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1.

Mr Wang had a total of $37 000 in his savings accounts with Action Bank, Bonus Bank and Champion Bank at the beginning of 2007. Saving accounts at Action Bank, Bonus Bank and Champion Bank enjoy interest rates of 1%, 0.5% and 0.3% per annum respectively. He had a total of $37 240 in the three banks at the end of 2007 and $37 481.89 at the end of 2008. Assuming that he did not deposit or withdraw any money in 2007 and 2008, find the amount of money he had in his Action Bank account at the end of 2008.

[4]

∫ x cos ( ln x ) dx .

2.

Solve

3.

Solve the inequality

4.

A curve is defined by the parametric equations x = t 2 and y = sin t , for −π < t ≤ π .

5.

[5]

x > ln x . Deduce the solution of x > 2 2 ln x. 2

[5]

(i)

Sketch the curve, indicating clearly the axial intercepts.

[2]

(ii)

Find the equation of the normal to the curve that is parallel to the y-axis.

[4]

The first three terms of a geometric progression are (a + ln b) , (a + ln b ) and 1 (a + ln 3 b ) , where b > 1. Show that a = − ln b . Find the common ratio and deduce 4 that the geometric progression is convergent. Given that b = e2, find the sum to infinity of the progression.

6.

[6]

2x2 + 3x – 1 Express (x – 1)(x2 + 1) in partial fractions. 2x2 + 3x – 1 Hence find the series expansion of (x – 1)(x2 + 1) in ascending powers of x, up to and including the term in x4. Find the coefficient of x2008.

[6]

x

7.

Find the distance between a point (x, y) on the curve y = e 2 and the point (1, 1) in x

terms of x. Hence find the coordinates of the point on the curve y = e 2 that is closest to the point (1, 1), giving your answer correct to 3 decimal places.

1

[6]

8.

A communicable disease is spreading within a small community with a population of 1000 people. A scientist proposes that the infected population, x, at time t days after the start of the spread of the disease, satisfies the differential equation dx = kx(1000 − x ) , where k is a positive constant. dt Initially one person in this community is infected and five days later, 12% of the population is infected. Find the time taken for half the population to contract the disease. State an assumption made by the scientist.

9.

[7]

A sequence is defined by u1 = 0 and (1 + n)u n +1 = n + u n for n ∈ + . Prove by mathematical induction that un = 1 −

1 for n ∈ + . n!

[5]

∞

Hence find the exact value of

10.

1 − ur . r r =1 2

∑

[2]

2

2

The sequence ur , r = 1, 2, 3, …, is defined by ur = [ r (r + 1)] − [ (r − 1)r ] . n

Find

∑u

r

.

[2]

r =1

2

n(n + 1) r3 = . ∑ 2 r =1 n n 1 By considering ∑ (r − 1) 3 , show that ∑ r 2 = n(n + 1)(2n + 1) . 6 r =1 r =1 n

(i) (ii)

11.

By simplifying ur , deduce that

[2] [5]

The diagram shows the graph of y = f(x): y y=x (1, 2)

1 y= 2 (–1, 0)

x

On separate clearly labelled diagrams, sketch the graphs of (i)

1 y = f(x) ,

(ii)

y = f(– | x | ),

(iii)

y = f ′(x).

[9]

2

12.

Given that y = x tan −1 x , prove that (1 + x 2 )

d2 y dy + 2x − 2 y − 2 = 0 . 2 dx dx

By repeated differentiation, show that the first 2 non-zero terms of the Maclaurin’s

x − tan −1 x 1 series for y is x 2 − x 4 . Hence evaluate lim . x →0 3 x3

13.

(i)

Use the substitution u = 2 x + 1 to find

∫x

[9]

2 x + 1 dx .

The region R is bounded by the curve y = x 2 x + 1 , the y-axis and line y =

[3]

1 . 2

(ii)

Find the exact area of region R using your result in part (i).

(iii)

Find the volume of the solid generated when R is rotated through four right angles about the x–axis.

14.

[3]

[3]

The functions f and g are defined as follows: f : x 3 − 2 x − x 2 , x ∈ , x ≤ k ,

g:x e

4− x

, 0 ≤ x ≤ 4.

State the largest value of k such that f −1 exists, and find f −1 in a similar form.

[4]

(i)

Show that the composite function gf does not exist.

[1]

(ii)

If h is a restriction of f, write down the maximal domain of h such that the composite function gh exists. Define gh in a similar form and state its range. [4]

(iii)

Find the set of values of x such that g–1g(x + 1) = g g–1(x + 1).

