MATHEMATICS 9740/01 Paper 1 14 September 2012
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1 NATIONAL JUNIOR COLLEGE PRELIMINARY EXAMINATIONS Higher MATHEMATICS 9740/0 Paper 4 September 0 hours Additional Materials: Answer Paper List of Formulae (MF5) Cover Sheet hours READ THESE INSTRUCTIONS FIRST Write your name, registration number, subject tutorial group, on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-eact numerical answers correct to significant figures, or decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are epected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the eamination, fasten all your work securely together. The number of marks is given in the brackets [ ] at the end of each question or part question. This document consists of 9 printed pages and blank pages. National Junior College NJC /0/0
2 The diagram shows the graph of y = f '( ), where ( ) f is a cubic polynomial, has a minimum point at (.5, 9). f '( ) 0 7 (.5, 9) Given that f( 0) =, find f ( ). [4] State the range of values of for which f ''( ) < 0. [] Relative to the origin O, the position vectors of three points, A, B and P are, i j+ k, 5 + i k and ( λ) + i+ ( λ ) j+ k, where λ is a real parameter, λ. (i) Show that A, B and P are collinear. [] (ii) Find the value of λ such that P is on the line BA produced and area of triangle OAP is 6 5 square units. Give a reason for your choice. [4] NJC /0/0
3 Use the standard series for e and ( + ) n to find the Maclaurin series for f ( ), where ( ) f( ) = e +, up to and including the term in. [] (a) By substituting =, find an approimation for ( ) 9 5 e. [] (b) Find the series for f( ) up to and including the term in. Hence or otherwise, find the Maclaurin series for e +, up to and including the term in. [] 4 It is known that a particular type of bacteria grows very well under certain controlled conditions in a specially prepared Petri dish. The researcher believes that the growth rate of d such bacteria can be modeled by t t =, where milligrams is the amount of dt bacteria grown in the dish after t hours. (i) Using the substitution du d u t =. = ut, show that the differential equation can be reduced to [] (ii) Find in terms of t, given that there was 0. milligrams of bacteria after 5 minutes. Hence find the amount of this particular type of bacteria after 4 hours. [4] (iii) Eplain if this mathematical model is a realistic one. [] NJC /0/0 [Turn Over
4 4 5 A curve C has parametric equations = cos t, y = tan t, for π π < t <. (i) Sketch the curve C, indicating clearly any asymptotes, and aial intercept(s). [] (ii) The point P on the curve has parameter t = π. If the normal to the curve at P passes the through the point (b, 0), find the eact value of b. [] (iii) Show that the area bounded by the curve C, y-ais and y = 4, is π +. [4] tan 4 6 The planes p, p and p have equations =, + y + az = 5 and + y+ z = b, where a and b are real constants. Given that p and p intersect at the line l, show that the vector equation of l, in terms of a, is r = i+ ( λa) j+ λk, where λ is a real constant. [] (a) The acute angle between l and p is 60. Find the possible values of a. [] (b) Given that the shortest distance from origin to p is 6 and without solving for the value of b, determine the possible position vectors of the foot of perpendicular from the origin to p. [] (c) What can be said about a and b if p, p and p do not have any points in common? [4] NJC /0/0 [Turn Over
