Algebra y funciones [219 marks]

[4 marks]1a.

Let and .

Express in the form , where .

Markschemeattempt to apply rules of logarithms (M1)

e.g. ,

correct application of (seen anywhere) A1

e.g.

correct application of (seen anywhere) A1

e.g.

so

(accept ) A1 N1

[4 marks]

f(x) = 3 lnx g(x) = ln5x3

g(x) f(x) + lna a ∈ Z+

ln = b lnaab lnab = lna + lnb

ln = b lnaab

3 lnx = lnx3

lnab = lna + lnb

ln5 = ln5 + lnx3 x3

ln5 = ln5 + 3 lnxx3

g(x) = f(x) + ln5 g(x) = 3 lnx + ln5

[3 marks]1b. The graph of g is a transformation of the graph of f . Give a full geometric description of this transformation.

Markschemetransformation with correct name, direction, and value A3

e.g. translation by , shift up by , vertical translation of

[3 marks]

( )0ln5

ln5 ln5

2a. [5 marks]

In the expansion of , the term in can be expressed as .

(a) Write down the value of , of and of .

(b) Find the coefficient of the term in .

(3x − 2)12 x5 ( ) × (3x × (−212r

)p )q

p q r

x5

Markscheme(a) , , (accept ) A1A1A1 N3

[3 marks]

(b) correct working (A1)

eg , , , ,

coefficient of term in is A1 N2

Note: Do not award the final A1 for an answer that contains .

[2 marks]

Total [5 marks]

p = 5 q = 7 r = 7 r = 5

( ) × (3x × (−2127

)5 )7 792 243 −27 24634368

x5 −24634368

x

[3 marks]2b. Write down the value of , of and of .

Markscheme , , (accept ) A1A1A1 N3

[3 marks]

p q r

p = 5 q = 7 r = 7 r = 5

[2 marks]2c. Find the coefficient of the term in .

Markschemecorrect working (A1)

eg , , , ,

coefficient of term in is A1 N2

Note: Do not award the final A1 for an answer that contains .

[2 marks]

Total [5 marks]

x5

( ) × (3x × (−2127

)5 )7 792 243 −27 24634368

x5 −24634368

x

3a. [7 marks]

Let and .

(a) Find .

(b) Find .

(c) Find .

p = 6log3 q = 7log3

log3p2

( )log3p

q

(9p)log3

Markscheme(a) METHOD 1

evidence of correct formula (M1)

eg ,

A1 N2

METHOD 2

valid method using (M1)

eg , ,

A1 N2

[2 marks]

(b) METHOD 1

evidence of correct formula (M1)

eg ,

A1 N2

METHOD 2

valid method using and (M1)

eg , ,

A1 N2

[2 marks]

(c) METHOD 1

evidence of correct formula (M1)

eg ,

(may be seen in expression) A1

eg

A1 N2

METHOD 2

valid method using (M1)

eg ,

correct working A1

eg ,

A1 N2

[3 marks]

Total [7 marks]

log = n loguun 2 plog3

( ) = 12log3 p2

p = 36

(log3 36)2 log312 12 3log3

( ) = 12log3 p2

log( ) = logp − logqp

q6 − 7

( ) = −1log3p

q

p = 36 q = 37

( )log336

37log3−1 − 3log3

( ) = −1log3p

q

uv = u + vlog3 log3 log3 log9 + logp

9 = 2log3

2 + logp

(9p) = 8log3

p = 36

(9 × )log3 36 ( × )log3 32 36

9 +log3 log336 log338

(9p) = 8log3

[2 marks]3b. Find .log3p2

MarkschemeMETHOD 1

evidence of correct formula (M1)

eg ,

A1 N2

METHOD 2

valid method using (M1)

eg , ,

A1 N2

[2 marks]

log = n loguun 2 plog3

( ) = 12log3 p2

p = 36

(log3 36)2 log312 12 3log3

( ) = 12log3 p2

[2 marks]3c. Find .