END OF PAPER

3

[3]

2008 C1 H2 Mathematics Promotional Examination Solution: 1

2

Let a, b, c be the amount in his Action Bank, Bonus Bank and Champion Bank accounts at the beginning of 2007. a + b + c = 37 000 –––(1) 1.01a + 1.005b + 1.003c = 37 240 –––(2) 1.012a + 1.0052b + 1.0032c = 37 481.89 –––(3) From GC, a = 15 000, b = 12 000, c = 10 000. So at the end of 2008, he had 1.012 × 15 000 = $15 301.50 in his Action Bank account. u = cos ( ln x ) v' = x 1 u ' = − sin ( ln x ) x ∫ x cos ( ln x ) dx

v=

1 2 x 2

1 2 1 x cos ( ln x ) + ∫ x sin ( ln x ) dx 2 2 u = sin ( ln x ) v' = x =

1 cos ( ln x ) x ∫ x cos ( ln x ) dx

u'=

v=

1 2 x 2

1 2 1 1 1 x cos ( ln x ) + x 2 sin ( ln x ) − ∫ x cos ( ln x ) dx 2 2 2 2 5 1 1 x cos ( ln x ) dx = x 2 cos ( ln x ) + x 2 sin ( ln x ) ∫ 4 2 4 2 2 1 2 ∫ x cos ( ln x ) dx = 5 x cos ( ln x ) + 5 x sin ( ln x ) + C

=

3

From GC, 0 < x < 4.42806 or x > 13.706 i.e. 0 < x < 4.42 or x > 13.8 x2 2 2 Replacing x by x : 2 > ln x x > 2 2 ln x From above, 0 < x2 < 4.42806 or x2 > 13.706 0 < x < 2.1043 or x > 3.702 0 < x < 2.10 or x > 3.71

4

4(i)

π2

O

4(ii) dx dy = 2t = cos t dt dt dy dy dt cos t = × = dx dt dx 2t Normal // y-axis ⇒ Tangent // x-axis ⇒

dy =0 dx

cos t = 0 ⇒ cos t = 0 2t ∴t = − When t = ±

π 2

π 2

or

, x=

π 2

π2 4

Equation of normal is x =

5

6

π2 4

1 1 Since the 1st 3 terms are in G.P., (a + ln b)2 = (a + ln b)(a + ln b) 2 3 1 4 1 2 2 2 2 a + a ln b + (ln b) = a + a ln b + (ln b) 4 3 3 1 1 1 a = − ln b ⇒ a = − ln b 3 12 4 1 1 a + ln b ln b 1 2 4 Common ratio, r = = = 3 a + ln b ln b 3 4 1 Since r = < 1 , the G.P. is convergent. 3 1 1 1 Given that b = e2, a = − ln e 2 = − 2 ln e = − 4 4 2 2 a + ln e 3 1 9 S∞ = = (− + 2) = 1 2 2 4 1− 3 2x2 + 3x – 1 A Bx + C Let (x – 1)(x2 + 1) = x – 1 + x2 + 1 2x2 + 3x – 1 = A(x2 + 1) + (Bx + C)(x – 1) Let x = 1: 4 = 2A ⇒ A = 2 Let x = 0: –1 = 2 – C ⇒ C = 3 Compare coefficients of x2 : 2 = 2 + B ⇒ B = 0

5

2x2 + 3x – 1 2 3 (x – 1)(x2 + 1) = x – 1 + x2 + 1 = –2(1 – x)–1 + 3(1 + x2)–1 = –2(1 + x + x2 + x3 + x4 + ...) + 3(1 – x2 + x4 – ...) = 1 – 2x – 5x2 – 2x3 + x4 +... Coefficient of x2008 = –2 + 3 = 1

7

2

( x − 1) + ( y − 1)

Distance between (1, 1) and (x, y), W =

2

x

Since (x, y) lies on the curve, then y = e 2 Thus, W =

( x − 1) 2

2

(

(

)

x

2

+ e 2 −1 x

)

⇒ W 2 = ( x − 1) + e 2 − 1

2

)

(

x x dW = 2 ( x − 1) + e 2 e 2 − 1 dx x dW 1 x = ( x − 1) + e 2 e 2 − 1 dx 2 dW 1 x2 x2 When = 0, ( x − 1) + e e − 1 = 0 dx 2