5 5 7(a) A geometric series has common ratio r, and an arithmetic series has first term a and common difference d, where a and d are non-zero. The first three terms of the geometric series are equal to the ninth, fourth and second terms respectively of the arithmetic series. (i) Show that d =. a [] (ii) Deduce that the geometric series is convergent, and find in terms of a, the sum to infinity. [] (b) A mountaineer climbs a mountain of height metres. The amount of distance he climbs for the first hour is 00 metres. For each subsequent hour, he climbs 0 metres less than the previous hour. The number of whole hours that has passed just before he reaches the summit is n. (i) Write down an epression for the total distance climbed by the mountaineer and hence show that pn + qn, where p and q are constants to be determined. [] (ii) Deduce the value of n if = 500. [] NJC /0/0 [Turn Over
6 6 8 A calculator is not to be used in answering this question. The polynomial p( z ) is defined by ( ) p z = z + mz 7z + 5, where m is a real constant. It is given that ( z + ) is a factor of p( z ), find the value of m. Given that z, z and z are the roots of the equation p( z ) ( z ) ( z ) ( z ) π < arg < arg < arg π, [] = 0, where (i) find z, z and z in eact Cartesian form + iywhere y,. [] n n + (ii) show that z z is purely imaginary for n. [] (iii) eplain whether the locus of the comple number w satisfying the equation w = a, for some positive constant a, passes through all the points representing the comple numbers z, z and z. [] NJC /0/0 [Turn Over
7 9 A sequence u, u, u, is such that u = and 7 u = n u + n nn ( + )( n +, for all positive integers n. ) (i) Prove by mathematical induction that u n =. [4] nn ( + ) (ii) Hence, find N. [] r= r r+ r+ ( )( ) (iii) Deduce from your results that the series is less than 4. [] (iv) N Using the result in (ii), find, leaving your answer in the form r= 0 r( r )( r ) b+ f( N) where b is a constant and f( N ) is a function involving N. [] NJC /0/0 [Turn Over
8 8 0(a) The curve C has equation a+ 4a y = a, ± a, where a is a positive constant. (i) Find the equations of the asymptotes, leaving your answers in terms of a. [] (ii) Show that if C has two turning points, then 0< a < 6. [] For a = 4, sketch the curve C, stating the equations of any asymptotes, the coordinates of the turning points and aial intercept(s) if any. [] Hence, find the range of values of h such that the graph of ( 6 5) ( ) h + + y+ = h, where h is a positive integer, intersects C more than once. [] (b) The diagram shows the graph of y = f ( ), which has turning points at A( 5, 6) and B(, ). The horizontal and vertical asymptotes are y = 0 and = respectively. Sketch the graph of y =, f( ) showing clearly all relevant asymptotes, aial intercepts and turning point(s), where possible. [] y y = f( ) A( 5, 6) O B(, ) = NJC /0/0 [Turn Over