MarkschemeMETHOD 1

evidence of correct formula (M1)

eg ,

A1 N2

METHOD 2

valid method using and (M1)

eg , ,

A1 N2

[2 marks]

( )log3p

q

log( ) = logp − logqp

q6 − 7

( ) = −1log3p

q

p = 36 q = 37

( )log336

37log3−1 − 3log3

( ) = −1log3p

q

[3 marks]3d. Find .(9p)log3

MarkschemeMETHOD 1

evidence of correct formula (M1)

eg ,

(may be seen in expression) A1

eg

A1 N2

METHOD 2

valid method using (M1)

eg ,

correct working A1

eg ,

A1 N2

[3 marks]

Total [7 marks]

uv = u + vlog3 log3 log3 log9 + logp

9 = 2log3

2 + logp

(9p) = 8log3

p = 36

(9 × )log3 36 ( × )log3 32 36

9 +log3 log336 log338

(9p) = 8log3

[7 marks]4. The constant term in the expansion of , where is . Find .

Markschemeevidence of binomial expansion (M1)

eg selecting correct term,

evidence of identifying constant term in expansion for power (A1)

eg , 4 term

evidence of correct term (may be seen in equation) A2

eg ,

attempt to set up their equation (M1)

eg ,

correct equation in one variable (A1)

eg ,

A1 N4

[7 marks]

( + )x

a

a2

x

6a ∈ R 1280 a

+ ( ) + …( )x

a

6( )a2

x

0 61

( )x

a

5( )a2

x

1

6

r = 3 th

20 a6

a3( )6

3( )x

a

3( )a2

x

3

( ) = 128063

( )x

a

3( )a2

x

3= 1280a3

a

20 = 1280a3 = 64a3

a = 4

[1 mark]5a.

Write down the value of

(i) ;

Markscheme(i) A1 N1[1 mark]

27log3

27 = 3log3

[1 mark]5b. (ii) ;

Markscheme(ii) A1 N1

[1 mark]

log818

= −1log818

[1 mark]5c. (iii) .

Markscheme(iii) A1 N1

[1 mark]

4log16

4 =log1612

[3 marks]5d. 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]6a.

Consider the expansion of .

Write down the number of terms in this expansion.

Markscheme11 terms A1 N1[1 mark]

(x + 3)10

[4 marks]6b. Find the term containing .

Markschemeevidence of binomial expansion (M1)

eg , attempt to expand

evidence of choosing correct term (A1)

eg , ,

correct working (A1)

eg , ,

A1 N3[4 marks]

x3

( )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

8

7. [7 marks]Consider the expansion of . The constant term is .

Find .

Markschemevalid approach (M1)

eg ,

, Pascal’s triangle to line

attempt to find value of r which gives term in (M1)eg exponent in binomial must give

correct working (A1)eg

evidence of correct term (A1)

eg

equating their term and 16128 to solve for M1

eg

A1A1 N2 Note: If no working shown, award N0 for .

Total [7 marks]

x2(3 + )x2 k

x

816 128

k

( )8r

(3 )x2 8−r ( )k

x

r

+ ( ) ( ) + ( ) + …(3 )x2 8 81

(3 )x2 7 k

x

82

(3 )x2 6( )k

x

29th

x0

, =x−2 x2( )x2 8−r ( )k

x

r

x0

2(8 − r) − r = −2, 18 − 3r = 0, 2r + (−8 + r) = −2

( ), ( ) , r = 6, r = 282

86

(3 )x2 2( )k

x

2

k

( ) = 16128, =x2 86

(3 )x2 2( )k

x

6k6 16128

28(9)

k = ±2

k = 2

[2 marks]8a.

Let .

Find .

Markscheme A2 N2

Note: Award A1 if only 1 error. [2 marks]

f(x) = p + p + qxx3 x2

(x)f ′

(x) = 3p + 2px + qf ′ x2

[5 marks]8b. Given that , show that .

Markschemeevidence of discriminant (must be seen explicitly, not in quadratic formula) (M1)

eg

correct substitution into discriminant (may be seen in inequality) A1eg

then has two equal roots or no roots (R1)recognizing discriminant less or equal than zero R1eg

correct working that clearly leads to the required answer A1eg

AG N0[5 marks]

(x) ⩾ 0f ′ ⩽ 3pqp2

− 4acb2

(2p − 4 × 3p × q, 4 − 12pq)2 p2

(x) ⩾ 0f ′ f ′

Δ ⩽ 0, 4 − 12pq ⩽ 0p2

− 3pq ⩽ 0, 4 ⩽ 12pqp2 p2

⩽ 3pqp2

[3 marks]9a.

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]9b. Find the values of where the function is decreasing.

