Taller final de Klasse 11 [261 marks]

[1 mark]1a.

Write down the value of

(i) ;

Markscheme(i) A1 N1[1 mark]

27log3

27 = 3log3

[1 mark]1b. (ii) ;

Markscheme(ii) A1 N1

[1 mark]

log818

= −1log818

[1 mark]1c. (iii) .

Markscheme(iii) A1 N1

[1 mark]

4log16

4 =log1612

[3 marks]1d. Hence, solve .

Markschemecorrect equation with their three values (A1)eg

correct working involving powers (A1)eg

A1 N2[3 marks]

27 + − 4 = xlog3 log818

log16 log4

= x, 3 + (−1) − = x32

log412

log4

x = , =432 4

32 4 xlog4

x = 8

[1 mark]2a.

Consider the expansion of .

Write down the number of terms in this expansion.

Markscheme11 terms A1 N1[1 mark]

(x + 3)10

[4 marks]2b. Find the term containing .x3

Markschemeevidence of binomial expansion (M1)

eg , attempt to expand

evidence of choosing correct term (A1)

eg , ,

correct working (A1)

eg , ,

A1 N3[4 marks]

( )n

ran−rbr

term, r = 78th ( )107

(x (3)3 )7

( )107

(x (3)3 )7 ( )103

(x (3)3 )7

262440 (accept 262000 )x3 x3

3a. [3 marks]

Let .

(i) Show that .

(ii) Write down the domain of .

Markscheme(i) interchanging x and y (seen anywhere) M1

e.g.

correct manipulation A1

e.g. ,

AG N0

(ii) A1 N1

[3 marks]

f(x) = ex+3

(x) = lnx − 3f−1

f−1

x = ey+3

lnx = y + 3 lny = x + 3

(x) = lnx − 3f−1

x > 0

[4 marks]3b. Solve the equation .

Markschemecollecting like terms; using laws of logs (A1)(A1)

e.g. , , ,

simplify (A1)

e.g. ,

A1 N2

[4 marks]

(x) = lnf−1 1x

lnx − ln( ) = 31x

lnx + lnx = 3 ln( ) = 3x1x

ln = 3x2

lnx = 32

=x2 e3

x = (= )e32 e3−−√

4a. [2 marks]

Let . The vertex of the graph of is at and the graph passes through .

Write down the value of and of .

f(x) = a(x − h + k)2 f (2,3) (1,7)

h k

Markscheme A1A1 N2

[2 marks]

h = 2, k = 3

4b. [3 marks]Find the value of .

Markschemeattempt to substitute in any order into their (M1)

eg

correct equation (A1)eg

a = 4 A1 N2[3 marks]

a

(1,7) f(x)

7 = a(1 − 2 + 3, 7 = a(1 − 3 + 2, 1 = a(7 − 2 + 3)2 )2 )2

7 = a + 3

[3 marks]5a.

Let

Sketch the graph of .

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct sinusoidal shape.

Only if this A1 is awarded, award the following:

A1 for correct domain,

A1 for approximately correct range.

[3 marks]

f(x) = cos( x) + sin ( x) , for − 4 ⩽ x ⩽ 4.π

4π

4

f

[5 marks]5b. Find the values of where the function is decreasing.x

Markschemerecognizes decreasing to the left of minimum or right of maximum,

eg (R1)x-values of minimum and maximum (may be seen on sketch in part (a)) (A1)(A1)eg

two correct intervals A1A1 N5eg

[5 marks]

(x) < 0f ′

x = −3, (1, 1.4)

−4 < x < −3, 1 ⩽ x ⩽ 4; x < −3, x ⩾ 1

[3 marks]5c. The function can also be written in the form , where , and . Find the value of ;

Markschemerecognizes that is found from amplitude of wave (R1)y-value of minimum or maximum (A1)eg (−3, −1.41) , (1, 1.41)

A1 N3[3 marks]

f f(x) = asin ( (x + c))π

4a ∈ R 0 ⩽ c ⩽ 2 a

a

a = 1.41421

a = , (exact), 1.41,2√

[4 marks]5d. The function can also be written in the form , where , and . Find the value of .

MarkschemeMETHOD 1recognize that shift for sine is found at x-intercept (R1)attempt to find x-intercept (M1)eg

(A1) A1 N4

METHOD 2attempt to use a coordinate to make an equation (R1)eg

attempt to solve resulting equation (M1)eg sketch,

(A1) A1 N4

[4 marks]

f f(x) = asin ( (x + c))π

4a ∈ R 0 ⩽ c ⩽ 2 c

cos( x) + sin ( x) = 0, x = 3 + 4k, k ∈ Zπ

4π

4

x = −1

c = 1

sin ( c) = 1, sin ( (3 − c)) = 02√ π

42√ π

4

x = 3 + 4k, k ∈ Zx = −1

c = 1

[2 marks]6a.

Let . The equation has two equal roots.

Write down the value of the discriminant.

