2013 Bored of Studies Trial Examinations. Mathematics SOLUTIONS

Transcription

1 03 Bored of Studies Trial Examinations Mathematics SOLUTIONS

2 Section I. B 3. B 5. A 7. B 9. C. D 4. B 6. A 8. D 0. C Working/Justification Question We can eliminate (A) and (C), since they are not to 4 significant figures. By direct calculation e π = Question S n = n (n + ) as S n is an arithmetic series where n is an integer. By trial and error (and taking positive solutions of n), if S n = n = 6 S n = 55 n = 0 S n = 9 n = 3 S n = 55 n = + 4 which is not an integer Question 3 Solving directly using log laws: ( ) 36 x ln = ln 5 x 36 x = 5x (x + 9)(x 4) = 0 x = 4, 9 BUT from the original equation, we can only take the logarithms of positive values. In other words we have the restriction 36 x > 0 and x > 0. Only x = 4 satisfies this condition.

3 Question 4 π 0 sin kx dx = k [cos kx]π 0 = (cos kπ ) k We require (cos kπ ) = 0 k cos kπ = This is satified, when k = 0, ±, ±4, ±6, ±8,... (i.e. the even numbers) Question 5 Let B, G and R represent blue, green and red respectively. Calculating the relevant probabilities: P (BB) = P (RR) = P (GG) = 9 8 P (BG GB) = Note that P (BR RB) = (A) is true since P (BB) = P (RR) (B) is false since P (BB) P (GG) (C) is false since P (BG GB) P (RR) (D) is false since P (BR RB) P (GG)

4 Question 6 di dt = t 5 I(t) = t t 0 + c Assuming I(t) 0, note that (A) is true since I(0) = I(0) = c (B) is false since di dt actually decreases when t increases (C) is false since I(5) = 5 + c = 5 + I(0) > 0 (D) is false since at t = 5, di dt = 0 but d I dt = 5 < 0 (i.e. local maximum occurs at t = 5) Question 7 Since the circle contains dotted lines, then we can eliminate (A) and (C), leaving us with (B) and (D). Since the absolute value graph has solid lines, we can eliminate (D). Question 8 Note that (A) is true as per ratio of intercepts of parallel lines theorem (B) is true as per the converse of the ratio of intercepts of parallel lines theorem (or can use similar triangles argument) (C) is true since this leads to ABC ADE with ratio of corresponding sides : (D) is not always true as it is possible to construct DE to be twice of BC whilst not being parallel to BC which immediately violates the statement that B and C are midpoints of AD and AE. 3

5 Question 9 Deceleration implies dv < 0 so there should be negative slope at first. By changing direction, the velocity dt changes sign (thus eliminating (A) and (B)). Accelerating quickly to the original speed and remaining at that speed implies that the velocity must be relatively stable, which eliminates (D). Question 0 Note that (A) is not always true since f(x) = g(x) + c for some non-zero constant c satisfies f (x) = g (x) (B) is not always true since we can have f(x) = g(x)+c for some non-zero constant c at the stationary point (C) is always true since differentiating both sides of f (x) = g (x) leads to f (x) = g (x) for all real x (D) is not always true since at the stationary points f (x) g (x) = 0 4

6 Section II Question (a) ax bx square both sides a x ax + b x bx + (b a )x x(b a) 0 x(b a)((b + a)x ) 0 x 0 or x a + b Note that 0 < a < b then b a > 0 and a + b > 0 (b) ( ( )) d x ln = d ( ) ln( x ) ln( x + ) dx x + dx = x( x ) x( x + ) = ( x + ) ( x ) x x( x )( x + ) x = x x(x ) (c) 3k² y Consider the sketch of y = x k for k x k Range is k f(x) 3k k O k k x k² 5

7 (d)(i) y O π π 3π π x (d)(ii) First solve for sin x = for 0 x π or 0 x 4π sin x = x = π 6, 5π 6, 3π 6, 7π 6 x = π, 5π, 3π, 7π Reading from the graph, solutions are: 0 x π or 5π x 3π or 7π x π (e)(i) 3b y The general equation is (y y 0 ) = 4a(x x 0 ) where (x 0, y 0 ) is the vertex and a is the focal length. The y-coordinate of the vertex must also be the y- coordinate of the focus (b, 3b) hence y 0 = 3b. Also, the horizontal distance between the focus and the vertex is the focal a then x 0 = b + a. So the equation of the parabola is (y 3b) = 4a(x (b + a)). O b b + a k x 6