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

(x) < 0f ′

x = −3, (1, 1.4)

−4 < x < −3, 1 ⩽ x ⩽ 4; x < −3, x ⩾ 1

[3 marks]9c. 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]9d. The function can also be written in the form , where , and . Find the value of .f f(x) = asin( (x + c))π

4a ∈ R 0 ⩽ c ⩽ 2 c

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]

4

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]10a.

Let , where .

Write down the equations of the vertical and horizontal asymptotes of the graph of .

Markscheme (must be equations) A1A1 N2

[2 marks]

f(x) = 3x

x−qx ≠ q

f

x = q, y = 3

10b. [2 marks]The vertical and horizontal asymptotes to the graph of intersect at the point .

Find the value of .

Markschemerecognizing connection between point of intersection and asymptote (R1)eg

A1 N2[2 marks]

f Q(1,3)

q

x = 1q = 1

10c. [4 marks]The vertical and horizontal asymptotes to the graph of intersect at the point .

The point lies on the graph of . Show that .

f Q(1,3)

P(x, y) f PQ = +(x − 1) 2 ( )3x−1

2− −−−−−−−−−−−−−√

Markschemecorrect substitution into distance formula A1

eg

attempt to substitute (M1)

eg

correct simplification of (A1)

eg

correct expression clearly leading to the required answer A1

eg

AG N0

[4 marks]

√

+(x − 1) 2 (y − 3) 2− −−−−−−−−−−−−−−√

y = 3x

x−1

+(x − 1) 2 ( − 3)3x

x−1

2− −−−−−−−−−−−−−−−−√( − 3)3x

x−13x−3x(x−1)

x−1

, 3x−3x+3x−1

+(x − 1) 2 ( )3x−3x+3x−1

2− −−−−−−−−−−−−−−−−−√PQ = +(x − 1) 2 ( )3

x−1

2− −−−−−−−−−−−−−√

10d. [6 marks]The vertical and horizontal asymptotes to the graph of intersect at the point .

Hence find the coordinates of the points on the graph of that are closest to .

Markschemerecognizing that closest is when is a minimum (R1)eg sketch of ,

(seen anywhere) A1A1attempt to find y-coordinates (M1)eg

A1A1 N4[6 marks]

f Q(1,3)

f (1,3)

PQ

PQ (PQ (x) = 0)′

x = −0.73205 x = 2.73205

f(−0.73205)

(−0.73205,1.267949),(2.73205,4.73205)

(−0.732,1.27),(2.73,4.73)

[3 marks]11a.

Let . Part of the graph of is shown in the following diagram.

The graph crosses the -axis at the points and .

Find the -coordinate of and of .

f(x) = 5 − x2 f

x A B

x A B

Markschemerecognizing (M1)eg

A1A1 N3[3 marks]

f(x) = 0

f = 0, = 5x2

x = ±2.23606

x = ± (exact), x = ±2.245√

11b. [3 marks]The region enclosed by the graph of and the -axis is revolved about the -axis.

Find the volume of the solid formed.

Markschemeattempt to substitute either limits or the function into formulainvolving (M1)eg

volume A2 N3[3 marks]

f x 360∘ x

f2

π∫ dx, π ( − 10 + 25), 2π(5 − )x2 2 ∫ 2.24−2.24 x4 x2 ∫ 5√

0 f2

187.328= 187

[2 marks]12a.

Let and , for .

Find .

Markschemeinterchanging and (M1)eg

A1 N2

[2 marks]

f(x) = 3x − 2 g(x) = 53x

x ≠ 0

(x)f−1

x y

x = 3y − 2

(x) = (accept y = , )f−1 x+23

x+23

x+23

[2 marks]12b. Show that .

Markschemeattempt to form composite (in any order) (M1)

eg

correct substitution A1eg

AG N0

[2 marks]

(g ∘ )(x) =f−1 5x+2

g ( ) , x+23

+253x

3

5

3( )x+2

3

(g ∘ )(x) =f−1 5x+2

[2 marks]12c.

Let , for . The graph of h has a horizontal asymptote at .

Find the -intercept of the graph of .

h(x) = 5x+2

x ⩾ 0 y = 0

y h

Markschemevalid approach (M1)eg

A1 N2

[2 marks]

h(0), 50+2

y = (accept (0, 2.5))52

[3 marks]12d. Hence, sketch the graph of .