Markschemecorrect value , or A2 N2

[2 marks]

f(x) = 3 − 6x + px2 f(x) = 0

0 36 − 12p

[1 mark]6b. Hence, show that .p = 3

Markschemecorrect equation which clearly leads to A1eg

AG N0[1 mark]

p = 3

p = 3

36 − 12p = 0, 36 = 12p

p = 3

6c. [4 marks]The graph of has its vertex on the -axis.

Find the coordinates of the vertex of the graph of .

MarkschemeMETHOD 1valid approach (M1)eg

correct working A1eg

correct answers A1A1 N2eg

METHOD 2valid approach (M1)eg , factorisation, completing the square

correct working A1eg

correct answers A1A1 N2eg

METHOD 3valid approach using derivative (M1)eg

correct equation A1eg

correct answers A1A1 N2eg

[4 marks]

f x

f

x = − b

2a

− , x =(−6)

2(3)66

x = 1, y = 0; (1, 0)

f(x) = 0

− 2x + 1 = 0, (3x − 3)(x − 1), f(x) = 3(x − 1x2 )2

x = 1, y = 0; (1, 0)

(x) = 0, 6x − 6f ′

6x − 6 = 0

x = 1, y = 0; (1, 0)

6d. [1 mark]The graph of has its vertex on the -axis.

Write down the solution of .

Markscheme A1 N1

[1 mark]

f x

f(x) = 0

x = 1

6e. [1 mark]The graph of has its vertex on the -axis.

The function can be written in the form . Write down the value of .

Markscheme A1 N1

[1 mark]

f x

f(x) = a(x − h + k)2 a

a = 3

6f. [1 mark]The graph of has its vertex on the -axis.

The function can be written in the form . Write down the value of .

Markscheme A1 N1

[1 mark]

f x

f(x) = a(x − h + k)2 h

h = 1

6g. [1 mark]The graph of has its vertex on the -axis.

The function can be written in the form . Write down the value of .

Markscheme A1 N1

[1 mark]

f x

f(x) = a(x − h + k)2 k

k = 0

6h. [4 marks]The graph of has its vertex on the -axis.

The graph of a function is obtained from the graph of by a reflection of in the -axis, followed by a translation by the vector .

Find , giving your answer in the form .

Markschemeattempt to apply vertical reflection (M1)

eg , sketch

attempt to apply vertical shift 6 units up (M1)eg , vertex

transformations performed correctly (in correct order) (A1)eg

A1 N3[4 marks]

f x

g f f x ( )06

g g(x) = A + Bx + Cx2

−f(x), − 3(x − 1)2

−f(x) + 6 (1,6)

−3(x − 1 + 6, − 3 + 6x − 3 + 6)2 x2

g(x) = −3 + 6x + 3x2

[2 marks]7a.

The number of bacteria in two colonies, and , starts increasing at the same time.

The number of bacteria in colony after hours is modelled by the function .

Find the number of bacteria in colony after four hours.

Markschemecorrect substitution into formula (A1)eg

bacteria in the dish A1 N2[2 marks]

A B

A t A(t) = 12e0.4t

A

12e0.4(0)

12

[3 marks]7b. Find the number of bacteria in colony after four hours.A

Markschemecorrect substitution into formula (A1)eg

(A1) bacteria in the dish (integer answer only) A1 N3

[3 marks]

12e0.4(4)

59.4363

59

[3 marks]7c. How long does it take for the number of bacteria in colony to reach ?

Markschemecorrect equation (A1)eg

valid attempt to solve (M1)eg graph, use of logs

(hours) A1 N3[3 marks]

A 400

A(t) = 400, 12 = 400e0.4t

8.76639

8.77

7d. [3 marks]The number of bacteria in colony after hours is modelled by the function .

After four hours, there are bacteria in colony . Find the value of .

Markschemevalid attempt to solve (M1)eg , use of logs

correct working (A1)eg sketch of intersection,

(exact), A1 N3

[3 marks]

B t B(t) = 24ekt

60 B k

n(4) = 60, 60 = 24e4k

4k = ln2.5

k = 0.229072

k = ln 2.54

k = 0.229

7e. [4 marks]The number of bacteria in colony after hours is modelled by the function .

The number of bacteria in colony first exceeds the number of bacteria in colony after hours, where . Find the value of .

MarkschemeMETHOD 1setting up an equation or inequality (accept any variable for ) (M1)eg correct working (A1)eg sketch of intersection,

(accept ) (A1) (integer answer only) A1 N3

METHOD 2 (from earlier work)

A1A1valid reasoning (R1)eg and

(integer answer only) A1 N3[4 marks]

B t B(t) = 24ekt

A B n n ∈ Z n

n

A(t) > B(t), 12 = 24 , = 2e0.4n e0.229n e0.4n e0.229n

= 2e0.171n

4.05521 4.05349n = 5

A(4) = 59, B(4) = 60A(5) = 88.668, B(5) = 75.446

A(4) < B(4) A(5) > B(5)n = 5

8. [8 marks]The equation has two distinct real roots.

Find the possible values of .