8 (e)(ii) The horizontal distance between the vertex and the directrix is the focal length a. Hence k = b + a, but b < k 3b. Substituting k = b + a we get b < b + a 3b which gives 0 < a b (f) S = sin k θ Note that 0 < sin θ < since 0 < θ < π k=0 = + sin θ + sin 4 θ +... which is a limiting sum = sin θ = cos θ Similarly, C = cos k θ Note that 0 < cos θ < since 0 < θ < π k=0 = + cos θ + cos 4 θ +... which is a limiting sum = cos θ = sin θ LHS = S + C = cos θ + sin θ = = RHS 7

9 (g) Let A be the area bounded by y = x, x = a, x = b and the positive x axis. A = b a x dx = [ln x] b a = ln b ln a = ln ( b a ) If we similarly define A then we have A = ln But a b = a b b = b. a a ( ) b Hence A = ln so A = A. a ( b a ). 8

10 Question (a)(i) Let the volume of the solid be V. y π V = π 6 tan x dx 0 = π π 6 0 (sec x ) dx π O π 6 π x = π[tan x x] π 6 0 ( = π 3 π ) 6 cubic units (a)(ii) Let the volume of the cylinder which has radius 3 and height π 6 be denoted V C. We can see that π V C > V ( ) 3 π ( 6 > π 3 π ) 6 π 8 + π 6 > π > 3 3 Also, we know that V > 0 ( π 3 π ) > 0 6 π 6 < π < < π < 3 9

11 (b) (ax) ln a = (bx) ln b ln ( (ax) ln a) = ln ( ln (bx) b) ln a ln ax = ln b ln bx ln a(ln a + ln x) = ln b(ln b + ln x) ln x(ln a ln b) = (ln b) (ln a) ln x(ln a ln b) = (ln b ln a)(ln b + ln a) but a b ln a ln b ln ln x = ln b + ln a ( ) = ln ab x x = ab x = ab (c) y = f(x) dy dx = f (x) [f(x)] d y dx = f (x)[f(x)] + f(x)[f (x)] [f(x)] 4 Since f(x) has a minimum stationary point at x = α then f (α) = 0 and f (α) > 0, provided f(α) 0. 0

12 So when x = α for y = f(x) then d y dx = f (α)[f(α)] + f(α)[f (α)] [f(α)] 4 = f (α) [f(α)] < 0 since f (α) > 0 and [f(α)] > 0 Hence a maximum occurs at x = α for y = f(x) (d)(i) LHS = cos θ + sin θ + cos θ sin θ = cos θ ( ) ( sin θ) + ( + sin θ) ( + sin θ)( sin θ) = cos θ sin θ = cos θ cos θ = cos θ = sec θ = RHS

13 (d)(ii) π 4 0 sec θ dθ = π 4 0 ( cos θ + sin θ + cos θ ) dθ sin θ = [ln( + sin θ) ln( sin θ)] π 4 0 = = [ ( ln + ) ( ln )] [ ln ( ) ( )] + ln = [ ( ) ln + ln ( ) ln + ln ] ( = ) + + ln + = ln ( + ) ( ) = ln +

14 Question 3 (a) In order to win Game A, the person must pull any card other than the target number without replacement, in each turn for all (n ) turns. P (win Game A) = n n n n n 3 n... 3 = n In order to win Game B, the person only needs to select the target number once so P (win Game B) = n = P (win Game A) (b)(i) y = x dy dx = x = r at point R Gradient of P Q m P Q = p q p q = p + q But P Q is parallel to the tangent at R hence r = p + q r = p + q 3

15 (b)(ii) First obtain the equation of P Q. y p = (p + q)(x p) (p + q)x y pq = 0 Let d be the perpendicular distance from P Q to R. d = r(p + q) r pq (p + q) + but r = p + q = (p+q) (p+q) pq 4 (p + q) + = (p + q) 4pq 4 (p + q) + = p pq + q 4 (p + q) + (q p) = 4 (p + q) + We also need the distance P Q. P Q = (q p) + (q p ) = (q p) + (q p) (q + p) = (q p) + (p + q) noting that p < r < q q p > 0 Let A be the area of the triangle P QR A = d P Q = (q p) 4 (p + q) + (q p) + (p + q) = 8 (q p)3 square units 4

16 (c) Let the two roots be α and β. Since the two roots are real then the disciminant is non-negative. a 4b 0 a 4b a b noting that a > 0 and b > 0 Since the roots are no more than unit apart then α β. α β α + β αβ (α + β) 4αβ a 4b a 4b + 4b a + 4b Extracting the right hand side of the inequality a + 4b a + 4b noting that a > 0 and b > 0 b a + 4b (d)(i) B tan α = BD AD tan α = AD BD X Similarly, tan β = CD BD A D C 5