Differentiating w.r.t. x, 2

)

(

)

(

From GC, x = 0.70160785 dW dx

0.70160785-

0.70160785

0.70160785+

-ve

0

+ve

–

/

\ Thus, W is minimum when x = 0.70160785 0.70160785

8

2 When x = 0.70160785, y = −e = 1.4202 The point is (0.702, 1.420) [to 3 d.p.] dx = kx(1000 − x ) dt 1 ∫ x(1000 − x) dx = ∫ kdt 1 1 Method 1(Partial Fraction): ∫ + dx = kt + C 1000 x 1000(1000 − x) 1 [ln x − ln(1000 − x)] = kt + C 1000 1 x ln = kt + C 1000 1000 − x

6

Method 2(Complete the sq.):

1 dx = kt + C − ( x − 500) 2 1 x ln = kt + C 1000 1000 − x

∫ (500)

2

x = Ae1000 kt 1000 − x x = A(1000 − x)e1000 kt When t = 0, x = 1, A =

1 999

When t = 5, x = 120, k =

9

1 2997 ln 5000 22

When x = 500, 999 = e1000 kt ln 999 t= = 7.03 days = 168 hrs 39 mins 1000k Assumption: No one leaves and enters the community; birth rate = death rate. 1 Let Pn be the statement: un = 1 − for n ∈ + . n! To prove that P1 is true: When n = 1, LHS = u1 = 0 1 RHS = 1– = 0 = LHS 1! ∴ P1 is true. 1 Assume that Pk is true for some k ∈ + , i.e. uk = 1 − . k! 1 To prove that Pk+1 is true, i.e. to prove uk +1 = 1 − (k + 1)! u k When n = k + 1, LHS = u k +1 = + k 1+ k 1+ k 1 1 = k +1− 1+ k k ! 1 (k + 1)k !− 1 = 1 + k k! (k + 1)!− 1 1 = = 1− = RHS (k + 1)! (k + 1)! Thus, Pk is true ⇒ Pk+1 is true Since P1 is true and Pk is true ⇒ Pk+1 is true, by the Principle of Mathematical Induction, the statement is true for all n ∈ + . 1 ) ∞ ∞ 1 − (1 − 1 − ur r! = ∑ ∑ r 2r r =1 2 r =1 ∞ 1 =∑ r r =1 2 r !

= e1/ 2 − 1

7

10

n

∑u

r

= 12 (2 2 ) − 0

r =1

+2 2 (32 ) − 12 (22 ) +32 (4 2 ) − 22 (32 ) +... + n 2 (n + 1) 2 − (n − 1)2 n 2 = [n( n + 1) ]

2

10i Now, u r = r 2 (r + 1) 2 − (r − 1) 2 r 2 = r 2 ( r + 1 + r − 1)[ r + 1 − ( r − 1)] = r 2 (2r )(2) = 4r 3 Thus,

n

n

r =1

r =1

∑ u r = 4∑ r 3 = [n(n + 1)]

2

n(n + 1) ⇒ ∑r = 2 r =1 10ii Method 1: n

2

3

n

n

∑ (r − 1) 3 = ∑ (r 3 − 3r 2 + 3r − 1) r =1 n −1

r =1

∑ r3 =

n

∑ r 3 − 3∑ r 2 + 3∑ r − n

n

n

r =1

r =1

r =1

r =1

2

2

n(n + 1) (n − 1)n 3n 3∑ r 2 = − + (n + 1) − n 2 2 2 r =1 3n = n3 + (n + 1) − n 2 n n = (2n 2 + 3n + 1) = (n + 1)(2n + 1) 2 2 n 1 r 2 = n(n + 1)(2n + 1) ∑ 6 r =1 Method 2: n

n

n

r =1

r =1

∑ (r − 1) 3 = ∑ (r 3 − 3r 2 + 3r − 1) = n

n

n

r =1

r =1

n

n

r =1

r =1

∑ r 3 − 3∑ r 2 + 3∑ r − ∑1

n

n

r =1

r =1

3∑ r 2 = ∑ [ r 3 − ( r − 1)3 ] + 3∑ r − n r =1

= [13 – 03 + 23 – 13 : + n3 – (n – 1)3 ] + = n3 +

3n (n + 1) − n 2

3n (n + 1) − n 2

8

=

n n (2n 2 + 3n + 1) = (n + 1)(2n + 1) 2 2

n

1 n(n + 1)(2n + 1) 6 r =1 Method 3:

∑r

2

=

n

n

r =1 n

r =1

∑ (r − 1) 3 = ∑ (r 3 − 3r 2 + 3r − 1) (∑ r 3 ) − n3 = r =1 n

n

n

n

r =1

r =1

r =1

∑ r 3 − 3∑ r 2 + 3∑ r − n

3n 3∑ r 2 = n3 + ( n + 1) − n 2 r =1 n n = (2n 2 + 3n + 1) = (n + 1)(2n + 1) 2 2 n 1 r 2 = n(n + 1)( 2n + 1) ∑ 6 r =1

11i y=

1 f(x) y=2

(1, ½) (0, 0) x = –1

11ii y = f(– | x | ) y=

1 2 (–1, 0) (1, 0)

11iii y = f ′(x) y=1 –1 1

12

y = x tan −1 x

9

x dy = tan −1 x + dx 1 + x2 (1 + x2 ) ddyx = (1 + x2 ) tan −1 x + x dy d2 y 2 x + (1 + x 2 ) 2 = 2 x tan −1 x + 1 + 1 dx dx 2 (1 + x2 ) ddxy2 + 2 x ddyx − 2 y − 2 = 0 3 2 2 (1 + x2 ) ddxy3 + 2 x ddxy2 + 2 x ddxy2 + 2 ddyx − 2 ddyx = 0 3 d2 y 2 d y 1 + x + 4 x ( ) dx3 dx2 = 0 4 d3 y d2 y 2 d y 1 + x + 6 x + 4 ( ) dx4 dx3 dx2 = 0 When x = 0, y = 0 dy =0 dx d2 y =2 dx 2 d3 y =0 dx 3 d4 y = −8 dx 4 ( −8) x 4 + ... = x 2 − 1 x 4 + ... 2 Maclaurin’s Series for y is y = x 2 + 2! 4! 3 Method 1: x − tan −1 x x 2 − x tan −1 x lim = lim x →0 x →0 x3 x4 1 x 2 − x 2 − x 4 + ... 3 = lim x→0 x4 1 = lim + ... x→0 3 1 = 3 Method 2: 1 x − x − x 3 + ... −1 x − tan x 3 lim = lim x →0 x →0 x3 x3 1 1 = lim + ... = x →0 3 3 13i du u = 2x +1 ⇒ =2 dx 10

∫x

2 x + 1 dx = ∫ =

u −1 1 u du 2 2

1 1 32 u − u 2 du ∫ 4 5

3

1u 2 1u 2 = − +C 4 52 4 32 5 3 1 1 = (2 x + 1) 2 − (2 x + 1) 2 + C 10 6 13ii 1 1 Let x 2 x + 1 = . From GC, x = 2 2 y y = x 2x + 1

x 1 2

Area of R 1/2 1 1 = − ∫ x 2 x + 1 dx 2 2 0 1/ 2

5 3 2 ( 2 x + 1) 2 ( 2 x + 1) 2 = − − 4 10 6 0 2 2 2 2 1 1 = − − − ( − ) 4 5 3 10 6

= 13iii

11 2 1 − 60 15

=

1 (11 2 − 4) units 2 60 2

1 1 1 2 2 Volume of solid = π 2 − π ∫ 0 x ( 2 x + 1) dx 2

1/2

x 4 x3 = −π + 4 2 3 0 17π = = 0.556 units3 96

π

14 f(x) = 3 – 2x – x2 = 4 – (x + 1)2 For f −1 to exist, f must be one–one. ∴ k = −1 To find f −1 : Let y = 4 – (x + 1)2 ⇒ x = −1 ± 4 − y

(−1, 4)

−3

1

11

Since x ≤ −1, x = −1 − 4 − y Thus, f −1 : x − 1 − 4 − x , x ∈ (−∞, 4] 14i Range of f = (−∞, 4], domain of g = [0, 4] Since range of f ⊆/ domain of g, gf does not exist. 14ii Maximal range of h for gh to exist is [0, 4], therefore maximal domain of h is [−3, −1] 2

2

gh(x) = e 4 − 4 + ( x +1) = e ( x +1) = e| x +1| = e − x −1 since x ∈ [ –3, –1] ∴ gh : x e − x −1 , x ∈ [−3, −1] Range of gh = [1, e2] = [1, 7.39] 14iii Dg −1 g ( x +1) = [ -1, 3] and Dgg −1 ( x +1) = [0, e2 –1] Hence set of values of x = [0, 3]

12

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Mr Wang had a total of $37 000 in his savings accounts with Action Bank, Bonus Bank and Champion Bank at the beginning of 2007. Saving accounts at Action Bank, Bonus Bank and Champion Bank enjoy interest rates of 1%, 0.5% and 0.3% per annum respectively. He had a total of $37 240 in the three banks at the end of 2007 and $37 481.89 at the end of 2008. Assuming that he did not deposit or withdraw any money in 2007 and 2008, find the amount of money he had in his Action Bank account at the end of 2008.

