'A' level H1 & H2 Math 8863 & 9740 Papers TI-84 Plus

Tackling

‘A’ level H1 & H2 Math 8863 & 9740 Papers (Oct/Nov 2008) with

TI‐84 Plus

© Interactive Math Exploration Centre ALL RIGHTS RESERVED. No part of this document may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, or by any information storage and retrieval system, without permission in writing from the Publisher.

First Published August 2009 ISBN No. 978-981-08-3785-3 ISBN No. 978-981-08-3786-0

Interactive Math Exploration Centre Corresponding Office:

Blk 231, Bain Street, #04-39, Bras Basah Complex, Singapore 180231 Tel: (65) 6334 6067 • Fax: (65) 6334 0475 • Email: [email protected]

© InterActive Math Exploration Centre

Table of Content 1.

Solution to A level H2 Math Paper 1

2 –18

2.

Solution to A level H2 Math Paper 2

19 – 36

3.

Solution to A level H1 Math Paper

37 – 54

1


f Solution to A level H2 Math Paper 1 1

Since the two shaded areas are equal, we have

∫ ∫

2

∫

x 2 dx =

∫

1

2 1

y dx =

4 a 4 a

x dy y dy

2 ⎡ 2 ⎡ x3 ⎤ ⎢2y = ⎢ ⎥ ⎢ 3 ⎣ 3 ⎦1 ⎣ 3

4

⎤ ⎥ ⎥ ⎦a

3

3

2(4) 2 2(a) 2 23 13 – = – 3 3 3 3 3

7 2 × 8 2(a) 2 = – 3 3 3

2(a) 3

3 2

3 2

=

Thus, a =

2×8 7 9 – = 3 3 3 9 2 2

and so

⎛ 9 ⎞3 a = ⎜ ⎟ = 2.73 (to 3 sig. fig.) [Answer] ⎝2⎠

Note: If time permits, you could use the graphic calculator to check your answer by checking whether the value of

∫

4 2.73

y dy is approximately equal to 7 . 3

Using the graphic calculator, first press and select 9:fnInt( as shown on the left below. Next, key in the following command statements as shown on the right below and then press .

2


2

Let Pn be the statement Sn =

n 1 n(n + 1)(4n + 5) , where Sn = ∑ un and 6 n =1 un = n(2n+1) for n ≥ 1 .

To show P1 is true, LHS: S1 = u1 = 1 (2(1) + 1) =3 1 (1)(1 + 1)(4(1) + 5) 6 = 18 = 3 6

RHS: S1 =

Since LHS = RHS, P1 is true. Assume that Pk is true for some k ≥ 1 , that is, Sk = To prove that Pk+1 is true, that is Sk+1 =

1 k (k + 1)(4k + 5) . 6

1 (k + 1)[(k + 1) + 1][4(k + 1) + 5] , 6

LHS = Sk+1 = Sk + uk+1 1 = k (k + 1)(4k + 5) + (k +1) (2(k +1) + 1) 6 1 = k (k + 1)(4k + 5) + (k +1) (2k +3) 6 1 = (k + 1) [ k (4k + 5) + 6(2k + 3) ] 6 1 = (k + 1) ⎡⎣ 4k 2 + 5k + 12k + 18⎤⎦ 6 1 = (k + 1) ⎡⎣ 4k 2 + 17 k + 18⎤⎦ 6 1 = (k + 1)(k + 2)(4k + 9) 6 1 = (k + 1)[(k + 1) + 1][4(k + 1) + 5] 6 = RHS. Thus, if the statement is true for n = k, then it is true for n = k+1 as proven above. Since P1 is true and Pk +1 is true if Pk is true, by Mathematical Induction, Pn is true for all n 1 n ≥ 1 . That is, Sn = n(n + 1)(4n + 5) , where Sn = ∑ un and un = n(2n+1) for n ≥ 1 . 6 n =1 [Shown]

3


3

⎛5⎞ ⎛1⎞ ⎜ ⎟ Given OA = 4 , OB = ⎜ −1⎟ . ⎜ ⎟ ⎜ ⎟ ⎜0⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠

(i)

OP = OA + OB

(ii)

(since OAPB is a parallelogram.) ⎛1⎞ ⎛5⎞ ⎛ 6⎞ = ⎜ 4 ⎟ + ⎜ −1⎟ = ⎜ 3 ⎟ . [Answer] ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −3 ⎟ ⎜ 0 ⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛1⎞ ⎛5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 4 ⎟ i ⎜ −1 ⎟ ˆ = OA i OB = ⎜⎝ −3 ⎟⎠ ⎜⎝ 0 ⎟⎠ cos AOB OA OB ⎛1⎞ ⎛5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 4 ⎟ ⎜ −1 ⎟ ⎜ −3 ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠

=

1× 5 + 4 × (−1) + (−3) × 0 1 + 42 + (−3) 2 52 + (−1) 2 + 02 2

= 5−4+0

26 26

= 1 . 26 ˆ AOB = cos –1( 1 ) = 87.8o (to 3 sig. fig.). [Answer]

Thus,

26

⎛1⎞ ⎛5⎞

(iii) Area of parallelogram OAPB = OA × OB = ⎜ 4 ⎟ × ⎜ −1⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −3 ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠

⎛ −3 ⎞ = ⎜ −15 ⎟ ⎜ ⎟ ⎜ −21 ⎟ ⎝ ⎠

=

(−3) 2 + (−15) 2 + (−21) 2

= 675 = 15 3 units2. [Answer] Note: Check that the cross product obtained is correct by taking the dot product of your answer obtained with OA or OB to see whether it is zero. ⎛ −3 ⎞ ⎛ 1 ⎞

i.e. ⎜ −15 ⎟ i ⎜ 4 ⎟ = –3 – 60 + 63 = 0. ⎜ ⎟ ⎜ ⎟ ⎜ −21 ⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠

4


4(i)

dy 3x = 2 x +1 dx 3x ∫ dy = ∫ x 2 + 1 dx 3 2x ∫ dy = 2 ∫ x 2 + 1 dx

Thus, the general solution is: y = (ii)

3 ln ( x 2 + 1) + C. [Answer] 2

When x = 0, y = 2. Thus, 2 =

3 ln ( 02 + 1) + C 2

and so C = 2. The required particular solution: y = (iii)

3 ln ( x 2 + 1) + 2. [Answer] 2

dy 3x = 2 . x +1 dx

Dividing the numerator and denominator of the RHS by x2, we have 3 dy = x . dx 1 + 1 x2

dy → 0. [Answer] Therefore, when x → ± ∞ , both 3 and 12 tend to zero and so dx x x

(iv) Answer :

3 y = ln( x 2 + 1) + 2 2 3 y = ln( x 2 + 1) 2 3 y = ln( x 2 + 1) − 2 2

Note that the graphs are symmetrical about the y-axis and the graphs become flatter when |x| becomes larger. You can obtain the required family of solution curves by using the graphic calculator. First, key in the equations after pressing as seen on the right below. Note that Y1 can be found by pressing .

5


Next, press right below.