Markschemeevidence of discriminant (M1)eg correct substitution into discriminant (A1)eg correct discriminant A1eg recognizing discriminant is positive R1eg attempt to solve their quadratic in (M1)eg factorizing, correct working A1eg , sketch of positive parabola on the x-axiscorrect values A2 N4eg [8 marks]

+ (k + 2)x + 2k = 0x2

k

− 4ac, Δ = 0b2

(k + 2 − 4(2k), + 4k + 4 − 8k)2 k2

− 4k + 4, (k − 2k2 )2

Δ > 0, (k + 2 − 4(2k) > 0)2

k

k = 4± 16−16√2

(k − 2 > 0, k = 2)2

k ∈ R and k ≠ 2, R∖2, ]−∞, 2[ ∪ ]2, ∞[

[3 marks]9a.

Let , for .

Find .

MarkschemeMETHOD 1

attempt to set up equation (M1)

eg ,

correct working (A1)

eg ,

A1 N2

METHOD 2

interchanging and (seen anywhere) (M1)

eg

correct working (A1)

eg ,

A1 N2

[3 marks]

f(x) = x − 5− −−−−√ x ≥ 5

(2)f−1

2 = y − 5− −−−√ 2 = x − 5− −−−−√

4 = y − 5 x = + 522

(2) = 9f−1

x y

x = y − 5− −−−√

= y − 5x2 y = + 5x2

(2) = 9f−1

[3 marks]9b. Let be a function such that exists for all real numbers. Given that , find .g g−1 g(30) = 3 (f ∘ )(3)g−1

Markschemerecognizing (M1)

eg

correct working (A1)

eg ,

A1 N2

Note: Award A0 for multiple values, eg .

[3 marks]

(3) = 30g−1

f(30)

(f ∘ )(3) =g−1 30 − 5− −−−−√ 25−−√

(f ∘ )(3) = 5g−1

±5

10a. [6 marks]

The diagram below shows part of the graph of a function .

The graph has a maximum at A( , ) and a minimum at B( , ) .

The function can be written in the form . Find the value of

(a)

(b)

(c) .

f

1 5 3 −1

f f(x) = p sin(qx) + r

p

q

r

Markscheme(a) valid approach to find (M1)

eg amplitude ,

A1 N2

[2 marks]

(b) valid approach to find (M1)

eg period = 4 ,

A1 N2

[2 marks]

(c) valid approach to find (M1)

eg axis = , sketch of horizontal axis,

A1 N2

[2 marks]

Total [6 marks]

p

= max−min2

p = 6

p = 3

q

q = 2π

period

q = π

2

r

max+min2

f(0)

r = 2

[2 marks]10b.

Markschemevalid approach to find (M1)

eg amplitude ,

A1 N2

[2 marks]

p

p

= max−min2

p = 6

p = 3

[2 marks]10c.

Markschemevalid approach to find (M1)

eg period = 4 ,

A1 N2

[2 marks]

q

q

q = 2π

period

q = π

2

[2 marks]10d. .r

Markschemevalid approach to find (M1)

eg axis = , sketch of horizontal axis,

A1 N2

[2 marks]

Total [6 marks]

r

max+min2

f(0)

r = 2

11a. [3 marks]

The following diagram shows triangle ABC.

Find AC.

Markschemeevidence of choosing cosine rule (M1)eg

correct substitution into the right-hand side (A1)eg

A1 N2[3 marks]

A = A + B − 2(AB)(BC)cos(A C)C2 B2 C2 B

+ − 2(6)(10)cos62 102 100∘

AC = 12.5234

AC = 12.5 (cm)

[3 marks]11b. Find .

Markschemeevidence of choosing a valid approach (M1)

eg sine rule, cosine rule

correct substitution (A1)

eg

A1 N2

[3 marks]

B AC

= , cos(B A) =sin(B A)C6

sin 100∘

12.5C + −(AC)2 102 62

2(AC)(10)

B A = 28.1525C

B A =C 28.2∘

[3 marks]12a.

The population of deer in an enclosed game reserve is modelled by the function , where is in

months, and corresponds to 1 January 2014.

Find the number of deer in the reserve on 1 May 2014.

Markscheme (A1)

correct substitution into formula (A1)eg

969 (deer) (must be an integer) A1 N3[3 marks]

P(t) = 210 sin(0.5t − 2.6) + 990 t

t = 1

t = 5

210 sin(0.5 × 5 − 2.6) + 990, P(5)

969.034982…

[2 marks]12b. Find the rate of change of the deer population on 1 May 2014.

Markschemeevidence of considering derivative (M1)eg

(deer per month) A1 N2[2 marks]

P ′

104.475

104

[1 mark]12c. Interpret the answer to part (i) with reference to the deer population size on 1 May 2014.

Markscheme(the deer population size is) increasing A1 N1[1 mark]

[2 marks]13a.

The following diagram shows a circle with centre and radius .

The points , and lie on the circumference of the circle, and radians.

Find the length of the arc .