17 LHS = tan α + tan β = AC + CD BD = AC BD but AC = BD = = RHS (d)(ii) Using the similar approach to part (i) then tan(90 β) + tan β = cot β + cot β = but tan(90 β) = cot β tan β + cot β = (d)(iii) In part (ii) we showed that if we can construct DX such that DX = BC and DX BC then tan β + cot β =. But tan β + tan β = tan β tan β + = 0 The disciminant of this quadratic equation in tan β is 3 which is negative so there are no real roots for tan β. This means there cannot exist such a β where the condition in (ii) is satisfied. Hence the initial assumption that we find a point X where we can construct DX such that DX = BC and DX BC is false. 6

18 Question 4 (a) At some particular time say t = t 0 we have A(t 0 ) = B(t 0 ) and db dt = k da dt bp 0 e bt 0 = akp 0 e at 0 but A(t 0 ) = P 0 e at 0 and B(t 0 ) = P 0 e bt 0 bb(t 0 ) = aka(t 0 ) ba(t 0 ) = aka(t 0 ) k = b a noting that A(t) 0 (a)(ii) If b = a then k = 4. Given the condition A(t 0 ) = B(t 0 ) P 0 e at 0 = P 0 e bt 0 e at 0 = e at 0 e at 0 = t 0 = ln a (b)(i) f(x) = x x + f (x) = (x + ) x (x + ) = x (x + ) Stationary points occur when f (x) = 0 7

19 x (x + ) = 0 x = noting that x 0 f() = Using the first derivative test around the neighbourhood of the stationary point x f (x) Maximum turning point at (, ) (b)(ii) f(x) = x x + x x = x + x as x 0 then f(x) 0 since x 0 and x 0 (b)(iii) y ½ O x 8

20 (b)(iv) y = mx and y = f(x) enclose a region if they intersect more than once (assuming continuity of the curves across the region). This means we require multiple solutions to the equation: mx = x x + x(mx ( m)) = 0 The quadratic factor needs to have real roots for the original equation to have multiple solutions. So we require the discriminant to be non-negative m( m) 0 m(m ) 0 0 m BUT consider the boundary cases, when m = 0 or m = we only get the solution x = 0 in. This also is consistent graphically, as m = 0 for y = mx gives a flat horizontal line which will never enclose a bounded area with the curve and m = is actually where the line y = mx is tangent to the curve. Hence the condition is actually 0 < m <. (b)(v) Let the area be A. First find the points of intersection by solving. x(mx ( m)) = 0 x = 0, m m noting the given domain x 0 A = m m 0 ( ) x x + mx dx = [ ln(x + ) mx ] m m 0 = ( ) ln + m m = m ln m square units 9

21 (b)(vi) Note that A > 0 m ln m > 0 m > ln m e m > m Given m is some variable between 0 and we can say that e x > x where some variable x is used in place of m. (c) Let I = a a f(x) dx and I T be the trapezoidal rule approximation to I. I T = a (f(a) + f( a) + f(0)) = a (g(a) + k + g( a) + k + g(0) + k) but since g(x) is an odd function g( a) = g(a) = a(g(0) + k) but g( 0) = g(0) g(0) = 0 = ak But I = a a (g(x) + k) dx = a a a g(x) dx + k dx a but since g(x) is an odd function a a g(x) dx = 0 = k[x] a a = ak = I T the trapezoial rule approximation gives the exact value of the integral 0

22 Question 5 (a)(i) We know that ABC is common. BQ = BP BC but AB = BQ AB = BP BC AB BC = BC BP AP B ABC (two sides in proportion and included angle equal) (a)(ii) Let BAP = x and P AQ = y Since AB = BQ then BAQ is isosceles hence BQA = BAQ = x + y (equal angles opposite equal sides) BP A = P AQ + AQP = x + y (exterior angle equals sum of opposite interior angles of a triangle) Since ABC is common and AP B ABC, then we have either BAP = BAC or BAP = ACB for corresponding angles to be equal. However, BAC is clearly greater than BAP (as y > 0) so therefore BAP = ACB = x QAC + ACQ = AQP (exterior angle equals sum of opposite interior angles of a triangle) QAC = AQP ACQ = x + y x = y = P AQ AQ bisects P AC