5(i)

∫

1 3

0

to display the standard window with the 3 graphs as shown on the

1 dx = 1 + 9x2

∫

1

1 3

9 ( 19 + x 2 )

0

dx

1 13 1 dx 2 ∫ 9 0 1 + x2 3

=

( )

1

⎛ ⎞⎡ ⎛ ⎞⎤ 3 = 1 ⎜ 11 ⎟ ⎢ tan −1 ⎜ 1x ⎟ ⎥ 9⎝ 3⎠ ⎝ 3 ⎠⎦0 ⎣

= 1 ⎡⎣ tan −1 ( 3 x ) ⎤⎦ =

3 1⎡

Note that: −1 ∫ 2 1 2 dx = 1a tan ax + c a +x

1 3

0

( )

tan −1 3 − tan −1 (0) ⎤ ⎥⎦

3 3 ⎢⎣ = 1 ⎡⎢ π − 0 ⎤⎥ 3⎣3 ⎦

=

π

9

. [Answer]

(ii) Let u = ln x, dv = xn dx. n+1 So, du = 1x dx , v = x (as n ≠ –1) . n +1

Applying the method of integration by parts, we have e n +1 e n ⎡ x n +1 ⎤ – e ⎛ x ⎞ 1 dx x ln x d x = ∫1 ∫ 1 ⎜⎝ n + 1 ⎟⎠ x ⎢(ln x) ⎥ n + 1⎦ 1 ⎣ n +1 n +1 = (ln e) e – (ln1) 1 –

n +1

n +1

n +1

e

e 1 x n dx ∫ n +1 1

– 1 ⎡⎢ x ⎤⎥ n +1 n + 1 ⎣ n + 1⎦ 1

= e

n +1

n +1 n +1 = e – e − 12

( n + 1)

n +1

=

(n + 1)e

n +1

− en +1 + 1

( n + 1)

2

n +1 = ne +21 . [Answer]

( n + 1)

6


6(a)

Using the cosine rule, AC 2 = AB 2 + BC 2 – 2(AB)(BC) cos θ = 12 + 32 – 2(1)(3) cos θ = 10 – 6 cos θ . 2 When θ is a small angle, cos θ ≈ 1 – θ .

So,

(

2

)

2 AC ≈ 10 – 6 1 − θ2 , if θ is a small angle

2

= 4 + 3θ 2 .

( 4 + 3θ )

Next, AC ≈

1

2 2

Hence, AC ≈

(

2 4(1 + 3θ4 )

. [Shown]

)=4 ( 1 2

1 2

2 1 + 3θ4

)

1 2

( ( ) )

2 = 2 1 + 12 3θ4 + .....

= 2 + 3 θ 2 + …. 4 Thus, AC ≈ a + b θ 2 , where a = 2, b = 3 . [Shown] 4

(b)

f(x) = tan (2x + π4 )

f '(x) = 2 sec2 (2x + π4 )

(

f ''(x) = 4 sec (2x + π4 ) 2 tan(2x + π4 ) sec(2x + π4 ) = 8 tan (2x + π4 ) sec2 (2x + π4 ).

)

Thus, f(0) = tan ( π4 ) = 1 f '(0) = 2 sec2 ( π4 ) = 4

f ''(0) = 8 tan ( π4 ) sec 2( π4 ) = 16.

Hence, f(x) = f(0) + f '(0) x + f ''(0) x2 + ….. = =

2! 16 2 1 + 4x + x + … 2!

1 + 4x + 8x2 + …. [Answer]

7


7

= y + x + y = 2y + x. ⎛ x⎞= π . Total length of semi-circular part = 1 Total length of straight parts

2

(2π ) ⎜ ⎟ ⎝2⎠

2

x

As total time taken is given to be 180, we have 180 = 3(x + 2y) + 9 ⎛⎜ π x ⎞⎟

⎝2 ⎠ 180 = 3x + 6y + 9π x 2 60 = x + 2y + 3π x 2 3π x 2y = 60 – x – 2 x y = 30 – – 3π x .------------------(1) 2 4

Let A be the area of the flower-bed.

( )

1 x 2 π 2 2 x 3π x π x2 = x [30 – – ]+ 2 4 8 2 2 x 3π x π x2 = 30x – – + 2 8 4 2 2 x = 30x – – 5π x 2 8 1 5π ⎞ 2 ⎛ = 30x – ⎜ + ⎟ x ⎝2 8 ⎠ dA = 30 – ⎛⎜1 + 5π ⎞⎟ x . dx 4 ⎠ ⎝

A = xy +

Thus,

dA =0. dx

Let We have So

30 = ⎛⎜1 + 5π ⎞⎟ x . ⎝

x=

4 ⎠

30 ≈ 6.0889 = 6.09 (to 3 sig fig). 5π 1+ 4

Substitute x = 6.0889 into (1), we have y = 12.6 (to 3 sig fig). As

d2 A = – ⎛⎜1 + 5π ⎞⎟ < 0, A is maximum when x = 30 . 5π dx 2 4 ⎠ ⎝ 1+ 4

That is, when x ≈ 6.09 and y ≈ 12.6, the area of the flower-bed is maximum. [Answer]

8


8(i)

As |1 + 3 i| =

12 +

( 3)

2

we have z1 = 1 + 3 i = 2 e and so

( )

= 4 = 2 and arg(1 + 3 i) = tan −1 13 = π , 3 i

π 3

i

π

z13 = (2 e 3 )3 = 23eiπ = – 8. [Answer]

If time permits, you could use your graphic calculator to check the answer, as shown on the screen shot on the right. (ii)

Note that complex roots occur in conjugate pairs. Thus, 1 – 3 i is also a root of the equation. Since (z – (1 + 3 i))(z – (1 – 3 i)) = ((z – 1) – 3 i))((z – 1) + 3 i) = (z – 1)2 – ( 3 i )2 = z2 – 2z + 4, we have

2z3 + az2 + bz + 4 = (z2 – 2z + 4)(Az + B).

As the coefficient of z3 is 2, A = 2. Also, B = 1 as the constant term is 4. Thus, 2z3 + az2 + bz + 4 = (z2 – 2z + 4)(2z + 1).-------------------(1) Next, expanding the RHS of (1), (z2 – 2z + 4)(2z + 1) = 2z3 + z2 – 4z2 – 2z + 8z + 4 = 2z3 – 3z2 + 6z + 4. Comparing the coefficient of z2, a = – 3. Comparing the coefficient of z, b = 6.

[Answer] [Answer]

Alternatively, to find the values of a and b, Put z = 1 into (1): 2 + a + b + 4 = (12 – 2 + 4)(2 + 1) a + b = 3 ------------------------------(2) and so Put z = –1 into (1): –2 + a – b + 4 = (12 + 2 + 4)(–2 + 1) a – b = –9 -----------------------------(3) and so Solving (2) and (3) simultaneously, we have a = –3 and b = 6. [Answer] (iii) From 8(ii) above , we know that the two roots of 2z3– 3z2 + 6z + 4 = 0 are 1 + 3 i and 1 – 3 i. As 2z3– 3z2 + 6z + 4 = (z2 – 2z + 4)(2z + 1), the third root is − 1 . 2

That is, the roots are 1 + 3 i, 1 – 3 i and − 1 . [Answer] 2

9


The Argand diagram of the roots is shown on the right.

9(i)

f(x) =

ax + b , cx + d

f '(x) =

( cx + d ) a − (ax + b)c 2 ( cx + d )

= =

abcd ≠ 0.

acx + ad − acx − bc

( cx + d ) ad − bc

( cx + d )

2

2

.

Since ad – bc ≠ 0, f '(x) ≠ 0 for all real values of x. Thus, the graph of y = f(x) has no turning points. [Shown] (ii)

Dividing ax+ b by cx + d, we have the following. a c

ad

b− ax + b c = a+ Thus, f(x) = cx + d c cx + d

cx + d ax + b ax + ad c b − ad c

= a − ad − bc . c

a

c (cx + d )

ad

(cx + d ) + b − ax + b c = c Alternatively, f(x) = cx + d cx + d ad b− a c = + c cx + d

= a − ad − bc . c

c (cx + d )

When ad – bc = 0, f(x) = a . Thus, the graph is a horizontal line with equation y = a . c

c

[Answer]

10


9(iii) Let f(x) =

3x − 7 . Here a = 3, b = – 7, c = 2, d = 1. 2x +1

We have ad – bc = (3)(1) – (– 7)(2) = 17. ad − bc 17 From the working of 9(i), f '(x) = = > 0 for x ≠ − 1 . 2 2 2 ( cx + d ) ( 2 x + 1) 3x − 7 Thus, the graph of y = has a positive gradient at all points (except x = − 1 ) of the 2 2x +1 graph. [Shown] (iv)(a) Answer :

Asymptotes: x = − 1 and y = 3 . 2 2 Points of intersection: (0, -7) and ( 7 , 0) 3

We can get the graph by using the graphic calculator. We can also use the calculator to find the x-intercept and y-intercept. First, we key in the equations into the graphic calculator and press press to obtain the y-intercept.