Markscheme

A1A2 N3

Notes: Award A1 for approximately correct shape (reciprocal, decreasing, concave up).

Only if this A1 is awarded, award A2 for all the following approximately correct features: y-intercept at , asymptotic to x-

axis, correct domain .

If only two of these features are correct, award A1.

[3 marks]

h

(0,2.5)

x ⩾ 0

[1 mark]12e. For the graph of , write down the -intercept;

Markscheme A1 N1

[1 mark]

h−1 x

x = (accept (2.5, 0))52

[1 mark]12f. For the graph of , write down the equation of the vertical asymptote.

Markscheme (must be an equation) A1 N1

[1 mark]

h−1

x = 0

[3 marks]12g. Given that , find the value of .(a) = 3h−1 a

MarkschemeMETHOD 1attempt to substitute into (seen anywhere) (M1)eg

correct equation (A1)eg

A1 N2[3 marks]METHOD 2attempt to find inverse (may be seen in (d)) (M1)eg

correct equation, (A1)

A1 N2[3 marks]

3 h

h(3), 53+2

a = , h(3) = a53+2

a = 1

x = , = − 2, + 25y+2

h−1 5x

5x

− 2 = 35x

a = 1

[3 marks]13a.

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]13b. Let be a function such that exists for all real numbers. Given that , find .

Markschemerecognizing (M1)

eg

correct working (A1)

eg ,

A1 N2

Note: Award A0 for multiple values, eg .

[3 marks]

g g−1 g(30) = 3 (f ∘ )(3)g−1

(3) = 30g−1

f(30)

(f ∘ )(3) =g−1 30 − 5− −−−−√ 25−−√

(f ∘ )(3) = 5g−1

±5

[2 marks]14a.

Consider .

Find the value of .

Markschemesubstitute into (M1)

eg ,

A1 N2

[2 marks]

f(x) = ln( + 1)x4

f(0)

0 f

ln(0 + 1) ln1

f(0) = 0

[5 marks]14b. Find the set of values of for which is increasing.

Markscheme (seen anywhere) A1A1

Note: Award A1 for and A1 for .

recognizing increasing where (seen anywhere) R1

eg , diagram of signs

attempt to solve (M1)

eg ,

increasing for (accept ) A1 N1

[5 marks]

x f

(x) = × 4f ′ 1+1x4

x3

1+1x4

4x3

f (x) > 0f ′

(x) > 0f ′

(x) > 0f ′

4 = 0x3 > 0x3

f x > 0 x ≥ 0

14c. [5 marks]

The second derivative is given by .

The equation has only three solutions, when , .

(i) Find .

(ii) Hence, show that there is no point of inflexion on the graph of at .

(x) =f ′′ 4 (3− )x2 x4

( +1)x4 2

(x) = 0f ′′ x = 0 ± 3√4 (±1.316…)

(1)f ′′

f x = 0

Markscheme(i) substituting into (A1)

eg ,

A1 N2

(ii) valid interpretation of point of inflexion (seen anywhere) R1

eg no change of sign in , no change in concavity,

increasing both sides of zero

attempt to find for (M1)

eg , , diagram of signs

correct working leading to positive value A1

eg , discussing signs of numerator and denominator

there is no point of inflexion at AG N0

[5 marks]

x = 1 f ′′

4(3−1)

(1+1) 24×2

4

(1) = 2f ′′

(x)f ′′

f ′

(x)f ′′ x < 0

(−1)f ′′ 4 (3− )(−1)2 (−1)4

( +1)(−1)4 2

(−1) = 2f ′′

x = 0

14d. [3 marks]There is a point of inflexion on the graph of at .

Sketch the graph of , for .

Markscheme

A1A1A1 N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

Only if this A1 is awarded, then award the following:

A1 for curve through ( , ) , A1 for increasing throughout.

Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

f x = 3√4 (x = 1.316…)

f x ≥ 0

0 0

−1

15a. [3 marks]

The velocity of a particle in ms is given by , for .

On the grid below, sketch the graph of .

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct shape crossing x-axis with .

Only if this A1 is awarded, award the following:

A1 for maximum in circle, A1 for endpoints in circle.

[3 marks]

−1 v = − 1esin t 0 ≤ t ≤ 5

v

3 < x < 3.5

[1 mark]15b. Find the total distance travelled by the particle in the first five seconds.