[4]

∫ x cos ( ln x ) dx .

2.

Solve

3.

Solve the inequality

4.

A curve is defined by the parametric equations x = t 2 and y = sin t , for −π < t ≤ π .

5.

[5]

x > ln x . Deduce the solution of x > 2 2 ln x. 2

[5]

(i)

Sketch the curve, indicating clearly the axial intercepts.

[2]

(ii)

Find the equation of the normal to the curve that is parallel to the y-axis.

[4]

The first three terms of a geometric progression are (a + ln b) , (a + ln b ) and 1 (a + ln 3 b ) , where b > 1. Show that a = − ln b . Find the common ratio and deduce 4 that the geometric progression is convergent. Given that b = e2, find the sum to infinity of the progression.

6.

[6]

2x2 + 3x – 1 Express (x – 1)(x2 + 1) in partial fractions. 2x2 + 3x – 1 Hence find the series expansion of (x – 1)(x2 + 1) in ascending powers of x, up to and including the term in x4. Find the coefficient of x2008.

[6]

x

7.

Find the distance between a point (x, y) on the curve y = e 2 and the point (1, 1) in x

terms of x. Hence find the coordinates of the point on the curve y = e 2 that is closest to the point (1, 1), giving your answer correct to 3 decimal places.

1

[6]

8.

A communicable disease is spreading within a small community with a population of 1000 people. A scientist proposes that the infected population, x, at time t days after the start of the spread of the disease, satisfies the differential equation dx = kx(1000 − x ) , where k is a positive constant. dt Initially one person in this community is infected and five days later, 12% of the population is infected. Find the time taken for half the population to contract the disease. State an assumption made by the scientist.

9.

[7]

A sequence is defined by u1 = 0 and (1 + n)u n +1 = n + u n for n ∈ + . Prove by mathematical induction that un = 1 −

1 for n ∈ + . n!

[5]

∞

Hence find the exact value of

10.

1 − ur . r r =1 2

∑

[2]

2

2

The sequence ur , r = 1, 2, 3, …, is defined by ur = [ r (r + 1)] − [ (r − 1)r ] . n

Find

∑u

r

.

[2]

r =1

2

n(n + 1) r3 = . ∑ 2 r =1 n n 1 By considering ∑ (r − 1) 3 , show that ∑ r 2 = n(n + 1)(2n + 1) . 6 r =1 r =1 n

(i) (ii)

11.

By simplifying ur , deduce that

[2] [5]

The diagram shows the graph of y = f(x): y y=x (1, 2)

1 y= 2 (–1, 0)

x

On separate clearly labelled diagrams, sketch the graphs of (i)

1 y = f(x) ,

(ii)

y = f(– | x | ),

(iii)

y = f ′(x).

[9]

2

12.

Given that y = x tan −1 x , prove that (1 + x 2 )

d2 y dy + 2x − 2 y − 2 = 0 . 2 dx dx

By repeated differentiation, show that the first 2 non-zero terms of the Maclaurin’s

x − tan −1 x 1 series for y is x 2 − x 4 . Hence evaluate lim . x →0 3 x3

13.

(i)

Use the substitution u = 2 x + 1 to find

∫x

[9]

2 x + 1 dx .

The region R is bounded by the curve y = x 2 x + 1 , the y-axis and line y =

[3]

1 . 2

(ii)

Find the exact area of region R using your result in part (i).

(iii)

Find the volume of the solid generated when R is rotated through four right angles about the x–axis.

14.

[3]

[3]

The functions f and g are defined as follows: f : x 3 − 2 x − x 2 , x ∈ , x ≤ k ,

g:x e

4− x

, 0 ≤ x ≤ 4.

State the largest value of k such that f −1 exists, and find f −1 in a similar form.

[4]

(i)

Show that the composite function gf does not exist.

[1]

(ii)

If h is a restriction of f, write down the maximal domain of h such that the composite function gh exists. Define gh in a similar form and state its range. [4]

(iii)

Find the set of values of x such that g–1g(x + 1) = g g–1(x + 1).