, then

for CALC menu and press To find the x-intercept, press 5:intersect. Move the cursor nearer to the x-intercept and press

to select three times.

11


9(iv)(b) Answer :

Asymptotes: x = − 1 , 2 y = − 32 and y =

3 2

.

Points of intersection: ( 7 , 0) 3

Using the graphic calculator, key in the equations as shown on the left below and press to get the required graph, which is shown on the right below. Note that you could . get Y1 by pressing

10.(i)

Observe that 10, 13, 16, 19, … is an arithmetic progression. Its common difference is 3. Let a = 10, d = 3 n Sn= (2a + (n – 1)d) > 2000 2 n (2(10) + 3(n – 1)) > 2000 2 n (20+3n -3) > 2000 2 n (17+3n) > 2000 2 17n +3n2 > 4000 3n 2 + 17 n − 4000 > 0 To solve the quadratic inequality, we first find the roots of the quadratic equation 3n 2 + 17 n − 4000 = 0 . To find the roots of the quadratic equation 3n 2 + 17 n − 4000 = 0 , we may use the graphic calculator. 12


Step 1: Press

and select 5:PlySmlt2 as shown on the left below by pressing .

Step 2: Press any key to see the MAIn MEnu shown on the right above. Select option 1:POLY ROOT FINDER by pressing or . Step 3: Select the parameters as seen on the left screen shot below.

Step 4: Go to nEXT by pressing shot above.

and type in the values as shown on the right screen

Step 5: Go to SOLVE by pressing

to see the answer shown below.

Thus, (n + 39.46)(n – 33.79) >0. As

+

– −39.46

+ 33.79

We have n < – 39.46 or n > 33.79. Therefore, the least value of n is 34 months or 2 years 10 months. Hence, she will first save over $2 000 on 1st Oct 2011. [Answer]

13


10(ii)(a)

Compound interest = 0.02(10) + 0.02(1.02×10) + 0.02(1.022×10) + …. + 0.02(1.0223×10) = 0.2 (1 + 1.02 + … + 1.0223 ) =

0.2 (1.0224 − 1) 0.02

= $6.08 (to 3 sig. fig.) [Answer] (b)

End of 1st month, she has (1.02)(10) in total. End of 2nd month, she has (1.02)[10(1.02) + 10] = (1.02)2(10) + (1.02)(10) in total. End of 3rd month, she has (1.02) [ (1.02)2(10) + (1.02)(10) + 10] = (1.02)3(10) + (1.02)2(10) + (1.02)(10) in total. Following the pattern, at the end of two years (which is 24 months), the total amount she has is (1.02)24(10) + (1.02)23(10) + … + (1.02)(10) = 10 [1.0224 + 1.0223 + … + 1.02] ⎡1.02(1.0224 − 1) ⎤ ⎥ 1.02 − 1 ⎣ ⎦

= 10 ⎢

= $310 (to 3 sig. fig.) [Answer] (c)

Total amount at the end of nth month = (1.02)n(10) + (1.02)n –1(10)+ … + (1.02)(10) = 10[1.02n + 1.02n – 1 + … + 1.02] ⎡1.02(1.02n − 1) ⎤ = 10 ⎢ ⎥ ⎣ 1.02 − 1 ⎦ = 510 (1.02n – 1).

Let 510 (1.02n – 1) > 2000. We have

⎞ 1.02n > ⎛⎜ + 1⎟ 510 ⎝ ⎠ ⎛ 200 ⎞ n lg 1.02 > lg ⎜ + 1⎟ ⎝ 51 ⎠ n > 80.476. 2000

Thus, the least value of n is 81 months. [Answer] 14


11

p1 : 2x – 5 y + 3 z = 3 p2 : 3x + 2 y – 5 z = – 5 p3 : 5x – 20.9 y + 17z = 16.6 To solve this system of equations, we may use the graphic calculator. and select 5:PlySmlt2 as shown on the left below by pressing

Step 1: Press .

Step 2: Press any key to see the MAIn MEnu shown on the right above. Select option 2:SIMULT Eqn SOLVER by pressing . Step 3: Select the parameters as seen on the left screen shot below.

Step 4: Go to nEXT by pressing shot above. Step 5: Go to SOLVE by pressing

and type in the values as shown on the right screen to obtain the solution as shown below.

Coordinates of the intersecting point = ⎛⎜ − 4 , − 4 , 7 ⎞⎟ . [Answer] ⎝ 11

11 11 ⎠

15


11(i)

p1 : 2x – 5y + 3z = 3 p2 : 3x + 2y – 5z = –5 The coordinates of the points on the line l satisfy the system of the two equations above. To solve this system of equations, we could use the graphic calculator. and select 5:PlySmlt2 as shown on the left below by pressing

Step 1: Press .

.

Step 2: Press any key to see the MAIn MEnu shown on the right above. Select option 2:SIMULT Eqn SOLVER by pressing . Step 3: Select the parameters as seen on the left screen shot below.


and type in the values as shown on the right screen to obtain the solution as shown below.

⎛ x ⎞ ⎛ −1 + α ⎞ Thus, ⎜ y ⎟ = ⎜ −1 + α ⎟ , α ∈ R . ⎟ ⎜ ⎟ ⎜ ⎜z⎟ ⎜ α ⎟ ⎠ ⎝ ⎠ ⎝

Hence, a vector equation of the line l is ⎛ 1⎞ ⎛ −1 + α ⎞ ⎛ −1⎞ ⎜ ⎟ ⎜ ⎟ r = −1 + α = −1 + α ⎜1⎟ , α ∈ R . [Answer] ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 1⎟ ⎜ α ⎟ ⎜0⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

16


Alternatively, we could also solve the system of equations by finding x and y in terms of z by using the method of elimination. 2x – 5y = 3 – 3z ----------(1) 3x + 2y = –5 + 5z ----------(2) 2 × (1): 5 × (2): (3) + (4) : and so

4x – 10y = 6 – 6z ---------(3) 15x + 10y = –25 + 25z---------(4) 19x = –19 + 19z x = –1 + z.

Substituting x = –1 + z into (1), we have 2(–1 + z) – 5y = 3 – 3z and so y = –1 + z. ⎛ x ⎞ ⎛ −1 + α ⎞ Thus, ⎜ y ⎟ = ⎜ −1 + α ⎟ , α ∈ R . ⎜ ⎟ ⎜z⎟ ⎝ ⎠

11(ii)

⎜ ⎜ α ⎝

⎟ ⎟ ⎠

⎛5⎞

p3 : 5x + λ y + 17z = μ or r . ⎜ λ ⎟ = μ ⎜ ⎟ ⎜17 ⎟ ⎝ ⎠

Since all three planes meet in the line l, the plane p3 contains the line l. Therefore, the coordinates of all the points on l must satisfy the equation of the plane p3. Thus,

⎛ −1 + α ⎞ ⎛ 5 ⎞ ⎜ ⎟ . ⎜ ⎟ = μ , for all values of α. ⎜ −1 + α ⎟ ⎜ λ ⎟ ⎜ α ⎟ ⎜17 ⎟ ⎝ ⎠ ⎝ ⎠

–5 + 5α – λ + λ α +17α = μ (22 + λ )α = μ + 5 + λ ----------------(*) Let α = 1, then 22 + λ = μ + 5 + λ and so μ = 17. Let α = 0, then 0= μ +5+ λ As μ = 17, we have λ = –22. Hence, to have all three planes meet in the line l, λ = –22 and μ = 17. [Answer]

(iii) If three planes have no point in common, then the line l does not meet the plane p3. Therefore, there is no value of α that satisfies the equation (*) in 11(ii).