Markscheme (exact), A1 N1

[1 mark]

t = π 3.14

[4 marks]15c. Write down the positive -intercept.t

Markschemerecognizing distance is area under velocity curve (M1)

eg , shading on diagram, attempt to integrate

valid approach to find the total area (M1)

eg , , ,

correct working with integration and limits (accept or missing ) (A1)

eg , ,

distance (m) A1 N3

[4 marks]

s = ∫ v

area A + area B ∫ vdt − ∫ vdt vdt+ vdt∫ 3.140 ∫ 5

3.14 ∫ |v|

dx dt

vdt+ vdt∫ 3.140 ∫ 3.14

5 3.067… + 0.878… − 1∫ 50 ∣∣esin t ∣∣

= 3.95

16a. [6 marks]

Let and be functions such that .

(a) The graph of is mapped to the graph of under the following transformations:

vertical stretch by a factor of , followed by a translation .

Write down the value of

(i) ;

(ii) ;

(iii) .

(b) Let . The point A( , ) on the graph of is mapped to the point on the graph of . Find .

Markscheme(a) (i) A1 N1

(ii) A1 N1

(iii) A1 N1

[3 marks]

(b) recognizing one transformation (M1)

eg horizontal stretch by , reflection in -axis

is ( , ) A1A1 N3

[3 marks]

Total [6 marks]

f g g(x) = 2f(x + 1) + 5

f g

k ( )p

q

k

p

q

h(x) = −g(3x) 6 5 g A′ h A′

k = 2

p = −1

q = 5

13

x

A′ 2 −5

16b. [3 marks]The graph of is mapped to the graph of under the following transformations:

vertical stretch by a factor of , followed by a translation .

Write down the value of

(i) ;

(ii) ;

(iii) .

f g

k ( )p

q

k

p

q

Markscheme(i) A1 N1

(ii) A1 N1

(iii) A1 N1

[3 marks]

k = 2

p = −1

q = 5

[3 marks]16c. Let . The point A( , ) on the graph of is mapped to the point on the graph of . Find .

Markschemerecognizing one transformation (M1)

eg horizontal stretch by , reflection in -axis

is ( , ) A1A1 N3

[3 marks]

Total [6 marks]

h(x) = −g(3x) 6 5 g A′ h A′

13

x

A′ 2 −5

[1 mark]17a.

Let . Part of the graph of is shown below.

Write down .

Markscheme (exact), A1 N1

[1 mark]

f(x) = 100(1+50 )e−0.2x

f

f(0)

f(0) = 10051

1.96

[2 marks]17b. Solve .

Markschemesetting up equation (M1)

eg , sketch of graph with horizontal line at

A1 N2

[2 marks]

f(x) = 95

95 = 1001+50e−0.2x

y = 95

x = 34.3

[3 marks]17c. Find the range of .

Markschemeupper bound of is (A1)

lower bound of is (A1)

range is A1 N3

[3 marks]

f

y 100

y 0

0 < y < 100

[5 marks]17d. Show that .

MarkschemeMETHOD 1

setting function ready to apply the chain rule (M1)

eg

evidence of correct differentiation (must be substituted into chain rule) (A1)(A1)

eg ,

correct chain rule derivative A1

eg

correct working clearly leading to the required answer A1

eg

AG N0

METHOD 2

attempt to apply the quotient rule (accept reversed numerator terms) (M1)

eg ,

evidence of correct differentiation inside the quotient rule (A1)(A1)

eg ,

any correct expression for derivative ( may not be explicitly seen) (A1)

eg

correct working clearly leading to the required answer A1

eg ,

AG N0

[5 marks]

(x) =f ′ 1000e−0.2x

(1+50 )e−0.2x 2

100(1 + 50e−0.2x )−1

= −100(1 + 50u′ e−0.2x )−2 = (50 )(−0.2)v′ e−0.2x

(x) = −100(1 + 50 (50 )(−0.2)f ′ e−0.2x )−2 e−0.2x

(x) = 1000 (1 + 50f ′ e−0.2x e−0.2x )−2

(x) =f ′ 1000e−0.2x

(1+50 )e−0.2x 2

v −uu′ v′

v2

u −vv′ u′

v2

(x) =f ′ (1+50 )(0)−100(50 ×−0.2)e−0.2x e−0.2x

(1+50 )e−0.2x 2

100(−10) −0e−0.2x

(1+50 )e−0.2x 2

0

−100(50 ×−0.2)e−0.2x

(1+50 )e−0.2x 2

(x) =f ′ 0−100(−10)e−0.2x

(1+50 )e−0.2x 2

−100(−10)e−0.2x

(1+50 )e−0.2x 2

(x) =f ′ 1000e−0.2x

(1+50 )e−0.2x 2

[4 marks]17e. Find the maximum rate of change of .f

MarkschemeMETHOD 1

sketch of (A1)