Markschemecorrect substitution into arc length formula (A1)eg arc length (cm) A1 N2[2 marks]

O 5 cm

A rmB rmC A C = 0.7O

ABC

0.7 × 5= 3.5

[2 marks]13b. Find the perimeter of the shaded sector.

Markschemevalid approach (M1)

eg

perimeter (cm) A1 N2

[2 marks]

3.5 + 5 + 5, arc + 2r

= 13.5

[2 marks]13c. Find the area of the shaded sector.

Markschemecorrect substitution into area formula (A1)eg

A1 N2[2 marks]

(0.7)(512

)2

area = 8.75 (c )m2

[4 marks]14a.

In triangle , and . The area of the triangle is .

Find the two possible values for .

Markschemecorrect substitution into area formula (A1)eg

correct working (A1)eg

;

; A1A1 N3(accept degrees ie ; )

[4 marks]

ABC AB = 6 cm AC = 8 cm 16 cm2

A

(6)(8)sin A = 16, sin A =12

1624

A = arcsin ( )23

A = 0.729727656… ,2.41186499… ( , )41.8103149∘ 138.1896851∘

A = 0.730 2.41

41.8∘ 138∘

[3 marks]14b. Given that is obtuse, find .

Markschemeevidence of choosing cosine rule (M1)eg

correct substitution into RHS (angle must be obtuse) (A1)eg ,

A1 N2[3 marks]

A BC

B = A + A − 2(AB)(AC) cosA, + − 2ab cosCC2 B2 C2 a2 b2

B = + − 2(6)(8)cos2.41, + − 2(6)(8)cosC2 62 82 62 82 138∘

BC = 171.55− −−−−√

BC = 13.09786

BC = 13.1 cm

[2 marks]15a.

Consider points A( , , ) , B( , , ) and C( , , ) . The line passes through C and is parallel to .

Find .

1 −2 −1 7 −4 3 1 −2 3 L1 AB− →−−

AB− →−−

Markschemevalid approach (M1)

eg , ,

A1 N2

[2 marks]

−⎛⎝⎜

7−43

⎞⎠⎟

⎛⎝⎜

1−2−1

⎞⎠⎟ A − B = +AB

− →−−AO− →−−

OB− →−−

=AB− →−− ⎛

⎝⎜6

−24

⎞⎠⎟

[2 marks]15b. Hence, write down a vector equation for .

Markschemeany correct equation in the form (accept any parameter for )

where and is a scalar multiple of A2 N2

eg , ,

Note: Award A1 for , A1 for , A0 for .

[2 marks]

L1

r = a + tb t

a =⎛⎝⎜

1−23

⎞⎠⎟ b AB

− →−−

r = + t⎛⎝⎜

1−23

⎞⎠⎟

⎛⎝⎜

6−24

⎞⎠⎟ (x,y,z) = (1,−2,3) + t(3,−1,2) r =

⎛⎝⎜

1 + 6t

−2 − 2t

3 + 4t

⎞⎠⎟

a + tb = a + tbL1 r = b + ta

[3 marks]15c.

A second line, , is given by .

Given that is perpendicular to , show that .

Markschemerecognizing that scalar product (seen anywhere) R1

correct calculation of scalar product (A1)

eg ,

correct working A1

eg ,

AG N0

[3 marks]

L2 r = + s⎛⎝⎜

−12

15

⎞⎠⎟

⎛⎝⎜

3−3p

⎞⎠⎟

L1 L2 p = −6

= 0

6(3) − 2(−3) + 4p 18 + 6 + 4p

24 + 4p = 0 4p = −24

p = −6

[7 marks]15d. The line intersects the line at point Q. Find the -coordinate of Q.L1 L2 x

Markschemesetting lines equal (M1)

eg ,

any two correct equations with different parameters A1A1

eg , ,

attempt to solve their simultaneous equations (M1)

one correct parameter A1

eg ,

attempt to substitute parameter into vector equation (M1)

eg ,

(accept (4, -3, 5), ignore incorrect values for and ) A1 N3

[7 marks]

=L1 L2 + t = + s⎛⎝⎜

1−23

⎞⎠⎟

⎛⎝⎜

6−24

⎞⎠⎟

⎛⎝⎜

−12

15

⎞⎠⎟

⎛⎝⎜

3−3−6

⎞⎠⎟

1 + 6t = 1 + 3s −2 − 2t = 2 − 3s 3 + 4t = 15 − 6s

t = 12

s = 53

+⎛⎝⎜

1−23

⎞⎠⎟ 1

2

⎛⎝⎜

6−24

⎞⎠⎟ 1 + × 61

2

x = 4 y z

16a. [1 mark]

The line passes through the points and .

Show that

Markschemecorrect approach A1

eg , ,

AG N0

[1 mark]

L1 A(2,1,4) B(1,1,5)

=AB− →−− ⎛

⎝⎜−101

⎞⎠⎟

−⎛⎝⎜

115

⎞⎠⎟

⎛⎝⎜

214

⎞⎠⎟ AO + OB b − a

=AB− →−− ⎛

⎝⎜−101

⎞⎠⎟

16b. [1 mark]Hence, write down a direction vector for ;

Markschemecorrect vector (or any multiple) A1 N1

eg d =

[1 mark]

L1

⎛⎝⎜

−101

⎞⎠⎟

16c. [2 marks]Hence, write down a vector equation for .L1

16c.