However, when λ ≠ –22, from the equation (*), α =

μ +5+λ exists. 22 + λ

Thus, λ = – 22. From the conclusion in 11(ii), we deduce that the three planes have no common point when λ = – 22, μ ≠ 17. [Answer]

17


Alternatively, when three planes have no point in common, the line l is parallel to the plane p3 and it does not meet the plane p3. When a normal vector to the plane p3 is perpendicular to a direction vector of l, the line l is parallel to the plane p3. ⎛ 5 ⎞ ⎛ 1⎞ Thus, ⎜ λ ⎟ . ⎜1⎟ = 0 and so λ = –22. ⎜ ⎟ ⎜ ⎟ ⎜17 ⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ From the equation (*) in 11(ii), when λ = –22 and μ = 17, the line l meets the plane p3. Therefore, when λ = –22, μ ≠ 17, the line l does not meet the plane p3.

11(iv)

⎛ 1 ⎞ ⎛ −1⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ = ⎜0⎟ ⎜ −1⎟ − ⎜ −1⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎜ 0 ⎟ ⎜ 3⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ 2 ⎞ ⎛ 1⎞ ⎛ −3 ⎞ n = ⎜⎜ 0 ⎟⎟ × ⎜⎜1⎟⎟ = ⎜⎜ 1 ⎟⎟ ⎜ 3 ⎟ ⎜ 1⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Thus, the vector equation of the plane is ⎛ −3 ⎞ ⎛ 1 ⎞ ⎛ −3 ⎞ r . ⎜⎜ 1 ⎟⎟ = ⎜⎜ −1⎟⎟ . ⎜⎜ 1 ⎟⎟ = 2 ⎜2⎟ ⎜3⎟ ⎜2⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ and so, the Cartesian equation of the required plane is –3x + y + 2z = 2. [Answer]

18


f Solution to A level H2 Math Paper 2 1(i)

Answer:

You could obtain the graph of y = f(x) by using your graphic calculator before sketching on your answer script. Step 1: Press below.

and type in the equation as follows, shown on the left screen shot

Step 2: As the domain of the function considered is given as −3 ≤ x ≤ 3 , we may set the window by pressing and then set Xmin = –3, Xmax = 3. Need not key in the values for Ymin and Ymax. The graphic calculator will calculate the appropriate are pressed. The graph is shown on the values of Ymin and Ymax when right screen shot below.

(ii)

f(x) = ex sin x

x 2 x3 x3 + + …) (x – + …) 2 6 6 3 3 x x x2 x3 (x – + …) + … = 1(x – + …) + x (x – + …) + 6 6 2 6 3 3 x x = x – + x2 + + … 6 2 3 x = x + x2 + + … [Answer] 3 = (1 + x +

19


1(iii) Let g(x) = x + x2 +

x3 . 3

Answer:

You could obtain the graph of y = g(x) by using the graphic calculator. Step 1: Press

and type in the equation as follows, shown on the left screen shot below.

Step 2: Press

and both graphs are shown on the right screen shot below.

However, if are pressed instead of pressing shown below with the range of g being considered.

(iv) Let

, then both graphs are

| f(x) – g(x) | < 0.5

|ex sin x – (x + x2 +

x3 ) | < 0.5 3

Using the graphic calculator, the solution is – 1.96 < x < 1.56 (to 3 sig. fig.) [Answer]

20


To solve the inequality by using the graphic calculator : Step 1: Press and key in the equations as shown below. Note that abs( could be . obtained from the CATALOG menu by pressing .

Step 2: From the graphs obtained in part (iii), we observe that between x = –2 and x = 2, the two graphs are quite close to each other. Thus, we set the window with Xmin = –2 and Xmax = 2 and press for 0:ZoomFit. The graphs are shown below.

Step 3: Press

Step 4: Press

to display the CALC menu and select 5:intersect.

and the cursor will appear on the line y = .5.

Step 5: Find the points of intersection by moving the cursor nearer to the respective intersecting points and press three times.

21


2(i)

1

y2 = x(1 − x) 2 1

1

and so y = x(1 − x) 2 or y = – x(1 − x) 2 . 1

x(1 − x) 2 for the part of the curve which is above the x-axis.

Choose y =

As the curve above the x-axes is symmetrical to the curve below the x-axis, the area under the curve and above the x-axis is equal to that of the area below the x-axis. Thus, area of R = 2 ∫

1 0

1

x(1 − x) 2 dx [Answer]

Using the graphic calculator, area of R = 0.99888 = 0.999 units 2 (to 3 sig. fig.). [Answer] Alternatively, the equation of the part of the curve below the x-axis is 1

y = – x(1 − x) 2 . So, the area of R is the area of the region bounded by the curve with 1

1

equation y = x(1 − x) 2 , the curve with equation y = – x(1 − x) 2 , x = 0 and x = 1. Therefore, area of R =

∫

1 0

1 1 x(1 − x) 2 − ⎛⎜ − x(1 − x) 2 ⎝

⎞dx = 2 1 x(1 − x) 12 dx . ⎟ ∫0 ⎠

To obtain the numerical value of the integral, we may use the graphic calculator. Step 1: Press

and select option 9:fnInt( as shown on the left below.

Step 2: Press and key in the command statements as shown on the right above and then press . Next, type to obtain the required answer. Alternatively, you could also key in the command statements as follows.

(ii)

1

Volume = π ∫ y 2 dx 0

1

= π ∫ x (1 − x) dx 0 Let u = 1 – x. Then x = 1 – u and

dx = −1 . du

When x = 0, u = 1 – 0 = 1. When x = 1, u = 1 – 1 = 0.

22


Thus, 0 volume = π ∫ (1 − u ) u (−du ) 1 1

1

= π ∫ u 2 (1 − u ) du 0 1

1

1 2

=π ∫ u − u

3 2

0

⎡ u 32 u 52 ⎤ du = π ⎢ − ⎥ 3 5 2 ⎥⎦ 0 ⎣⎢ 2

= π ⎡ 2 − 2 − 0⎤ ⎢3 5 ⎥ ⎣ ⎦

= 4π units3. [Answer] 15

y2 = x 1 − x

2(iii)

Differentiate the above equation with respect to x, we have dy −1 2y = x 1 (1 – x) 2 (– 1) + 1 − x 2 dx 1 −1 = – 1 x (1 – x) 2 + (1 – x) 2 2 = (– 1 x + (1 – x) ) 1 2 1− x

( ) ( ) ( )

= 2 − 3x .

2 1− x dy 2 − 3x

= . dx 4 y 1− x dy 2 − 3x When = 0, = 0. dx 4 y 1− x So, 2 – 3x = 0 Thus,

x= 2.

3 dy For y > 0, > 0 when x = 23 dx

( )

−

and

( )

+ dy < 0 when x = 23 . dx

Thus, the x-coordinate of the maximum point of C is x = 2 . [Answer] 3

3.(a) Let w = reiθ and so w* = re − iθ . w reiθ Thus, p = = − iθ = ei (θ −( −θ )) = ei (2θ ) . w* re Hence, | p | = 1 and arg (p) = 2 θ . [Answer]

Alternatively, using the facts that |w| = | w*| and arg(w*) = – arg(w) = – θ , we have | w| = 1 and arg (p) = arg (w) – arg (w*) = θ – (– θ ) = 2 θ . | p| = | w*| 23


p5 = ei (2θ )5 = ei (10θ ) = cos(10 θ ) + i sin(10 θ ), 0 < θ < Given that 0 < θ
0. For this to happen,10 θ must lie in the 1st and 4th quadrants. Thus, 10 θ = 2 π , 4 π π 2π and so θ = , . 5 5 3(b)

Answer :

Note that the mid point between 0 and 8 + 6i is 1 (0+8+6i) = 4 + 3i. 2 Least value of arg(z) = α. From the diagram on the right, α = arg (4 + 3i) – cos-1 5 6 -1 3 -1 5 = tan – cos 4 6 ≈ 0.0578 rad = 0.058 rad (correct to 3 d.p.) [Answer]

24


Greatest value of arg(z) = β. From the diagram on the right, β = arg (4 + 3i) + cos-1 5 6 -1 3 5 = tan + cos 4 6 ≈ 1.2292 rad = 1.229 rad (correct to 3 d.p.) [Answer] Alternatively, it can also be solved by first finding the intersecting points of the circle and the line by using the graphic calculator. Note that the equation of the circle is x2 + y2 = 62 and the equation of the line is y − 3 = − 4 ( x − 4) . After the point of intersection is found, say 3 −1 d (c, d), then arg(z) = tan c .