23 (b)(i) ẍ = t + ẋ = t + dt = ln( + t) + c When t = 0 then ẋ = 0 c = 0 ẋ = ln( + t) The distance travelled from time zero to time t = T where the displacement is x = d is d = T 0 ln( + t) dt (b)(ii) Consider y = e x. Let A be the area bounded by the curve, the y-axis and the line y = T. Let A be the area bounded by the curve y = e x, the x-axis and the line x = ln(t + ). y y = e x e x = + y T A x = ln( + y) O A ln (T + ) x A = T 0 ln( + y) dy But we can also evaluate A by considering A. A = ln(t +) 0 (e x ) dx = [e x ln(t +) x] 0 = e ln(t +) ln(t + ) = T ln(t + )

24 A + A = Area of rectangle = T ln(t + ) A = T ln(t + ) T + ln(t + ) = (T + ) ln(t + ) T d = T 0 ln( + t) dt = (T + ) ln(t + ) T (c)(i) and (ii) y m AP = r sin α 0 r cos α r (0, r) P (rcos α, rsin α) = sin α cos α A ( r, 0) O A (r, 0) x m A P = r sin α 0 r cos α + r (0, r) = sin α cos α + m AP m A P = sin α cos α sin α cos α + = sin α cos α = sin α sin α AP A P = 3

25 (d)(i) From part (c)(ii), OAR = 90 cos θ = OA OR OA = a cos θ but OR = a (d)(ii) The x-value of P is equal to the length of OD. The y-value of P is equal to the length of AC. Note that BD = OR = a and ROB = OAC = OBD = θ (alternate angles as OR AC DB). In OCA: cos θ = AC OA AC = OA cos θ y = a cos θ from part (i) In BDO: tan θ = OD BD OD = BD tan θ x = a tan θ (d)(iii) y = a cos θ = a sec θ = a + tan θ but x = a tan θ = a + x 4a = 8a3 4a + x 4

26 Question 6 (a)(i) s h s r First note that the circumference of the base of the cone equals the arc length of the sector. πr = sθ r = sθ π Also, by Pythagoras theorem: r + h = s h = s r = s s θ 4π = s 4π θ π Let V be volume of the cone. V = πr h 3 = π 3 s θ 4π s 4π θ π = s3 θ 4π 4π θ 5

27 dv dθ = s3 4π ( θ ) θ 3 4π θ 4π θ ( ) = s3 θ(4π θ ) θ 3 4π 4π θ ( ) = s3 8θπ 3θ 3 4π 4π θ Stationary points occur when dv dθ = 0 ( ) s 3 8θπ 3θ 3 = 0 4π 4π θ θ(8π 3θ ) = 0 θ = 0 or θ = 8π 3 but clearly θ = 0 does not maximise the volume so consider θ = 8π 3 θ = π as 0 < θ < π = π 6 3 Since we are given that d V < 0 at this value, and there are no other stationary points in the domain, dθ then the volume is maximised at this value of θ. 6

28 (a)(ii) We know that A = s θ and B = πr but volume is maximised at θ = π 6 3 ( 3B = 3π s π π ) and also r = sθ π. = 4π s 4 3 A = 4 s4 ( π ) 6 3 3B = A = 4π s 4 3 (b)(i) In ABC: By the sine rule A sin ABC AC = sin BAC BC sin ABC = AC sin CAB BC B C (b)(ii) Similarly for P QR sin P QR = sin ABC sin P QR P R sin RP Q QR = AC QR sin CAB P R BC sin RP Q = since AC = P R, BC = QR and CAB = RP Q sin CAB = sin RP Q 7

29 sin ABC = sin P QR since ABC = P QR or ABC + P QR = 80 sin(80 x) = sin x (c)(i) Let A j be the value of the j-th monthly deposit n months from present. A = MR n A = MR n A 3 = MR n A n = MR P = A + A + A A n = MR( + R + R R n ) which is a geometric series = MR ( ) R n R but S = Rn R = MRS (c)(ii) Let B j be the amount remaining in the account j months after retirement. B = RP km B = RB km = R P kmr km B 3 = RB km = R 3 P kmr kmr km

30 B m = R m P kmr m kmr m... kmr km B m = R m P km( + R + R R m ) = R m P km(rm ) R We expect B m = 0 where the account is exhausted R m P km(rm ) R = 0 R m P = km(rm ) R R m P (R ) = kmr m km R m (km P (R )) = km R m = km (km P (R )) ( ) ln(r m km ) = ln (km P (R )) but P = MRS ( ) km m ln R = ln (km MRS(R )) but S = Rn R m = ( ) ln R ln k k R(R n ) (c)(iii) When k R(R n ) then m is undefined which means there does not exist an m such that account gets reduced to zero. In other words, the account actually grows over time where the rate of interest gained exceeds the rate of monthly withdrawals (as an aside, one can formally show that B j > B j ). 9