END OF PAPER

3

[3]

2008 C1 H2 Mathematics Promotional Examination Solution: 1

2

Let a, b, c be the amount in his Action Bank, Bonus Bank and Champion Bank accounts at the beginning of 2007. a + b + c = 37 000 –––(1) 1.01a + 1.005b + 1.003c = 37 240 –––(2) 1.012a + 1.0052b + 1.0032c = 37 481.89 –––(3) From GC, a = 15 000, b = 12 000, c = 10 000. So at the end of 2008, he had 1.012 × 15 000 = $15 301.50 in his Action Bank account. u = cos ( ln x ) v' = x 1 u ' = − sin ( ln x ) x ∫ x cos ( ln x ) dx

v=

1 2 x 2

1 2 1 x cos ( ln x ) + ∫ x sin ( ln x ) dx 2 2 u = sin ( ln x ) v' = x =

1 cos ( ln x ) x ∫ x cos ( ln x ) dx

u'=

v=

1 2 x 2

1 2 1 1 1 x cos ( ln x ) + x 2 sin ( ln x ) − ∫ x cos ( ln x ) dx 2 2 2 2 5 1 1 x cos ( ln x ) dx = x 2 cos ( ln x ) + x 2 sin ( ln x ) ∫ 4 2 4 2 2 1 2 ∫ x cos ( ln x ) dx = 5 x cos ( ln x ) + 5 x sin ( ln x ) + C

=

3

From GC, 0 < x < 4.42806 or x > 13.706 i.e. 0 < x < 4.42 or x > 13.8 x2 2 2 Replacing x by x : 2 > ln x x > 2 2 ln x From above, 0 < x2 < 4.42806 or x2 > 13.706 0 < x < 2.1043 or x > 3.702 0 < x < 2.10 or x > 3.71

4

4(i)

π2

O

4(ii) dx dy = 2t = cos t dt dt dy dy dt cos t = × = dx dt dx 2t Normal // y-axis ⇒ Tangent // x-axis ⇒

dy =0 dx

cos t = 0 ⇒ cos t = 0 2t ∴t = − When t = ±

π 2

π 2

or

, x=

π 2

π2 4

Equation of normal is x =

5

6

π2 4

1 1 Since the 1st 3 terms are in G.P., (a + ln b)2 = (a + ln b)(a + ln b) 2 3 1 4 1 2 2 2 2 a + a ln b + (ln b) = a + a ln b + (ln b) 4 3 3 1 1 1 a = − ln b ⇒ a = − ln b 3 12 4 1 1 a + ln b ln b 1 2 4 Common ratio, r = = = 3 a + ln b ln b 3 4 1 Since r = < 1 , the G.P. is convergent. 3 1 1 1 Given that b = e2, a = − ln e 2 = − 2 ln e = − 4 4 2 2 a + ln e 3 1 9 S∞ = = (− + 2) = 1 2 2 4 1− 3 2x2 + 3x – 1 A Bx + C Let (x – 1)(x2 + 1) = x – 1 + x2 + 1 2x2 + 3x – 1 = A(x2 + 1) + (Bx + C)(x – 1) Let x = 1: 4 = 2A ⇒ A = 2 Let x = 0: –1 = 2 – C ⇒ C = 3 Compare coefficients of x2 : 2 = 2 + B ⇒ B = 0

5

2x2 + 3x – 1 2 3 (x – 1)(x2 + 1) = x – 1 + x2 + 1 = –2(1 – x)–1 + 3(1 + x2)–1 = –2(1 + x + x2 + x3 + x4 + ...) + 3(1 – x2 + x4 – ...) = 1 – 2x – 5x2 – 2x3 + x4 +... Coefficient of x2008 = –2 + 3 = 1

7

2

( x − 1) + ( y − 1)

Distance between (1, 1) and (x, y), W =

2

x

Since (x, y) lies on the curve, then y = e 2 Thus, W =

( x − 1) 2

2

(

(

)

x

2

+ e 2 −1 x

)

⇒ W 2 = ( x − 1) + e 2 − 1

2

)

(

x x dW = 2 ( x − 1) + e 2 e 2 − 1 dx x dW 1 x = ( x − 1) + e 2 e 2 − 1 dx 2 dW 1 x2 x2 When = 0, ( x − 1) + e e − 1 = 0 dx 2