4(i)

Answer:

You may use the graphic calculator to check your graph. Step 1: Press and key in the equation as shown on the left screen shot below. X > 4 is typed in as a part of the equation as shown because of the domain given in the question. The sign of greater than ( > ) can be obtained by pressing for TEST menu and then select 3:>. Step 2: Press and type in the respective values in the window settings as shown on the right screen shot below.

Step 3: Press right.

to see the graph as shown on the

25


4(ii)

Let

y = (x – 4)2 + 1, x > 4 y – 1 = (x – 4)2 x – 4 = ± y −1 x = 4 ± y −1 . As x > 4, x = 4 + y − 1 . Since range of f is (1, ∞ ), we have Df −1 = R f = (1, ∞ ). [Answer] Hence, f -1(x) = 4 + x − 1 , x > 1. [Answer]

(iii) Answer:

You may use the graphic calculator to check your graph. Step 1: Press below.

and type in the equation in Y2 as shown on the left screen shot

Step 2: Under the same window setting that is done in part (i), press to select 5:ZSquare so that the scales on both axes are of equal length. The graphs are shown on the right screen shot below.

Note: You can verify the inverse you found by using your graphic calculator. Press for DRAW menu and select 8:DrawInv followed by to activate the command DrawInv Y1. If the two graphs coincide, it means you have obtained the correct inverse function.

26


4(iv)

The required equation of the line of reflection is y = x. [Answer] Note that both graphs would meet at the line y = x. Thus, solving the equation f(x) = f -1(x) is equivalent to solving the equation f(x) = x. Let

f(x) = x, x > 4 (x – 4)2 + 1 = x x2 – 8x + 17 = x x2 – 9x + 17 = 0 2 x = −(−9) ± (−9) − 4(1)(17) = 9 ± 13

2

2(1)

i.e.

x = 9 + 13 or x = 9 − 13 (rejected as x > 4) 2

2

Hence, the exact solution of the equation f(x) = f -1(x) is x = 9 + 13 . 2

5

List out the names and number them in order from 1 to 950. Since the sample size is 50, we divide 950 by 50, i.e. 950 = 19. Select a number randomly, say k, from 1 to 19 50

inclusive and pick students that correspond to numbers which are (k + the multiple of 19). For example, if the starting number is 8, then students with the numbers 8, 27, 46, 65… etc will be picked. Alternatively, since the sample size is 50, we divide 950 by 50, i.e. 950 = 19. That is, group the students 50

into 50 blocks, where each block consists 19 students. Number the students from 1 to 19 in each block. Select a number randomly, say k, from 1 to 19 and then pick the kth student in each block. In stratified sampling, the student population is divided into non-overlapping representative groups of strata, such as different types of sports activities which use different sports facilities. Students are selected randomly from each stratum with the sample size proportional to the relative size of the stratum. Hence, this is more likely to give a representative sample of the student population and avoid any possible cyclical pattern from the systematic sampling.

6.

Let X be the random variable of “The mass of calcium in mg in a 1 litre bottle”. x=

∑x n

=

1026 = 68.4. 15

Unbiased estimate of the population variance is

2 2 ⎡ x) ⎤ 1026.0 ) ⎤ ( ( 1 ⎡ 1 ∑ 2 ⎢ ⎥ s = ∑ x − n ⎥ = 14 ⎢⎢77265.90 − 15 ⎥⎥ = 506.25. n −1 ⎢ ⎣ ⎦ ⎣ ⎦

2

27


H0 : μ = 78 H1 : μ ≠ 78 Assuming H0 is true. Test statistic t =

x − 78 ~ t(14). s2 n

Using 2-tailed t-test to test at 5% level of significance where x = 68.4, s = 506.25 and n = 15, we obtain from the graphic calculator that the p-value is 0.1207. As p-value is greater than 0.05, we do not reject H0 and conclude that there is insufficient evidence to say that the mean mass of calcium in a bottle has changed at 5% significance level. You may use your graphic calculator to help you with the hypothesis testing. Step 1: Press

.

Step 2: Go to TESTS sub-menu by using the left arrow key 2:T Test… as shown on the left screen shot below.

and select option

Step 3: Highlight the following options and key in the respective data as shown on the right screen shot below.

Sx = √506.25

Step 4: Select Calculate to get the t-test results as shown on the screen shot below, where the p-value can also be obtained.

28


A win

7

0.7

A win

0.6

0.4

0.2

0.3

B win

0.2

A win

B win

st

1 set

0.8

B win

0.7

A win

0.3 0.8

A win

B win

B win 2nd set

3rd set

(i)

The probability that A wins the second set = 0.6(0.7) + 0.4(0.2) = 0.5. [Answer]

(ii)

The probability that A wins the match

(iii)

Let F denotes the event that B won the first set and let W denotes the event that A wins the match. P( F | W) =

= 0.6(0.7) + 0.6(0.3)(0.2) + 0.4(0.2)(0.7) = 0.512. [Answer]

P( F ∩ W ) 0.056 0.4(0.2)(0.7) = = P(W ) 0.512 0.512 = 7 = 0.109 (to 3 sig. fig.) [Answer] 64

8(i)

Using the graphic calculator, the product moment correlation r = 0.970 (correct to 3 sig. fig.) [Answer] Since r is close to 1, we may say that x and t has a strong positive linear correlation. However, a linear model is only appropriate if a linear relationship between x and t can also be observed from the scatter plot of the data points. [Answer] You may get the value of r by using your graphic calculator. Step 1: Press

, and select option 1:Edit…, as seen below by pressing

.

.

29


Step 2: Key in the values into the list table as shown on the right below. Here, x, the independent variable, is represented by L1 and t, the dependent variable, is represented by L2. Step 3: Press and go to CALC sub-menu. Select 4: LinReg(ax+b) (or 8: LinReg(a+bx))

Step 4: Key in the list variables in the order of independent variable and then dependent variable, as shown on the left below. Press to get the equation of the regression line of x on t, including the product moment correlation coefficient.

Note: To get the value of r, DiagnosticOn must be selected. Press the CATALOG menu. Scroll to DiagnosticOn and press twice. 8(ii)

to display

Answer:

You may use your graphic calculator to get the scatter diagram. Step 1: Press below.

to display the STAT PLOT menu as shown on the left screen shot

30


Step 2: Press

and select the respective options as shown on the left below.

Step 3: Press

for 9:ZoomStat to show the graph on the right below.

P

8(iii) The incorrect point is P(4.8, 7.6). With point P removed, the remaining scatter diagram shows points on a curve that gradually increases in a manner that the increment of t deceases as x increases, which suggests that the data points may fit better with t = a + b ln x . [Answer] (iv) Using the graphic calculator, we have t = 1.4247 + 4.3966 ln x = 1.42 + 4.40 ln x (to 3 sig. fig.).

Thus, a = 1.42, b = 4.40. [Answer] You may get the least square estimates of a and b for the model t = a + b ln x by using your graphic calculator. Step 1: Press

, then press

to select the option 1:Edit…

Step 2: Use the down arrow key to move the cursor down along the list of L1 to the number 4.8 and delete the entry by pressing . Next, move the cursor to the right using the right arrow key and delete the number 7.6 under the L2 list. Note: Observe that the table is similar to the earlier one in 8(i), except that the point P(4.8, 7.6) has been removed from the list. Step 3: Move the cursor to the top of the next list, which is L3 and highlighted as shown. . The table is shown below. Key in (for ln L1) and press

31


Step 4: Press , then press and press .

to go to CALC sub-menu. Select 8: LinReg(a+bx))

Step 5: Key in the list variables in the order of independent variable and then dependent variable, as shown on the left below. Press to get the equation of the new regression line of ln x on t, including the product moment correlation coefficient.