eg

recognizing maximum on (M1)

eg dot on max of sketch

finding maximum on graph of A1

eg ( , ) ,

maximum rate of increase is A1 N2

METHOD 2

recognizing (M1)

finding any correct expression for (A1)

eg

finding A1

maximum rate of increase is A1 N2

[4 marks]

(x)f ′

(x)f ′

(x)f ′

19.6 5 x = 19.560…

5

(x) = 0f ′′

(x) = 0f ′′

(−200 )−(1000 )(2(1+50 )(−10 ))(1+50 )e−0.2x 2 e−0.2x e−0.2x e−0.2x e−0.2x

(1+50 )e−0.2x 4

x = 19.560…

5

[3 marks]18a.

Let , for .

Find .

Markscheme A1A1A1 N3

Note: Award A1 for each term.

[3 marks]

f(x) = sin x + − 2x12x2 0 ≤ x ≤ π

(x)f ′

(x) = cosx + x − 2f ′

[3 marks]18b.

Let be a quadratic function such that . The line is the axis of symmetry of the graph of .

Find .

g g(0) = 5 x = 2 g

g(4)

Markschemerecognizing gives the point ( , ) (R1)

recognize symmetry (M1)

eg vertex, sketch

A1 N3

[3 marks]

g(0) = 5 0 5

g(4) = 5

18c. [4 marks]

The function can be expressed in the form .

(i) Write down the value of .

(ii) Find the value of .

Markscheme(i) A1 N1

(ii) substituting into (not the vertex) (M1)

eg ,

working towards solution (A1)

eg ,

A1 N2

[4 marks]

g g(x) = a(x − h + 3)2

h

a

h = 2

g(x) = a(x − 2 + 3)2

5 = a(0 − 2 + 3)2 5 = a(4 − 2 + 3)2

5 = 4a + 3 4a = 2

a = 12

[6 marks]18d. Find the value of for which the tangent to the graph of is parallel to the tangent to the graph of .x f g

Markscheme

correct derivative of A1A1

eg ,

evidence of equating both derivatives (M1)

eg

correct equation (A1)

eg

working towards a solution (A1)

eg , combining like terms

A1 N0

Note: Do not award final A1 if additional values are given.

[6 marks]

g(x) = (x − 2 + 3 = − 2x + 512

)2 12x2

g

2 × (x − 2)12

x − 2

=f ′ g′

cosx + x − 2 = x − 2

cosx = 0

x = π

2

[2 marks]19a.

Let , for .

Show that .

MarkschemeMETHOD 1correct use of chain rule A1A1eg

Note: Award A1 for , A1 for .

AG N0

[2 marks]METHOD 2correct substitution into quotient rule, with derivatives seen A1

eg

correct working A1

eg

AG N0

[2 marks]

f(x) = (ln x) 2

2x > 0

(x) =f ′ ln x

x

× , 2 ln x

21x

2 ln x

2x

2 ln x

2x× 1

x

(x) =f ′ ln x

x

2×2 ln x× −0×1x

(ln x) 2

4

4 ln x× 1x

4

(x) =f ′ ln x

x

[3 marks]19b. There is a minimum on the graph of . Find the -coordinate of this minimum.

Markschemesetting derivative (M1)eg correct working (A1)eg

A1 N2[3 marks]

f x

= 0(x) = 0, = 0f ′ ln x

x

lnx = 0, x = e0

x = 1

[2 marks]19c.

Let . The following diagram shows parts of the graphs of and g.

The graph of has an x-intercept at .

Write down the value of .

Markschemeintercept when (M1)

A1 N2[2 marks]

g(x) = 1x

f ′

f ′ x = p

p

(x) = 0f ′

p = 1

19d. [3 marks]The graph of intersects the graph of when .

Find the value of .