Markschemeany correct equation in the form r = a + tb (accept any parameter for t)

where a is or , and b is a scalar multiple of A2 N2

eg r =

Note: Award A1 for a + tb, A1 for = a + tb, A0 for r = b + ta.

[2 marks]

1

⎛⎝⎜

214

⎞⎠⎟

⎛⎝⎜

115

⎞⎠⎟

⎛⎝⎜

−101

⎞⎠⎟

+ t , =⎛⎝⎜

115

⎞⎠⎟

⎛⎝⎜

−101

⎞⎠⎟

⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

2 − s

14 + s

⎞⎠⎟

L1

16d. [6 marks]

Another line has equation r = . The lines and intersect at the point P.

Find the coordinates of P.

Markschemevalid approach (M1)

eg ,

one correct equation in one parameter A1eg

attempt to solve (M1)eg

one correct parameter A1eg ,

attempt to substitute their parameter into vector equation (M1)

eg

P(4, 1, 2) (accept position vector) A1 N2[6 marks]

L2 + s⎛⎝⎜

47

−4

⎞⎠⎟

⎛⎝⎜

0−11

⎞⎠⎟ L1 L2

=r1 r2 + t = + s⎛⎝⎜

214

⎞⎠⎟

⎛⎝⎜

−101

⎞⎠⎟

⎛⎝⎜

47

−4

⎞⎠⎟

⎛⎝⎜

0−11

⎞⎠⎟

2 − t = 4,1 = 7 − s,1 − t = 4

2 − 4 = t,s = 7 − 1,t = 1 − 4

t = −2,s = 6,t = −3

+ 6⎛⎝⎜

47

−4

⎞⎠⎟

⎛⎝⎜

0−11

⎞⎠⎟

16e. [1 mark]Write down a direction vector for .

Markschemecorrect direction vector for A1 N1

eg ,

[1 mark]

L2

L2⎛⎝⎜

0−11

⎞⎠⎟

⎛⎝⎜

02

−2

⎞⎠⎟

16f. [6 marks]Hence, find the angle between and .L1 L2

16f.

Markschemecorrect scalar product and magnitudes for their direction vectors (A1)(A1)(A1)scalar product

magnitudes

attempt to substitute their values into formula M1eg

correct value for cosine, A1

angle is A1 N1

[6 marks]

1 2

= 0 × −1 + −1 × 0 + 1 × 1 (= 1)

= , ( , )+ +02 (−1)2 12− −−−−−−−−−−−−√ − + +12 02 12− −−−−−−−−−−√ 2√ 2√

, 0+0+1

( )×( )+ +02 (−1)2 12√ − + +12 02 12√

1×2√ 2√

12

(= )π

360∘

[2 marks]17a.

The line is parallel to the vector .

Find the gradient of the line .

Markschemeattempt to find gradient (M1)eg reference to change in is and/or is ,

gradient A1 N2

[2 marks]

L ( )32

L

x 3 y 2 32

= 23

[3 marks]17b.

The line passes through the point .

Find the equation of the line in the form .

Markschemeattempt to substitute coordinates and/or gradient into Cartesian equationfor a line (M1)eg correct substitution (A1)eg

A1 N2[3 marks]

L (9,4)

L y = ax + b

y − 4 = m(x − 9), y = x + b, 9 = a(4) + c23

4 = (9) + c, y − 4 = (x − 9)23

23

y = x − 2 (accept a = , b = −2)23

23

[2 marks]17c. Write down a vector equation for the line .

Markschemeany correct equation in the form r = a + tb (any parameter for t), where a indicates position eg or , and b is a scalar

multiple of

eg r = , r = 0i − 2 j + s(3i + 2 j) A2 N2

Note: Award A1 for a + tb, A1 for L = a + tb, A0 for r = b + ta.

[2 marks]

L

( )94

( )0−2

( )32

( ) + t ( ) ,( ) = ( )94

32

x

y

3t + 92t + 4

− →−− − →−− − →−− − →−−

[2 marks]18a.

In the following diagram, = p, = q and .

Express each of the following vectors in terms of p and q,

;

Markschemeappropriate approach (M1)eg

= p – q A1 N2[2 marks]

OP− →−−

OQ− →−−

=PT− →−−

12PQ− →−−

QP− →−−

= + , P − QQP− →−−

QO− →−−

OP− →−−

QP− →−−

[3 marks]18b. .

Markschemerecognizing correct vector for or (A1)eg (p – q), (q – p)

appropriate approach (M1)eg

(p + q) A1 N2

[3 marks]

OT− →−−

QT− →−−

PT− →−−

=QT− →−−

12

=PT− →−−

12

= + , + , +OT− →−−

OP− →−−

PT− →−−

OQ− →−−

QT− →−−

OP− →−−

12PQ− →−−

=OT− →−−

12

(accept )p+q

2

19a. [2 marks]

The following table shows the average weights ( y kg) for given heights (x cm) in a population of men.