8(v)

Using the graphic calculator, we have t = 1.4247 + 4.3966 ln x ---------------------(1) Substitute x = 4.8 into (1), we have t = 1.4247 + 4.3966 ln(4.8) ≈ 8.3213 = 8.32 (correct to 3 sig. fig.) [Answer]

(vi) Since x = 8 falls outside the range of data on which we obtained the regression line, extrapolation of the observed data points is not advisable and thus, the estimate of the value of t is not reliable when x = 8.0. [Answer]

9

Let X be the random variable of “ The number of grand pianos sold in a week”. Here, X ~ Po (1.8). P(X ≥ 4) = 1 – P(X ≤ 3) ≈ 1 – 0.89129 = 0.10871 = 0.109 (correct to 3 sig. fig.) [Answer]

32


To find the probability by using the graphic calculator: Step 1: Press

and under DISTR sub-menu, select D: poissoncdf(.

Step 2: Key in the data as shown on the right below.

Let Y be the random variable of “The number of upright pianos sold in a week”. Here, Y ~ Po (2.6). Let W be the random variable of “The total number of grand and upright pianos sold in a week”. Here, W = X +Y ~ Po (1.8 + 2.6) as X and Y are independent. Thus, W ~ Po (4.4) P(W = 4) ≈ 0.19174 = 0.192 (correct to 3 sig. fig.) [Answer] To find the probability by using the graphic calculator: Step 1: Type

and under DISTR sub-menu, choose C: poissonpdf( .

Step 2: Key in the following data as shown on the right below.

Let V be the random variable of “The number of grand pianos sold in a year of 50 weeks”. V ~ Po (50 × 1.8) That is, V ~ Po (90). Since λ = 90 ( >10) is large, we use a normal distribution as an approximation to the Poisson distribution with mean = variance = 90. That is, V ~ N(90, 90) approximately. P(V < 80) = P(V ≤ 79.5 ) ≈ 0.13419 = 0.134 (to 3 sig. fig.) (Answer)

33


To find the probability by using the graphic calculator: Step 1: Type

and under DISTR sub-menu, choose 2: normalcdf( .


The Poisson model assumes that events occur ‘uniformly’. That is, the expected number of events in a given time interval is proportional to the size of the interval. However, in real life, the mean number of grand pianos sold occur within a small fixed interval of time as it is a non-essential or luxury item, which could be sold in a seasonal manner. Thus, the Poisson distribution may not be a good model for the number of grand pianos sold in a year. [Answer]

10(i) Required number of ways = 3C2 × 4C3 × 5C3 = 120. [Answer] (ii) Required number of ways = 9C8 = 9. [Answer] (iii) Case 1: 4 diplomats from M

Then 4 diplomats have to be chosen from K and L. So, number of ways = 5C4 × 7C4 = 175 Case 2: 5 diplomats from M Then 3 diplomats have to be chosen from K & L. So, number of ways = 5C5 × 7C3= 35 Required number of ways = 175 + 35 = 210. [Answer] (iv) Required number of ways

= Number of unrestricted ways – (Number of ways chosen from L and M only) – (Number of ways chosen from K and M only) – (Number of ways chosen from K and L only) = 12C8 – 9C8 – 8C8 – 0 = 495 – 9 – 1 = 485. [Answer] 34


11(i)

X ~ N (50, 82) X1 + X2 ~ N (50 × 2, 82 × 2) as X1 + X2 are two independent observations of X. i.e.X1 + X2 ~ N (100, 128)

Thus, P(X1 + X2 >120) ≈ 0.0385499 = 0.0385 (correct to 3 sig. fig.) To calculate the probability by using the graphic calculator: Step 1: Press

and under DISTR sub-menu, choose 2:normalcdf(.


(ii)

X1 – X2 ~ N (50 – 50, 82 + 82)

i.e. X1 – X2 ~ N (0, 128). P(X1 > X2 + 15) = P(X1 – X2 > 15) ≈ 0.092449 = 0.0924 (correct to 3 sig. fig.) [Answer] To find the probability by using the graphic calculator: Step 1: Type

and under DISTR sub-menu, select 2:normalcdf(.


Since Y is a linear function of X and X is normal , Y is also normal. Since normal distribution is symmetrical about the mean and P(Y < 74) = P(Y > 146), 74 + 146 we have E(Y) = = 110. [Answer] 2

35


As P(Y < 74) = 0.0668, we have P(Z < 74 − 110 ) = 0.0668 where σ 2 = Var(Y). σ

Using the graphic calculator, 74 − 110 = –1.5000556 σ

so

σ = 23.9991.

Therefore, Var(Y) = σ 2 ≈ 575.96 = 576 (to 3 sig.fig). [Answer] To find the value of c when P(Z < c) = 0.0668 by using the graphic calculator: Step 1: Type

and under DISTR sub-menu, choose 3:invNorm(.


Note that E(Y) = a E(X) + b = 50a + b and

Var(Y) = a2 Var(X) + Var(b) = a2 Var(X) + 0 = 64 a2 .

Thus, 110 = 50a + b -------------------(1) 575.96 = 64 a2 ------------------- (2) Solving (2), a ≈ 2.9999 or – 2.9999 (rejected as a > 0) = 3 (to 3 sig. fig) (Answer) Substitute a = 2.9999 into (1), we have 110 = 50(2.9999)+ b b = – 39.995 = – 40 (to 3 sig.fig.) [Answer]

36


f Solutions to H1 Mathematics Paper 1

1

Answer:

You can obtain the graph from the graphic calculator: Step 1: Press

and type in the equation as shown on the left below.

Step 2: Press

and set the window as shown on the right below. We set Ymin = –2 and Ymax = 2 as −1 ≤ sin x ≤ 1 .

Step 3: Press

to see the graph shown on the right.

From the graph, when sin α = c, where 0 < α < 90o, (i) sin (2 π + α) = c [Answer] (ii) sin (3 π + α) = – c [Answer]

One possible value of x for which sin x = – c is either π + α or 2π − α . [Answer]

37


2

x + y = 20 ---------------------- (1) x2 + y2 = 300 ---------------------- (2)

Rearrange (1) y = 20 – x ---------------------- (3) Substitute y = 20 – x into (2), we have x2 + (20 – x)2 = 300 2 2 x + 400 – 40x + x = 300 2x2 – 40x + 100 = 0 x2 – 20x + 50 = 0

Alternatively, by completing the square, we have

2 x = 20 ± 20 − 4(1)(50)

2(1)

(x – 10)2 – 100 + 50 = 0

= 20 ± 200

(x – 10)2 – 50 = 0

= 20 ± 100 × 2

(x – 10)2 = 50

2

2 = 20 ± 10 2 2

= 10 ± 5 2

x = 10 ± 50

= 10 ± 25 × 2 = 10 ± 5 2 .

Substitute the values of x into (3), we have y = 20 – ( 10 ± 5 2 ) = 10 ∓ 5 2

Since x > y and 10 − 5 2 < 10 + 5 2 , we have x = 10 + 5 2 and y = 10 − 5 2 . [Answer]

3

y = 2x2 y = x2 + k2

------------- (1) ------------- (2)

Substitute y = 2x2 into (2): 2x2 = x2 + k2 x2 = k2 x = ± k2 = k or –k. Therefore, the x –coordinates of P and Q are k and –k respectively. [Shown]

38


Required area =

∫

k

−k

( x 2 + k 2 ) − (2 x 2 ) dx

k

= 2 ∫ k 2 − x 2 dx −k

= ⎡k 2 x − x3 ⎤ ⎢ 3⎥ ⎣

k

⎦−k

3 3 = ⎡ k 3 − k ⎤ − ⎡ −k 3 − (−k ) ⎤

3 ⎦ ⎣⎢

⎣

3

3 ⎦⎥

3

= 2k + 2k = 4k units2. [Answer] 3

3

3

3

Alternatively, as both graphs are symmetrical about the y–axis, k

required area = 2 ∫ ( x 2 + k 2 ) − (2 x 2 ) dx 0

k

k 3 3 3 3 = 2 ∫ k 2 − x 2 dx = 2 ⎡⎢ k 2 x − x ⎤⎥ = 2 ⎡⎢⎛⎜ k 3 − k ⎞⎟ − ⎛⎜ 03 − 0 ⎞⎟ ⎤⎥ = 4k units2. 0 3 3⎠ ⎝ 3 ⎠⎦ 3 ⎦0 ⎣ ⎣⎝

4(i)(a) Answer : Graph of y = f(x) = x2 – 1

To obtain the graph from the graphic calculator: Step 1: Press Step 2: Press

and type in the equation as shown on the left below. for 6:ZStandard and the graph is shown on the right below.