9 9 The functions f, g and h are defined by f : +,, 4 < <, g : ln( ),, k < < 4, h : 8 8,. (i) Sketch, on the same diagram, the graphs of f, f and f f, showing clearly the relationship between the graphs and indicating the aial intercepts. Find the eact value of for which f () =. [5] (ii) Find the minimum value of k such that f g eists. [] (iii) Using a non-graphical approach, find the set of eact values of for which f() < h(). [4] End of Paper NJC /0/0 [Turn Over
10 0 NATIONAL JUNIOR COLLEGE Preliminary Eaminations Higher MATHEMATICS 9740/0 Higher Paper 4 September 0, Friday hours Candidate Name: Subject Class: ma / IPma Registration No.: Subject Tutor: For office use over Page Question No. Marks Obtained TOTAL MARKS INSTRUCTIONS TO CANDIDATES Write your name, registration number, subject tutorial group, subject tutor s name and calculator model in the spaces provided on the cover sheet and attached it on top of your answer paper. Circle the questions you have attempted and arrange your answers in NUMERICAL ORDER. Write your calculator s model number(s) in the bo below. Scientific Calculator Model: Graphic Calculator Model: 0 Presentation / TOTAL 00 GRADE NJC /0/0 [Turn Over
11 ANNEX B 0 NJC H Maths Preliminary Eamination Paper Qn/No Topic Set System of Linear Equations Answers f ( ) = + ; < Vectors (ii) λ = 09 Since P is on BA produced, AP negative value of k. = k AB for a Maclaurin Series (a) 5 + (b) 4 Differential Equation 6t 56 (ii) = ; or.94 6t + 65 (iii) The particular solution of the DE suggests that the amount of bacteria in the Petri dish will grow indefinitely as time passes. 5 Differentiation and Application (ii) b = 4.5 of Integration involving Parametric Equations 6 Vectors (a) a = 7 or a = / / (b) ON = / or / / / 7 Arithmetic and Geometric Progressions (c) a =, b 7 (a)(ii) 5 a (b)(i) p = 5, q= 05 (ii) n = 9 8 Comple Numbers m = z = i, z = + i, z = (i) (iii) Since z = 5, z =, z z. The locus of comple numbers satisfying the
12 equation w = a, for some positive constant a, will not pass through all the points representing the comple numbers z, z and z. 9 Sequences and Series/ Method of Difference/ Mathematical (ii) ( N + )( N + ) Induction (iv) 44 N( N ) 0 Curve Sketching/ Graph Transformations (a) (i) Vertical asymptotes: Horizontal asymptote: y = = a or = a h Functions/ Inequalities (i) = (ii) k = ( e 9/8 )/ or.04 (iii) or.5 < 4
13 Qn 0 NJC H Math Prelim P Solutions Suggested Solution f = a + b + c + d where abcd,,,. ( ) Given f( 0) =, d =. f '( ) = a + b + c From the graph of y = f '( ), f '( ) = 0 a 4b+ c= 0 ( ) f '( 7) = 0 47a+ 4b+ c= 0 ( ) f '(.5) = a+ 5b+ c= 9 ( ) OR f ''(.5) = 0 5a+ b= 0 ( 4) Using GC to solve (), () & (): a =, b=, c= f( ) = For f ( ) is concave downwards ( ) <.5. (i) 6 AB = OB OA = 0 f '' < 0, + λ λ+ 6 ( λ + ) AP = + λ = λ+ = 0 0 Since λ + AP = AB for λ and A is a common point, this shows that A, B and P are collinear. Page of 6
14 0 NJC H Math Prelim P Solutions (ii) Given that area of triangle OAP is 6 5, OA AP = 6 5 λ + AB = 6 5 st part 6 λ = 0 λ + = λ + 4 = ( ) ( ) ( ) λ = 4 5 λ + 45 = 4 5 λ + = 08 λ+ = 08 or λ+ = 08 λ = 07 or λ = 09 Since P is on BA produced, AP = k AB for a negative value of k. Hence λ = 09. ( ) f( ) = e + ( ) ( )( ) ( ) = ( ) + ( ) + ( ) + = ( ) = = !! ( ) (a) When =, Page of 6