Differentiating w.r.t. x, 2

)

(

)

(

From GC, x = 0.70160785 dW dx

0.70160785-

0.70160785

0.70160785+

-ve

0

+ve

–

/

\ Thus, W is minimum when x = 0.70160785 0.70160785

8

2 When x = 0.70160785, y = −e = 1.4202 The point is (0.702, 1.420) [to 3 d.p.] dx = kx(1000 − x ) dt 1 ∫ x(1000 − x) dx = ∫ kdt 1 1 Method 1(Partial Fraction): ∫ + dx = kt + C 1000 x 1000(1000 − x) 1 [ln x − ln(1000 − x)] = kt + C 1000 1 x ln = kt + C 1000 1000 − x

6

Method 2(Complete the sq.):

1 dx = kt + C − ( x − 500) 2 1 x ln = kt + C 1000 1000 − x

∫ (500)

2

x = Ae1000 kt 1000 − x x = A(1000 − x)e1000 kt When t = 0, x = 1, A =

1 999

When t = 5, x = 120, k =

9

1 2997 ln 5000 22

When x = 500, 999 = e1000 kt ln 999 t= = 7.03 days = 168 hrs 39 mins 1000k Assumption: No one leaves and enters the community; birth rate = death rate. 1 Let Pn be the statement: un = 1 − for n ∈ + . n! To prove that P1 is true: When n = 1, LHS = u1 = 0 1 RHS = 1– = 0 = LHS 1! ∴ P1 is true. 1 Assume that Pk is true for some k ∈ + , i.e. uk = 1 − . k! 1 To prove that Pk+1 is true, i.e. to prove uk +1 = 1 − (k + 1)! u k When n = k + 1, LHS = u k +1 = + k 1+ k 1+ k 1 1 = k +1− 1+ k k ! 1 (k + 1)k !− 1 = 1 + k k! (k + 1)!− 1 1 = = 1− = RHS (k + 1)! (k + 1)! Thus, Pk is true ⇒ Pk+1 is true Since P1 is true and Pk is true ⇒ Pk+1 is true, by the Principle of Mathematical Induction, the statement is true for all n ∈ + . 1 ) ∞ ∞ 1 − (1 − 1 − ur r! = ∑ ∑ r 2r r =1 2 r =1 ∞ 1 =∑ r r =1 2 r !

= e1/ 2 − 1

7

10

n

∑u

r

= 12 (2 2 ) − 0

r =1

+2 2 (32 ) − 12 (22 ) +32 (4 2 ) − 22 (32 ) +... + n 2 (n + 1) 2 − (n − 1)2 n 2 = [n( n + 1) ]

2

10i Now, u r = r 2 (r + 1) 2 − (r − 1) 2 r 2 = r 2 ( r + 1 + r − 1)[ r + 1 − ( r − 1)] = r 2 (2r )(2) = 4r 3 Thus,

n

n

r =1

r =1

∑ u r = 4∑ r 3 = [n(n + 1)]

2

n(n + 1) ⇒ ∑r = 2 r =1 10ii Method 1: n

2

3

n

n

∑ (r − 1) 3 = ∑ (r 3 − 3r 2 + 3r − 1) r =1 n −1

r =1

∑ r3 =

n

∑ r 3 − 3∑ r 2 + 3∑ r − n

n

n

r =1

r =1

r =1

r =1

2

2

n(n + 1) (n − 1)n 3n 3∑ r 2 = − + (n + 1) − n 2 2 2 r =1 3n = n3 + (n + 1) − n 2 n n = (2n 2 + 3n + 1) = (n + 1)(2n + 1) 2 2 n 1 r 2 = n(n + 1)(2n + 1) ∑ 6 r =1 Method 2: n

n

n

r =1

r =1

∑ (r − 1) 3 = ∑ (r 3 − 3r 2 + 3r − 1) = n

n

n

r =1

r =1

n

n

r =1

r =1

∑ r 3 − 3∑ r 2 + 3∑ r − ∑1

n

n

r =1

r =1

3∑ r 2 = ∑ [ r 3 − ( r − 1)3 ] + 3∑ r − n r =1

= [13 – 03 + 23 – 13 : + n3 – (n – 1)3 ] + = n3 +

3n (n + 1) − n 2

3n (n + 1) − n 2

8

=

n n (2n 2 + 3n + 1) = (n + 1)(2n + 1) 2 2

n

1 n(n + 1)(2n + 1) 6 r =1 Method 3:

∑r

2

=

n

n

r =1 n

r =1

∑ (r − 1) 3 = ∑ (r 3 − 3r 2 + 3r − 1) (∑ r 3 ) − n3 = r =1 n

n

n

n

r =1

r =1

r =1

∑ r 3 − 3∑ r 2 + 3∑ r − n

3n 3∑ r 2 = n3 + ( n + 1) − n 2 r =1 n n = (2n 2 + 3n + 1) = (n + 1)(2n + 1) 2 2 n 1 r 2 = n(n + 1)( 2n + 1) ∑ 6 r =1

11i y=

1 f(x) y=2

(1, ½) (0, 0) x = –1

11ii y = f(– | x | ) y=

1 2 (–1, 0) (1, 0)

11iii y = f ′(x) y=1 –1 1

12

y = x tan −1 x

9

x dy = tan −1 x + dx 1 + x2 (1 + x2 ) ddyx = (1 + x2 ) tan −1 x + x dy d2 y 2 x + (1 + x 2 ) 2 = 2 x tan −1 x + 1 + 1 dx dx 2 (1 + x2 ) ddxy2 + 2 x ddyx − 2 y − 2 = 0 3 2 2 (1 + x2 ) ddxy3 + 2 x ddxy2 + 2 x ddxy2 + 2 ddyx − 2 ddyx = 0 3 d2 y 2 d y 1 + x + 4 x ( ) dx3 dx2 = 0 4 d3 y d2 y 2 d y 1 + x + 6 x + 4 ( ) dx4 dx3 dx2 = 0 When x = 0, y = 0 dy =0 dx d2 y =2 dx 2 d3 y =0 dx 3 d4 y = −8 dx 4 ( −8) x 4 + ... = x 2 − 1 x 4 + ... 2 Maclaurin’s Series for y is y = x 2 + 2! 4! 3 Method 1: x − tan −1 x x 2 − x tan −1 x lim = lim x →0 x →0 x3 x4 1 x 2 − x 2 − x 4 + ... 3 = lim x→0 x4 1 = lim + ... x→0 3 1 = 3 Method 2: 1 x − x − x 3 + ... −1 x − tan x 3 lim = lim x →0 x →0 x3 x3 1 1 = lim + ... = x →0 3 3 13i du u = 2x +1 ⇒ =2 dx 10

∫x

2 x + 1 dx = ∫ =

u −1 1 u du 2 2

1 1 32 u − u 2 du ∫ 4 5

3

1u 2 1u 2 = − +C 4 52 4 32 5 3 1 1 = (2 x + 1) 2 − (2 x + 1) 2 + C 10 6 13ii 1 1 Let x 2 x + 1 = . From GC, x = 2 2 y y = x 2x + 1

x 1 2

Area of R 1/2 1 1 = − ∫ x 2 x + 1 dx 2 2 0 1/ 2

5 3 2 ( 2 x + 1) 2 ( 2 x + 1) 2 = − − 4 10 6 0 2 2 2 2 1 1 = − − − ( − ) 4 5 3 10 6

= 13iii

11 2 1 − 60 15

=

1 (11 2 − 4) units 2 60 2

1 1 1 2 2 Volume of solid = π 2 − π ∫ 0 x ( 2 x + 1) dx 2

1/2

x 4 x3 = −π + 4 2 3 0 17π = = 0.556 units3 96

π

14 f(x) = 3 – 2x – x2 = 4 – (x + 1)2 For f −1 to exist, f must be one–one. ∴ k = −1 To find f −1 : Let y = 4 – (x + 1)2 ⇒ x = −1 ± 4 − y

(−1, 4)

−3

1

11

Since x ≤ −1, x = −1 − 4 − y Thus, f −1 : x − 1 − 4 − x , x ∈ (−∞, 4] 14i Range of f = (−∞, 4], domain of g = [0, 4] Since range of f ⊆/ domain of g, gf does not exist. 14ii Maximal range of h for gh to exist is [0, 4], therefore maximal domain of h is [−3, −1] 2

2

gh(x) = e 4 − 4 + ( x +1) = e ( x +1) = e| x +1| = e − x −1 since x ∈ [ –3, –1] ∴ gh : x e − x −1 , x ∈ [−3, −1] Range of gh = [1, e2] = [1, 7.39] 14iii Dg −1 g ( x +1) = [ -1, 3] and Dgg −1 ( x +1) = [0, e2 –1] Hence set of values of x = [0, 3]

12

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