Heights (x cm) 165 170 175 180 185

Weights (y kg) 67.8 70.0 72.7 75.5 77.2

The relationship between the variables is modelled by the regression equation .

Write down the value of and of .

Markscheme (exact) A1 N1

(exact), A1 N1[2 marks]

y = ax + b

a b

a = 0.486

b = −12.41 −12.4

19b. [2 marks]The relationship between the variables is modelled by the regression equation .

Hence, estimate the weight of a man whose height is 172 cm.

y = ax + b

Markschemecorrect substitution (A1)eg

(kg) A1 N2[2 marks]

0.486(172) − 12.41

71.182

71.2

[1 mark]19c. Write down the correlation coefficient.

Markscheme

A1 N1[1 mark]

r = 0.997276

r = 0.997

19d. [2 marks]State which two of the following describe the correlation between the variables.strong zero positivenegative no correlation weak

Markschemestrong, positive (must have both correct) A2 N2[2 marks]

[1 mark]20a.

The weights in grams of 80 rats are shown in the following cumulative frequency diagram.

Do NOT write solutions on this page.

Write down the median weight of the rats.

Markscheme50 (g) A1 N1[2 marks]

[3 marks]20b. Find the percentage of rats that weigh 70 grams or less.

Markscheme65 rats weigh less than 70 grams (A1)attempt to find a percentage (M1)eg

81.25 (%) (exact), 81.3 A1 N3[2 marks]

, × 1006580

6580

20c. [2 marks]The same data is presented in the following table.

Weights grams

Frequency

Write down the value of .

Markscheme A2 N2

[2 marks]

w 0 ⩽ w ⩽ 30 30 < w ⩽ 60 60 < w ⩽ 90 90 < w ⩽ 120

p 45 q 5

p

p = 10

20d. [2 marks]The same data is presented in the following table.

Weights grams

Frequency

Find the value of .

Markschemesubtracting to find (M1)eg

A1 N2[2 marks]

w 0 ⩽ w ⩽ 30 30 < w ⩽ 60 60 < w ⩽ 90 90 < w ⩽ 120

p 45 q 5

q

q

75 − 45 − 10

q = 20

20e. [3 marks]The same data is presented in the following table.

Weights grams

Frequency

Use the values from the table to estimate the mean and standard deviation of the weights.

Markschemeevidence of mid-interval values (M1)eg

(exact), (exact) A1A1 N3[3 marks]

w 0 ⩽ w ⩽ 30 30 < w ⩽ 60 60 < w ⩽ 90 90 < w ⩽ 120

p 45 q 5

15,45,75,105

= 52.5x σ = 22.5

20f. [2 marks]Assume that the weights of these rats are normally distributed with the mean and standard deviation estimated in part (c).Find the percentage of rats that weigh 70 grams or less.

Markscheme0.781650

78.2 (%) A2 N2[2 marks]

Markschemerecognize binomial probability (M1)

eg ,

valid approach (M1)

eg

A1 N2

[3 marks]

X ∼ B(n, p) ( )5r

× ×0.782r 0.2185−r

P(X ⩽ 3)

0.30067

0.301

[2 marks]21a.

Let and be independent events, where and .

Find .

Markschemecorrect substitution (A1)eg

A1 N2[2 marks]

A B P(A) = 0.3 P(B) = 0.6

P(A ∩ B)

0.3 × 0.6

P(A ∩ B) = 0.18

[2 marks]21b. Find .

Markschemecorrect substitution (A1)eg

A1 N2[2 marks]

P(A ∪ B)

P(A ∪ B) = 0.3 + 0.6 − 0.18

P(A ∪ B) = 0.72

21c. [1 mark]On the following Venn diagram, shade the region that represents .

Markscheme

A1 N1

A ∩ B′

[2 marks]21d. Find .

Markschemeappropriate approach (M1)eg

(may be seen in Venn diagram) A1 N2[2 marks]

P(A ∩ )B′

0.3 − 0.18, P(A) × P( )B′

P(A ∩ ) = 0.12B′

22a. [3 marks]

A forest has a large number of tall trees. The heights of the trees are normally distributed with a mean of metres and a standard

deviation of metres. Trees are classified as giant trees if they are more than metres tall.

A tree is selected at random from the forest.

Find the probability that this tree is a giant.

Markschemevalid approach (M1)eg

A1 N2[3 marks]

53

8 60

P(G) = P(H > 60, z = 0.875, P(H > 60) = 1 − 0.809, N (53, )82

0.190786

P(G) = 0.191

22b. [3 marks]A tree is selected at random from the forest.

Given that this tree is a giant, find the probability that it is taller than metres.

Markschemefinding (seen anywhere) (A1)recognizing conditional probability (R1)eg

correct working (A1)eg

A1 N3[6 marks]

70

P(H > 70) = 0.01679

P(A |B), P(H > 70 |H > 60)

0.016790.191

0.0880209

P(X > 70 |G) = 0.0880

[2 marks]22c. Two trees are selected at random. Find the probability that they are both giants.