15 0 NJC H Math Prelim P Solutions ( ) e e = 9 (b) e = 9 e 9 = 9 f '( ) = Using f '( ) = + + +, = + + f( ) e f ( ) = f( ) + e + = ( ) = = + + e + = f ( ) f( ) e + d d 9 5 Alternative method: e + ( )( ) = e = = + + ( ) = Page of 6
16 0 NJC H Math Prelim P Solutions 4(i) 4(ii) 4(iii) = ut d du = tu + t dt dt du t tu + t t ( ut ) + ( ut ) = 0 dt 4du 4 t = ut dt du = u ( shown) dt d u = dt u = + t C' u t = t C' t = where C = C' t+ C Since there was 0. milligrams of bacteria after 5 minutes, then ( 0.5) 0. = C = C C = 6 6t = 6t + 6( 4) 56 When t = 4, = = or ( ) 6t As t,. 6t + The particular solution of the DE suggests that the amount of bacteria in the Petri dish will grow indefinitely as time passes. Hence the model is not a realistic one. Page 4 of 6
17 0 NJC H Math Prelim P Solutions 5(i) y ( 0, ) 0 ( 0, ) = 5(ii) = cos t, y = tan t d dt = sint, dy dt = sec t dy dy d = = d dt dt sec t sin t When t = π, P( 0.5, ) and dy d = π sec = 4 π sin So, b = 4 b = 4.5 5(iii) y y = 4 ( 0, ) 0 ( 0, ) = = cos t = 0 t = 4 π y = tan 4 π = Page 5 of 6
18 0 NJC H Math Prelim P Solutions Area = = 4 y d or tan 4 π 4 costsec tt d tan 4 = π ( t ) = 4 cos sec d tan 4 π sec d 4 = [ t tan ] tan 4 t π 4 tt = (tan 4 4) + ( π ) = + π tan 4 4 dy tt 6 = --- () + y + az = () Substitute () into (): + y + az = 5 y = az Let z = λ, Hence, r = y = aλ, where λ is a real number. (Shown) z λ 6(a) The angle between l and p is 60. This implies sin 60 = 0 a ( a )( ) a + = 6 ( a + )( ) ( a a ) ( a ) = a a+ = a + a 6a = a 8a = ( a )( a ) = 0 a= 7 or a = Page 6 of 6
19 0 NJC H Math Prelim P Solutions 6(b) Since ON is parallel to the normal vector of p, ON 6 = ± = ± 6 / / = / or / / / Alternative Method: Let N be the foot of perpendicular from the origin to p. Since ON is parallel to the normal vector of p, and 6 ON = 6 α + 4α + α = 6 6α =, 6 6α = 9 α = ± / / ON = / or / / / 6(c) Given that p, p and p do not have common point, then line l must be parallel to p. Hence 0 a = 0 Page 7 of 6
20 0 NJC H Math Prelim P Solutions a + = 0 a = Also, a point (,, 0) in l must not lie in p. Hence b 0 b 7 7 (a)(i) a+ d a+ d r = = a+ d a+ 8d ( a+ d)( a+ 8 d) = ( a+ d) a + 9ad + 8d = a + 6ad + 9d = d ad 0 dd ( a) = 0 d = 0 (rej.) or d = a (shown) Alternatively, let b be the first term of the geometric series. Then b br br br d = = 5 b br = 5br 5br r r+ = (5r )( r ) = 0 r = or r = (rej because otherwise d = 0) 5 7 (a) (ii) Hence b b 5 d = = b= ( a+ 8 d) d = a 4d d = a (shown) a+ d a+ a 4a r = = = =. a+ d a+ 9a 0a 5 Since <, the geometric series is convergent. 5 Page 8 of 6
21 0 NJC H Math Prelim P Solutions a+ 8d Sum to infinity = r a+ 4a = 5 7(b) (i) 7(b) (ii) 5 = (5 a ) 5 = a The distance the mountaineer climbs for each hour follows an arithmetic progression with first term 00 metres and common difference ( 0) metres. Total distance travelled after n hours n [ (00) + ( n )( 0) ] n (600 0n+ 0) n (60 0 n) n(05 5 n) 5n + 05 n (shown) p= 5, q= 05 If = 500, then 5n + 05n 500 5n + 05n n or n 5.4 Hence n = 9. 8 p( ) = 0 ( ) m( ) ( ) 8(i) = m = 0 m = z z 7z+ 5 = 0 ( z )( z z ) = 0 ( 4) 6 4( )( 5) ( ) ± z = or z = 4± 4 = = ± i z = i, z = + i, z = Page 9 of 6
