0) sm0 cos0 = 3 (~) 32 = 96 7 The polynomial P(x) is of degree three and has a constant term 20. The roots of the equation P(x) = 0 are 1, -2 and k. When P(x) is divided by (x + 1 ), the remainder is 24. (i) Show that the value of k is 5. (4] P(x) = a(x- l)(x + 2)(x - k) P(0) = 20 2ak = 20 ak= 10 ... (1) P(-1) = 24 a(-2)(-1 - k) = 24 a+ ak= 12 .. . (2) Sub (1) into (2), a+ 10 = 12 a=2 k= 5 (shown) (ii) Find the remainder when P(x) is divided by (x-2). [11 P(x) = 2 (x- l)(x + 2)(x- 5) Remainder= P(2) = 2 (4) (-3) =-24

7 (a) Express each of 2r - 6x + 9 and -x' - 6x - 8 (ii) Using your value of k from part (i), find the 10 (a) Solve the equation lg(x-4)+2lg3=1+1g(j) [3] in the form a(x + b)2 + c, where a, band care time taken by the mouse to find the cheese on lg(x - 4) + 2 lg 3 = 1 + lg(f) constants. [3] the sixth attempt. [l] 2x2 - 6x + 9 = 2 (x2 - 3x) + 9 T = 1.6 e -0.605n + 3 lg(x-4)+lg3 2 =lgl O+lg(j) ={(x- ½J -¾}9 When n = 6, T= l.6e - 0· 605(6) + 3 Ig[(x-4)x9 ]==lglO(j) = 3.04 minutes = 2 (x- ½} -f +9 (iii) Will the mouse be able to find the cheese :. 9(x-4) = 5x within 1.5 minutes? Explain your answer . [2] 4x= 36 = 2(x- ½f +f It is not impossible that the mouse can x=9 find the cheese within 1.5 minutes since Solve the equation (iogg1 x )( loL 3) = 6¼ [4] -x2 - 6x - 8 = - (x2 + 6x) - 8 T = 1.6 e -0.605n + 3 and e -0.605n > 0 (b) · = -[(x+3}2-9]-8 :. T> 3 minutes. (1og81 x)( n) = 6¼ = -(x+3)2+9-8 ogx 2 9 Given that y=5xe 2x, =-(x+3) +1 (i) find the value of x when the tangent is parallel ('og,x t J j 25 (b) Use your answers from part (a) to explain to the x-axis. [3] log 3 81 log3 3 = 4 why the curve with equations y = 2x2 - 6x + 9 dy = 5e2x + 5x(2e 2x ) = 5e2x(1+2x) log 3 x and y = -x2 - 6x - 8 will not intersect. [2] dx Let u = log3 x, then the equation becomes From (a), we know that the graph of When the tangent is parallel to the x-axis , then ~lfr21 y = 2x2 - 6x + 9 has a minimum point at gradient of tangent= 0. (l .2.) while the graph of y = -x2 - 6x - 8 :. 5e2x(1+2x)= 0 2'2 has a maximum point at (-3, 1). Thus the Since e2x>O, then x = -½. : . u2 = 25 two graphs will not meet. find the gradient of the curve when x = ln 4, u=± 5 (ii) log3 x= ± 5 8 In an experiment on the cognitive ability of animals , giving your answer in the fonn of 5 -5 80(a In 2 + b). [1] x = 3 or 3 a scientist recorded the time taken by a mouse to dy 2ln4( 4) 1 find a piece of cheese in a series of n attempts. The When x = In 4, dx = 5e 1+2ln x = 243 or x = 243 time T, in minutes, the mouse took is modelled by = 5 (42) (1 + 2 ln22) the equation T=l .6e-kn +3 where k is a constant. It = 80 (1 + 4 In 2l 11 (a) Solve = 3-5x. [3] took the mouse 3.142 minutes to find the cheese on y is decreasing at a rate of 5(4ln2 + 1) units/s when = 3-5x and 3 -5x~ 0 the fourth attempt. x = In 4. . [ 2 ] Squaring both sides, we obtain (i) Find the value of k giving your answer correct (iii) Find the rate of change of x at this instant. to 3 significant figures . [2] 4x = 9 - 30x + 25x2 dy = dy x dx 25x2 - 34x + 9 = 0 When n = 4, T= 3.142. dt dx dt (25x - 9)(x - 1) = 0 l.6e-4k +3 = 3.142 dx x = -2._ or x= 1 .. -5(4 In 2 + I)= 80 (1 + 4 ln 2) x dt 25 e-4k _ 3,142- 3 = 0.088 75 . dx __ ~= _ ..Lumits/s But X $ ¾' :. X = is - 1.6 :. -4k- in 0.088 75 .. dt- 80 16 k b.605 (3s.f.) -