39


4(i)(b) Answer : Graph of y = gf(x) = g(x2 – 1) = | x2 – 1|

To obtain the graph from the graphic calculator: Step 1: Press

and type in the equation as shown on the left below. Note that abs( menu by pressing .

could be obtained from the CATALOG

Step 2: Press to see the graph plotted in the standard window setting as shown on the right below.

(c) Answer : Graph of y = fg(x) = f (|x|) = |x|2 – 1.

Note that since |x|2 = x2, we have y = fg(x) = x2 – 1= f(x) and so, the graph of y = fg(x) is identical to the graph of y = f(x).

40


Alternatively, you can also obtain the graph from the graphic calculator: Step 1: Press and key in the equation as shown on the left below. Note that abs( . could be obtained from the CATALOG menu by pressing Step 2: Press

4(ii)

to see the graph shown on the right below.

For f -1 to exit, f has to be a one-one function. From the graph in 4(i)(a), f is a one-one function with domain restricted to x ≤ a where a ≤ 0 . Therefore, the greatest possible value of a is 0. [Answer] From the graph, range of f is (–1, ∞ ). Thus, we have Df −1 = R f = (–1, ∞ ). [Answer] Let

y = x2 – 1, x ≤ 0 y + 1 = x2 x = ± y +1

As x ≤ 0, we have x = – y + 1 . Thus, f –1(x) = – x + 1 , x ≥ –1. [Answer]

5

x = t 3 – 12t2 + kt dx = 3t2 – 24t + k. dt dx > 0, x increases. When dt Let 3t2 – 24t + k > 0 for t ≥ 0 .

In order to have 3t2 – 24t + k > 0 for all t, its discriminant < 0. So, (– 24)2 – 4(3)(k) < 0 576 < 12k k > 48. [Answer]

41


Given that k = 36, we have x = t3 – 12 t2 + 36 t. 5(i)

Answer: Graph of x = t 3 – 12 t2 + 36 t.

To obtain the graph by using the graphic calculator: Step 1: Press

and key in the equation as shown on the left below.

Step 2: As the independent variable t is positive, press to set the window such that Xmin = 0, Xmax = 10. Need not key in the values for Ymin and Ymax. The graphic calculator will calculate the appropriate values of Ymin and Ymax when are pressed. The graph is shown on the right below. You may check that the curve passes through the point (6, 0) when are pressed.

(ii)

375 = t3 – 12 t2 + 36 t t – 12 t + 36 t – 375 = 0 3

2

Using the graphic calculator, t = 11.7 seconds (to 1 d.p.). [Answer] To solve the cubic equation by using the graphic calculator: and select 5:PlySmlt2 as shown on the left below by pressing

Step 1: Press .

Step 2: Press any key to see the MAIn MEnu shown on the right above. or . Select option 1:POLY ROOT FINDER by pressing

42


Step 3: Select the parameters as seen on the left screen shot below.


6.

and type in the values as shown on the right screen to see the answer shown below.

y = ln (2x + 4) dy = 2 = 1 . dx x+2 2x + 4

When x = 1, dy = dx

2 1 1 = . Thus, the gradient of the tangent to C at x =1 is . 2(1) + 4 3 3

Equation of the tangent to C at P(1, ln 6): y – y1 = m (x – x1) y – ln 6 = 1 (x – 1) 3 1 y = x – 1 + ln 6. 3 3

As the y-coordinate of T is 0, we let y = 0. So, 0 = 1 x – 1 + ln 6 3 3 1 x = 1 – ln 6 3 3

x = l – 3 ln 6.

Therefore, the exact x–coordinate of T is 1 – 3 ln 6. [Shown] 1 Gradient of the normal to C at x =1 is − 1 = – 3. 3

43


Thus, equation of the normal to C at P(1, ln 6): y – y1 = m (x – x1) y – ln 6 = – 3(x – 1) y = –3x + 3 + ln 6. As the y-coordinate of N is 0, we let y = 0. So, 0 = –3x + 3 + ln 6 3x = 3 + ln 6 x = 1 + ln 6 . 3

Therefore, the exact x–coordinate of N is 1 + ln 6 . [Answer]

3 ln 6 ln Length of TN = 1 + – (1 – 3 ln 6) = 1 + 6 + 3 ln 6 – 1 = 10 ln 6 . 3 3 3

Note that the height of triangle PTN is the y-coordinate of the point P, which is ln 6. Thus, area of triangle PTN = 1 × base × height

2 1 = × 10 ln 6 × ln 6 2 3 5 = ( ln 6 )2 units2. [Answer] 3

7

If it is normally distributed, the probability of the marks that are beyond 100 is more than 0.033 (refer to the screen shot on the right), which is not negligible. This means that at least 3% of the marks would be above 100, which is impossible. Therefore, normal distribution will not give a good approximation to the distribution of marks. Let X be the random variable of “The marks of the candidates”. As n = 50 is relatively large, by Central Limit Theorem, we have 2 X ~ N (72.1, 15.2 ) approximately. 50

Using the graphic calculator,

P (70.0 < X < 75.0) = 0.747 (to 3 sig fig). [Answer]

To find the probability by using the graphic calculator: Step 1: Press

to display DISTR menu. Under DISTR sub-menu, choose .

2: normalcdf( by pressing

Step 2: Key in the data according to the sequence as follows: lower bound, upper bound, mean and standard deviation, as shown on the right below.

44


8

Let X be the random variable denoting the number of loaves out of 6 which are ‘crusty’. Note that X ~ B (6, 0.6) So P(X = 3) = 0.27648 = 0.276 (to 3 sig. fig.). [Answer] To find the probability by using the graphic calculator: Step 1: Press

to display DISTR menu. Under DISTR sub-menu, choose

A:binompdf(.

Step 2: Key in the data according to the sequence as shown on the right below.

Let Y be the random variable denoting the number of loaves out of 40 which are ‘crusty’. Note that Y ~ B (40, 0.6). Since n = 40 is large, np = 40(0.6) = 24 > 5, nq = 40 (1 – 0.6) = 16 > 5, we have Y ~ N (np, npq) approximately. That is, Y ~ N( 24, 9.6) approximately. Thus, using the graphic calculator, P(Y ≥ 20) = P(Y > 19.5) = 0.927 (to 3 sig. fig.) [Answer] To find the probability by using the graphic calculator: Step 1: Press

for DISTR menu. Under DISTR sub-menu, choose

2: normalcdf(.

Step 2: Key in the following data according to the sequence as follows: lower bound, upper bound, mean and standard deviation, as shown on the right below.

Let W be the random variable denoting the mass of a loaf of bread in kg. Given that W ~ N (1.24, σ 2 ) P(W < 1) = 0.04

P( Z < 1 − 1.24 ) = 0.04. σ

45


Using the graphic calculator, we have − 0.24

σ

and so

= –1.7507

σ = − 0.24 = 0.137 (to 3 sig fig). [Answer] −1.7507

You can use the graphic calculator to get the value of c when P( Z < c) = 0.04 by simply pressing and select 3:invNorm(, then key in 0.04 and press .