Markschemeattempt to square their (M1)eg

A1 N2[2 marks]

P(G)

0.1912

0.0363996

P(both G) = 0.0364

22d. [3 marks] trees are selected at random.

Find the expected number of these trees that are giants.

100

Markschemecorrect substitution into formula for (A1)eg

A1 N2[3 marks]

E(X)

100(0.191)

E(G) = 19.1 [19.0, 19.1]

22e. [3 marks] trees are selected at random.

Find the probability that at least of these trees are giants.

Markschemerecognizing binomial probability (may be seen in part (c)(i)) (R1)eg

valid approach (seen anywhere) (M1)eg

correct working (A1)eg

A1 N2[3 marks]

100

25

X ∼ B(n, p)

P(X ⩾ 25) = 1 − P(X ⩽ 24), 1 − P(X < a)

P(X ⩽ 24) = 0.913… , 1 − 0.913…

0.0869002

P(X ⩾ 25) = 0.0869

[2 marks]23a.

The following is a cumulative frequency diagram for the time t, in minutes, taken by 80 students to complete a task.

Find the number of students who completed the task in less than 45 minutes.

Markschemeattempt to find number who took less than 45 minutes (M1)eg line on graph (vertical at approx 45, or horizontal at approx 70)

70 students (accept 69) A1 N2[2 marks]

[3 marks]23b. Find the number of students who took between 35 and 45 minutes to complete the task.

Markscheme55 students completed task in less than 35 minutes (A1)subtracting their values (M1)eg 70 – 55

15 students A1 N2[3 marks]

[2 marks]23c. Given that 50 students take less than k minutes to complete the task, find the value of .

Markschemecorrect approach (A1)eg line from y-axis on 50

A1 N2[2 marks]

k

k = 33

24a. [5 marks]

Jar A contains three red marbles and five green marbles. Two marbles are drawn from the jar, one after the other, withoutreplacement.

Find the probability that

(i) none of the marbles are green;

(ii) exactly one marble is green.

Markscheme(i) attempt to find (M1)

eg , ,

A1 N2

(ii) attempt to find (M1)

eg , ,

recognizing two ways to get one red, one green (M1)

eg , ,

A1 N2

[5 marks]

P(red) × P(red)

×38

27

×38

38

×38

28

P(none green) = 656

(= )328

P(red) × P(green)

×58

37

×38

58

1556

2P(R) × P(G) × + ×58

37

38

57

× × 238

58

P(exactly one green) = 3056

(= )1528

[3 marks]24b. Find the expected number of green marbles drawn from the jar.

Markscheme (seen anywhere) (A1)

correct substitution into formula for A1

eg ,

expected number of green marbles is A1 N2

[3 marks]

P(both green) = 2056

E(X)

0 × + 1 × + 2 ×656

3056

2056

+3064

5064

7056

(= )54

24c. [2 marks]

Jar B contains six red marbles and two green marbles. A fair six-sided die is tossed. If the score is or , a marble is drawn from jarA. Otherwise, a marble is drawn from jar B.

(i) Write down the probability that the marble is drawn from jar B.

(ii) Given that the marble was drawn from jar B, write down the probability that it is red.

1 2

Markscheme(i) A1 N1

(ii) A1 N1

[2 marks]

P(jar B) = 46

(= )23

P(red| jar B) = 68

(= )34

[6 marks]24d. Given that the marble is red, find the probability that it was drawn from jar A.

Markschemerecognizing conditional probability (M1)

eg , , tree diagram

attempt to multiply along either branch (may be seen on diagram) (M1)

eg

attempt to multiply along other branch (M1)

eg

adding the probabilities of two mutually exclusive paths (A1)

eg

correct substitution

eg , A1

A1 N3

[6 marks]

P(A|R) P(jar A and red)

P(red)

P(jar A and red) = ×13

38

(= )18

P(jar B and red) = ×23

68

(= )12

P(red) = × + ×13

38

23

68

P(jar A|red) =×1

338

× + ×13

38

23

68

18

58

P(jar A|red) = 15

25a. [8 marks]

A bag contains four gold balls and six silver balls.

Two balls are drawn at random from the bag, with replacement. Let be the number of gold balls drawn from the bag.

(i) Find .

(ii) Find .

(iii) Hence, find .