12 (b) A toy train moved at a speed of (2 + cm per second from point A to point B. Given that the distance covered was ( - 10) cm, find the time taken to move from point A to B in the form +q , where p and q are constants. [3] Time= Distance = Speed = X _ - 4-3 = = - 77 cm (i) Write down the first three terms in the expansion, in ascending powers of x, of (3-f r , where n is a positive integer. [2] (3-tr = 3n+(;)3n-) (-t~G}n-2(-t r _ 3n 3n-l( x) n(n-1) 3n-2(x2J - +nx -- +--- --5 2 25 = 3n _ nx3n-l x + n(n-l)x3n- 2 x2 5 50 The first two terms in the expansion, in ascending powers of x, of ( 3+ x )(3-f r are p + qi'-, where p and q are constants. (ii) Show that n - 5. [3] (3+x)(3-tf = (3 + x) [ 3n _ nx3n-l x + n(n-l)x3n- 2 x2] 5 50 Since coefficient of x = 0, then 3x ( _n xr-l J + 3n = 0 3n(- ~+l~ 0 (iii) 13 (a) (b) :. n = 5 (shown) Hence find the value of p and q. [3] So, p = 3 x 35 = 36 = ll!l q = coefficient of i'- = (-5x3 4 ) + 3 x 5x4x3 3 5 50 =-Sl + 8lx2 5 _ 243 - --5- Find J l 5 2 dx . +cos x [3] f 5 dx- f 5 dx l+cos2x - 1+2cos 2 x-1 (i) = f ,f sec 2 x dx = ~tanx+c 2 Show that dxd ln(--i-) = l _ cot x [2] smx x Jx 1n(sf:x) = Jx [21nx- ln(sinx)] = 1, _ cosx x sinx = l _ cot x (shown) X (ii) Hence evaluate r ¾cotx dx. [3] Integrating both sides of (i), we have f l-cotx dx = ln(-J:---) + c x smx 2 lnx-J cotxdx= ln(--i-)+c smx f cot x dx = 2 In x - In (--i-) + c smx f 2cotxdx=.1[21nx-ln(-i-)] 2 1 4 4 sm x I 14 :. f 2 lcotxdx= .1[21n 2-In(~)J_ [o-ln - .1 ] 1 4 4 sm 2 sm 1 = 0.0581 y "'- The diagram above shows part of the curve with the normal of the curve at Q. Given that the gradient of the tangent at Q is 1- , (i) find the coordinates of Q. dy _ 4 _ 2 dx - - Gradient of tangent at Q = f _2_= l_ 3 1 + 4x= 9 4x=8 x=2 : . y= 3 :. 0 = (2. 3). [3] (ii) find the coordinates of P, [2] Gradient of normal at Q = - ½ Equation of normal at Q is y-3 = - 2 (x-2) 2 y=- 2 x+6 2 When x = 0, y = 6 :. P= <O. 6).

(iii) Calculate the area bounded by the y-axis, line PQ and the curve. [ 4] -- Area of Area under Required area= -trapezium the curve = ½(6+3)x2- s: dx = 9 -l(1+4x)- ½ ]2 {(4) 0 = 9- ! =9- ¼[#- 1] = 9-1(27-1) 6 = 9 _.Ll. 3 = li units2 3 15 The gradient at any point on a particular curve is given by the expression x 2 + 1 , where x > 0. X Given that the curve passes through the point P(4, 18), find (i) the equation of the normal to the curve at P, dy = x2 + 1~ d.x X At P, gradient of tangent = 42 + 1 4 = 17 [3] Gradient of normal = - 1 \ :.Equation of the normal is y - 18 = - ...L (x - 4) 17 y=-_l__x+310 [3] 17 17 (ii) the equation of the curve, [3] y = f x2+.l§__ d.x x2 = L+16 L_ +c 3 ( -1) 3 -1 x3 16 =---+c 3 X Whenx=4,y= 18 18 = 64 _li+c 3 4 • C =1_ . . 3 3 Equation of the curve is y = ; - 1; + 1 (iii) the coordinates of the point on the curve where the gradient