Tan’s

9(i)

Mui’s 2 8

3 8

Red

5 8

Red

Blue

1 8

Green 3 8

5 8

Red

4 8

Blue

Blue

1 8

Green (ii)

The probability that Mui’s pen is blue, given that Tan’s pen is red = 5 . [Answer] 8

(iii) The probability that Mui’s pen is red = (iv)

3 2 5 3 × + × = 21 . [Answer] 8 8 8 8 64

Let A denotes the event that Tan’s pen is red and let B denotes the event that Mui’s pen is blue. 3 5

× P(A ∩ B) 8 8 P( A | B) = = 3 5 5 4 P( B) × + × 8 8 8 8

= 3 . [Answer] 7

46


10

Let X be the random variable denoting the number of hours of the lifetime of a particular type of battery. x =

∑x

n 10317 = ≈ 147.39 70

2 ⎛ x) ⎞ ( 1 ∑ 2 ⎜∑ x − ⎟ Unbiased estimate of population variance = n −1 ⎜ n ⎟ ⎝ ⎠ 2 ⎛ ⎞ = 1 ⎜1540231 − (10317 ) ⎟ 69 ⎜ 70 ⎟ ⎝ ⎠

≈ 284.82

H0: μ = 150 H1: μ < 150 As n = 70 is large, by Central Limit Theorem, X is approximately normally distributed. Thus, Z-test is used with x = 147.39, s =

284.82 and n = 70. Assuming H0 is true, we

obtain from the graphic calculator that the p-value = 0.0978 (to 3 sig. fig.) [Answer] To find the p-value using the graphic calculator: Step 1: Press pressing

, go to TESTS, choose 1: Z‐Test... as shown on the left below by .

Step 2: Choose the Stats option and key in the respective data as shown on the right below.

Step 3: Move the cursor to Calculate and press

to see the p-value.

Alternatively, we could calculate the p-value as follows:

47


Under H0, X ~ N(150, 284.82 ) approximately by Central Limit Theorem as n = 70 is large. 70

Thus, by using the graphic calculator, p-value = P( X < x ) = P( X < 10317 ) 70

= 0.0978 (to 3 sig.fig.). p–value is the lowest significant level at which the null hypothesis would be rejected on the basis of the observed outcome. Hence, the p-value in this question refers to the lowest significance level at which the claim that the mean lifetime of the battery is equal to 150 hours is rejected on the basis of the observed outcome. Let W be the random variable denoting the number of hours of lifetime of each battery in the combined sample. n = 70 + 50 = 120 ∑ w = ∑ x + ∑ y = 10317 + 7331 = 17 648

∑w

2

=

w =

∑x +∑ y 2

2

= 1540231 + 1100565 = 2640796

17648 120

≈ 147.067

2 ⎛ w) ⎞ ( 1 ⎜ ∑ 2 Unbiased estimate of population variance = ∑ w − n ⎟⎟ n −1 ⎜ ⎝ ⎠ 2 ⎛ 17648 ) ⎞ ( 1 = ⎜ 2640796 − ⎟ 119 ⎜ 120 ⎟ ⎝ ⎠

≈ 381.2056

H0 : μ = 150 H1 : μ < 150 Assuming H0 is true. As n = 120 is large, by the Central Limit Theorem, w − 150 ~ N (0, 1) approximately. test statistic z = s/ n Using 1-tail Z-test to test at 10% level of significance, where s =

381.2056 , w =

17648 120

and n = 120, we obtain from the graphic calculator that the p-value = 0.0499(to 3 sig.fig.). Since p-value < 0.1, we reject H0 and conclude that there is sufficient evidence that the population mean is less than 150 at 10% significance level.

48


You may use the graphic calculator to help you with the hypothesis testing. Step 1: Press

.

Step 2: Go to TESTS sub-menu by pressing the left arrow key 1: Z‐Test… as shown on the left below.

and select option

Step 3: Choose the Stats option and key in the respective data as shown on the right below.

Step 4: Move the cursor to Calculate and press below.

11(i)

to get the Z-test results as shown

Answer :

49


To obtain the scatter plot of the data by using the graphic calculator : Step 1: Press

, then press

to select the option 1:Edit…

Step 2: Key in the list table as shown on the right below.

Step 3: Press to display STAT PLOT menu as shown on the left below. Press and highlight the appropriate settings as shown on the right below by using the arrow and keys.

Step 4: Press below.

for 9:ZoomStat to see the appropriate scatter diagram shown

11(ii) Using the graphic calculator, x = 17, y = 343.75= 344 (to 3 sig. fig.) [Answer]

Answer:

50


To find the values of x and y by using the graphic calculator:

Step 1: Press , then press to go to CALC sub-menu. Select option 2:2‐Var Stats as seen on the left below by pressing . .

Step 2: Key in the appropriate list variables in the order as shown on the right above. Step 3: Press

and the value of x is shown. Press

to see the value of y .

11(iii) Using the graphic calculator, the equation of the regression line of y on x is

y = 53.3 + 17.1 x (to 3 sig.fig.) [Answer]

Note that the line must be drawn to pass through the point ( x , y ) .

51


To find the equation of the regression line of y on x by using the graphic calculator: Step 1: Press , then press to go to CALC sub-menu. Select 4:LinReg(ax+b) and press . . Step 2: Key in the list variables in the order of independent variable, dependent variable and Y1 as shown on the middle screen shot below. Press to get the equation of the regression line of y on x, including the product moment correlation coefficient r. Here, Y1 is typed in so that the regression line could be drawn on the screen later.

Note: To get the value of r, DiagnosticOn must be selected. Press CATALOG menu and scroll to DiagnosticOn, then press Step 3: Press below.

to display twice.

for 9:ZoomStat to see the graph of the regression line as shown

11(iv) Using the graphic calculator, the product moment correlation coefficient r is 0.969 (to 3 sig. fig.). [Answer] Note that the value of r can be found on the screen shot on the extreme right above.

From the scatter plot, we observe that the data points are close to the regression straight line. In addition, as the value of r is close to 1, it shows a strong positive linear correlation between x and y. (v) When x = 20, y = 53.333 + 17.083(20) = 394.993. The corresponding profit = $395 000 (to 3 sig. fig.) [Answer] (vi) As x = 40 is outside the range of data on which we obtained the regression line, the variables may not be linearly correlated. Hence, the estimate may not be reliable and extrapolation must be applied with caution. [Answer]

52


12

Let X be the random variable denoting the mass of an apple in kg. Here, X ~ N (0.234, 0.0252) Let S be the random variable denoting the mass in kg, of a small bag containing 5 apples. Then S = X1 + X2 + X3 + X4 + X5 ~ N (5× 0.234, 5× 0.0252). Thus, S ~ N(1.17, 0.003125) Let L be the random variable denoting the mass in kg, of a large bag containing 10 apples. Then L = X1 + X2 + X3 +……+ X9 + X10 ~ N (10× 0.234, 10× 0.0252). Thus, L ~ N(2.34, 0.00625) P(S > 1.2) ≈ 0.29575 = 0.296 ( to 3 sig. fig.) [Answer] Using the graphic calculator to find P(S > 1.2): Step 1: Type



S1+ S2 – L ~ N( 2(1.17) – 2.34, 2(0.003125) + 0.00625 ) Thus, S1+ S2 – L ~ N(0, 0.0125)

P(| S1+ S2 – L| < 0.2) = P( – 0.2 < S1+ S2 – L < 0.2) ≈ 0.92636 = 0.926 (to 3 sig. fig.) [Answer] Using the graphic calculator to find P(|S1+ S2 – L | < 0.2): Step 1: Type



53


Let M be the random variable denoting the amount of money in $ that Lee pays more than Foo. Thus, M = 1.5 (S1+ S2) – 1.2 L . As E(M) = 1.5(1.17+1.17) – 1.2(2.34) = 3.51 – 2.808 = 0.702 and Var(M) = 1.52(0.003125 + 0.003125) + 1.22(0.00625) = 0.0140625 +0.009 = 0.0230625 we have M ~ N (0.702, 0.0230625). Using the graphic calculator, P (M ≥ 0.50) ≈ 0.90826 = 0.908 (to 3 sig. fig). [Answer] To find the probability by using the graphic calculator: Step 1: Type



54