Markschemeequating functions (M1)eg

correct working (A1)eg

A1 N2

[3 marks]

g f ′ x = q

q

= g, =f ′ ln x

x1x

lnx = 1

q = e (accept x = e)

19e. [5 marks]The graph of intersects the graph of when .

Let be the region enclosed by the graph of , the graph of and the line .

Show that the area of is .

g f ′ x = q

R f ′ g x = p

R 12

Markschemeevidence of integrating and subtracting functions (in any order, seen anywhere) (M1)eg

correct integration A2

substituting limits into their integrated function and subtracting (in any order) (M1)

eg

Note: Do not award M1 if the integrated function has only one term.

correct working A1eg

AG N0

Notes: Candidates may work with two separate integrals, and only combine them at the end. Award marks in line with the

markscheme.

[5 marks]

2

( − ) dx, ∫ − g∫ e

q1x

ln x

xf ′

lnx − (ln x) 2

2

(lne − ln1) − ( − )(ln e) 2

2(ln 1) 2

2

(1 − 0) − ( − 0) , 1 −12

12

area = 12

[3 marks]20a.

Let .

Find the -intercepts of the graph of .

Markschemevalid approach (M1)eg , sketch of parabola showing two -intercepts

A1A1 N3[3 marks]

f(x) = (x − 1)(x − 4)

x f

f(x) = 0 x

x = 1, x = 4 (accept (1, 0), (4, 0))

20b. [3 marks]The region enclosed by the graph of and the -axis is rotated about the -axis.

Find the volume of the solid formed.

Markschemeattempt to substitute either limits or the function into formula involving (M1)eg

A2 N3[3 marks]

f x 360∘ x

f2

dx, π∫∫ 41 (f(x)) 2 ((x − 1)(x − 4)) 2

volume = 8.1π (exact), 25.4

21a. [3 marks]

A particle moves along a straight line such that its velocity, , is given by , for .

On the grid below, sketch the graph of , for .

Markscheme

A1A2 N3

Notes: Award A1 for approximately correct domain .

The shape must be approximately correct, with maximum skewed left. Only if the shape is approximately correct, award A2 for all

the following approximately correct features, in circle of tolerance where drawn (accept seeing correct coordinates for the maximum,

even if point outside circle):

Maximum point, passes through origin, asymptotic to -axis (but must not touch the axis).

If only two of these features are correct, award A1.

[3 marks]

v ms−1 v(t) = 10te−1.7t t ⩾ 0

v 0 ⩽ t ⩽ 4

0 ⩽ t ⩽ 4

t

[2 marks]21b. Find the distance travelled by the particle in the first three seconds.

Markschemevalid approach (including and ) (M1)eg , area from to (may be shaded in diagram)

A1 N2[2 marks]

0 3

10t dt, f(x)∫ 30 e−1.7t ∫ 3

0 0 3

distance = 3.33 (m)

[3 marks]21c. Find the velocity of the particle when its acceleration is zero.

Markschemerecognizing acceleration is derivative of velocity (R1)eg , attempt to find , reference to maximum on the graph of valid approach to find when (may be seen on graph) (M1)eg

A1 N3 Note: Award R1M1A0 for if velocity is not identified as final answer

[3 marks]

a = dv

dt

dv

dtv

v a = 0= 0, 10 − 17t = 0, t = 0.588dv

dte−1.7t e−1.7t

velocity = 2.16 (m )s−1

(0.588,216)

[1 mark]22a.

Let . The following diagram shows part of the curve of f .

The curve crosses the x-axis at the point P.

Write down the x-coordinate of P.

Markscheme (accept ) A1 N1

[1 mark]

f(x) = − 2x − 4x3

x = 2 (2, 0)

[2 marks]22b. Write down the gradient of the curve at P.

Markschemeevidence of finding gradient of f at (M1)

e.g.

the gradient is 10 A1 N2

[2 marks]

x = 2

(2)f ′

[3 marks]22c. Find the equation of the normal to the curve at P, giving your equation in the form .y = ax + b

Printed for Colegio Aleman de Barranquilla

© International Baccalaureate Organization 2016 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markschemeevidence of negative reciprocal of gradient (M1)

e.g. ,

evidence of correct substitution into equation of a line (A1)

e.g. ,

(accept , ) A1 N2

[3 marks]

−1(x)f ′

− 110

y − 0 = (x − 2)−110

0 = −0.1(2) + b

y = − x +110

210

a = −0.1 b = 0.2