X

P(X = 0)

P(X = 1)

E(X)

MarkschemeMETHOD 1

(i) appropriate approach (M1)

eg , ,

A1 N2

(ii) multiplying one pair of gold and silver probabilities (M1)

eg , , 0.24

adding the product of both pairs of gold and silver probabilities (M1)

eg ,

A1 N3

(iii)

(seen anywhere) (A1)

correct substitution into formula for (A1)

eg ,

A1 N3

METHOD 2

(i) evidence of recognizing binomial (may be seen in part (ii)) (M1)

eg ,

correct probability for use in binomial (A1)

eg , ,

A1 N3

(ii) correct set up (A1)

eg

A1 N2

(iii)

attempt to substitute into (M1)

eg

correct substitution into (A1)

eg

A1 N3

[8 marks]

×610

610

×610

59

×610

510

P(X = 0) = = 0.36925

×610

410

×610

49

× × 2610

410

× + ×610

49

410

69

P(X = 1) = = 0.481225

P(X = 2) = 0.16

E(X)

0 × 0.36 + 1 × 0.48 + 2 × 0.16 0.48 + 0.32

E(X) = = 0.845

X ∼ B(2,0.6) ( ) (0.4 (0.620

)2 )0

p = 0.4 X ∼ B(2,0.4) (0.4 (0.62C0 )0 )2

P(X = 0) = = 0.36925

(0.4 (0.62C1 )1 )1

P(X = 1) = = 0.481225

np

2 × 0.6

np

2 × 0.4

E(X) = = 0.845

[3 marks]25b. Hence, find .E(X)

MarkschemeMETHOD 1

(seen anywhere) (A1)

correct substitution into formula for (A1)

eg ,

A1 N3

METHOD 2

attempt to substitute into (M1)

eg

correct substitution into (A1)

eg

A1 N3

[3 marks]

P(X = 2) = 0.16

E(X)

0 × 0.36 + 1 × 0.48 + 2 × 0.16 0.48 + 0.32

E(X) = = 0.845

np

2 × 0.6

np

2 × 0.4

E(X) = = 0.845

[2 marks]25c.

Fourteen balls are drawn from the bag, with replacement.

Find the probability that exactly five of the balls are gold.

MarkschemeLet be the number of gold balls drawn from the bag.

evidence of recognizing binomial (seen anywhere) (M1)

eg ,

A1 N2

[2 marks]

Y

(0.4 (0.614C5 )5 )9 B(14,0.4)

P(Y = 5) = 0.207

[2 marks]25d. Find the probability that at most five of the balls are gold.

Markschemerecognize need to find (M1)

A1 N2

[2 marks]

P(Y ≤ 5)

P(Y ≤ 5) = 0.486

25e. [3 marks]Given that at most five of the balls are gold, find the probability that exactly five of the balls are gold. Give the answer correctto two decimal places.

MarkschemeLet be the number of gold balls drawn from the bag.

recognizing conditional probability (M1)

eg , , ,

(A1)

(to dp) A1 N2

[3 marks]

Y

P(A|B) P(Y = 5|Y ≤ 5) P(Y=5)

P(Y≤5)0.2070.486

P(Y = 5|Y ≤ 5) = 0.42522518

P(Y = 5|Y ≤ 5) = 0.43 2

[4 marks]26a.

Samantha goes to school five days a week. When it rains, the probability that she goes to school by bus is 0.5. When it does not rain,

the probability that she goes to school by bus is 0.3. The probability that it rains on any given day is 0.2.

On a randomly selected school day, find the probability that Samantha goes to school by bus.

Markschemeappropriate approach (M1)eg , tree diagram,

one correct multiplication (A1)eg

correct working (A1)eg

A1 N3[4 marks]

P(R ∩ B) + P( ∩ B)R′

0.2 × 0.5, 0.24

0.2 × 0.5 + 0.8 × 0.3, 0.1 + 0.24

P(bus) = 0.34(exact)

[3 marks]26b. Given that Samantha went to school by bus on Monday, find the probability that it was raining.

Markschemerecognizing conditional probability (R1)eg

correct working A1eg

A1 N2

[3 marks]

P(A|B) = P(A∩B)

P(B)

0.2×0.50.34

P(R|B) = , 0.294517

[2 marks]26c. In a randomly chosen school week, find the probability that Samantha goes to school by bus on exactly three days.

Markschemerecognizing binomial probability (R1)

eg ,

A1 N2[2 marks]

X ∼ B(n, p) ( )53

(0.34 , (0.34 (1 − 0.34)3 )3 )2

P(X = 3) = 0.171

26d. [5 marks]After school days, the probability that Samantha goes to school by bus at least once is greater than . Find the smallest

value of .

n 0.95

n

Printed for Colegio Aleman de Barranquilla

© International Baccalaureate Organization 2015 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

MarkschemeMETHOD 1evidence of using complement (seen anywhere) (M1)eg

valid approach (M1)eg

correct inequality (accept equation) A1eg

(A1) A1 N3

METHOD 2valid approach using guess and check/trial and error (M1)eg finding for various values of n

seeing the “cross over” values for the probabilities A1A1

recognising (R1) A1 N3

[5 marks]

1 − P (none), 1 − 0.95

1 − P (none) > 0.95, P (none) < 0.05, 1 − P (none) = 0.95

1 − (0.66 > 0.95, (0.66 = 0.05)n )n

n > 7.209 (accept n = 7.209)

n = 8

P(X ⩾ 1)

n = 7, P(X ⩾ 1) = 0.9454, n = 8, P(X ⩾ 1) = 0.939

0.9639 > 0.95

n = 8