22 0 NJC H Math Prelim P Solutions 8(ii) n * * z z are comple conjugates, ( ) = ( ) Since and n n z z = z. z and z are comple conjugates as well. n n Thus * ( ) n ( i) Im ( z ) z z = z z n n n n = which is purely imaginary. Alternative method: z n n n ( i) ( i) z = + n i( tan 0.5) i( tan 0.5) = 5e 5e = 5 n i( tan 0.5) n i( tan 0.5 ) n = 5 e e = n n n Since ( z z ) n ( n ) ( n ) ( n ) ( n ) ( ) 5 isin ( n tan 0.5) n cos tan 0.5 isin tan 0.5 cos tan 0.5 isin tan 0.5 n n Re = 0, z z is purely imaginary. n 8(iii) Since z = 5, z =, z z. The locus of comple numbers satisfying the equation w = a, for some positive constant a, will not pass through all the points representing the comple numbers z, z and z. 9(i) Let P n be the statement When n =, LHS = u = (Given) RHS = = ()(+ ) Hence, P is true. u n + =, where n. nn ( + ) + Suppose/assume P k is true for some k, i.e. u k = kk ( + ). Page 0 of 6
23 0 NJC H Math Prelim P Solutions Need to show that P k + is also true, i.e. u = k + ( k + )( k + ) u = k u + k kk ( + )( k+ ) = kk ( + ) kk ( + )( k+ ) ( k + ) = kk ( + )( k+ ) k = kk ( + )( k+ ) = ( k+ )( k+ ) Hence P k + is true if P k is true. Since P is true and P k is true implies that P k + is true, by mathematical induction, u n = for all positive integers n. nn ( + ) 9(ii) 9(iii) N N = r ( + )( r+ ) r r r= r= = ( u u + u u + u u 4 + u u N N+ ) ( u u ) = ( u un + ) = ( N + )( N + ) N + + = lim 4 5 N r = r + ( ) r+ + For r, ( r+ ) > rr ( + )( r+ ) OR < OR ( r+ ) rr ( + )( r+ ) N N < = < r= ( r+ ) r= rr ( + )( r+ ) 4 ( N+ )( N+ ) 4 Hence N lim < lim = N n r= r ( ) Page of 6
24 0 NJC H Math Prelim P Solutions 9(iv) Let r = j. Then r = j+ and r = j+. N j+ = N = r r r j j+ j+ ( )( ) j ( )( ) r= 0 + = 0 = j= N j= 8 ( + )( j+ ) j j j= N j= 7 = j j j j j j ( + )( + ) j ( + )( + ) j= = = N( N ) (8)(9) = 44 N( N ) 0 (a)(i) a+ 4a y = a By long division, 6a y = + a a Vertical asymptotes: Horizontal asymptote: y = = a or = a 0 (a) (ii) 6a a From y = +, a ( a)( a) ( 6a a)( ) 0 ( ) ( a )( + a + ) ( a ) ( a )( + a ) ( a ) dy = + d a = = Page of 6
25 dy = 0 d 0 NJC H Math Prelim P Solutions ( a )( + a ) ( a ) = 0 + a = 0 Given C has two turning points, > ( ) ( )( a ) b 4ac 0 4 > 0 a a > 6 < 0 ( a )( a ) < 0 6< a < 6 Since a is a positive constant, 0< a < 6. (Shown) 0 y 0 ( 0, 4) ( 6 + 5,.6) ( 6 5,.6) y = = 4 = 4 ( 6 5) ( ) y + ( ) h + + y+ = h ( ) = h It is an ellipse centred at ( 6 5, ). Page of 6
26 0 NJC H Math Prelim P Solutions Maimum turning point of C occurs at ( 6 5,.6). y 0 ( 6 5, ) ( 6 5,.6) 6+ 5 y = = 4 = 4 For the ellipse to intersect the curve C more than once, h >.6.. Since h is a positive integer, h. 0 (b) y y = f ( ) A 5, 6 0 y = 0 B, = y Page 4 of 6
27 0 NJC H Math Prelim P Solutions (i) ( 4,) y = f ( ) ( ) 0 9, 4 8, 4 4 y = f f (, ) 9, 8 4 y = f ( ) (, 4) f () = f () = f () f () = + + = = = or Since, = 4. (ii) R g = ( ln, ln( k )) ( 4,9/8] = D f For least k, ln( k ) = 9/8 k = ( e 9/8 )/ or.0 (iii) f() < h() + < 8 8 Consider > + < 8( ) > 0 ( 5)( + ) > 0 or.5 Then >.5 Page 5 of 6
28 0 NJC H Math Prelim P Solutions Consider < + < 8( + ) > 0 ( 8.5)( ) > 0 or 8.5 Then or.5 < 4 Page 6 of 6
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