Calculus

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Pranit_Calculus

[387 marks]

[5 marks]

1a.

Let .

Show that and deduce that f is an increasing function.

Markscheme

EITHER

derivative of is M1A1 M1A1 AG

(for all ) so the function is increasing R1 OR

M1A1 A1 M1 AG

(for all ) so the function is increasing R1 [5 marks]

Examiners report

Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates.

f(x) =√− −_1−x−x , 0 < x < 1 (x) = (1 − x f′ 1₂x−12 )− 3 2 x 1−x (1−x)−x(−1) (1−x)2 (x) = f′ 1₂( x ) 1−x −1 2 1 (1−x)2 =1 (1 − x 2x −1 2 )− 3 2 (x) > 0 f′ 0 < x < 1 f(x) = x 1 2 (1−x)12 (x) = f′ ( )− (−1) (1−x)12 1 2x −1 2 1 2x 1 2 (1−x)−12 1−x =1 (1 − x + (1 − x 2x −1 2 )− 1 2 1 2x 1 2 )− 3 2 =1 (1 − x [1 − x + x] 2x −1 2 )− 3 2 =1 (1 − x 2x −1 2 )− 3 2 (x) > 0 f′ 0 < x < 1 [6 marks]

1b. Show that the curve has one point of inflexion, and find its coordinates.

Markscheme

M1A1

changes sign at hence there is a point of inflexion R1 A1

the coordinates are [6 marks] y= f(x) (x) = (1 − x f′ 1₂x−12 )− 3 2 ⇒f′′(x) = −1₄x−32(1 − x)− + (1 − x 3 2 3 4x −1 2 )− 5 2 = −1 (1 − x [1 − 4x] 4x −3 2 )− 5 2 (x) = 0 ⇒ x = f′′ 1 4 (x) f′′ x=1₄ x= ⇒ y =1 4 √13 ( , )1 4 1 3 √

(2)

Examiners report

1c. Use the substitution to show that . [11 marks]

Markscheme

M1A1 M1A1 A1 M1A1 A1 A1 M1A1 hence AG [11 marks]

Examiners report

x=sin2θ ∫ f(x)dx = arcsin√x−√−x−−−− x−2 + c

x=sin2θ⇒dx= 2 sin θ cosθ

dθ ∫ x dx = ∫ 2 sin θ cosθdθ 1−x − −− √ sin2θ 1−sin2θ −−−−− √ = ∫ 2sin2θdθ = ∫ 1 − cos2θdθ = θ − sin 2θ + c1 2 θ= arcsin x√

sin 2θ = sin θ cosθ = =

1 2 √x 1 − x − −−−− √ _√−x−−−− x−2 ∫ x dx = arcsin − + c 1−x − −− √ _√x _√−x−−−− x−2

2. Given that the graph of has exactly one point at which the gradient is zero, find the value of k . [5 marks]

Markscheme

M1A1

For use of discriminant or completing the square (M1) (A1)

Note: Accept trial and error, sketches of parabolas with vertex (2,0) or use of second derivative. A1

[5 marks]

Examiners report

Generally candidates answer this question well using a diversity of methods. Surprisingly, a small number of candidates were successful in answering this question using the discriminant of the quadratic and in many cases reverted to trial and error to obtain the correct answer. y=x3− 6x2+ kx − 4 = 3 − 12x + k dy dx x2 − 4ac = 0 b2 3(x − 2 + k − 12)2 144 − 12k = 0 k= 12 3a. [1 mark]

The function is defined for . Write down the coordinates of the minimum point on the graph of f .

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Markscheme

A1

[1 mark]

Examiners report

Candidates answered parts (a) and (b) of this question well and, although many were also successful in part (c), just a few candidates gave answers to the required level of accuracy. Part d) was rather challenging for many candidates. The most common errors among the candidates who attempted this question were the confusion between tangents and normals and incorrect final answers due to premature rounding.

(3.79,−5)

3b. The points and , lie on the graph of . [2 marks]

Find p and q .

Markscheme

A1A1 [2 marks]

Examiners report

Candidates answered parts (a) and (b) of this question well and, although many were also successful in part (c), just a few candidates gave answers to the required level of accuracy. Part d) was rather challenging for many candidates. The most common errors among the candidates who attempted this question were the confusion between tangents and normals and incorrect final answers due to premature rounding.

P(p, 3) Q(q, 3), q > p y= f(x)

p= 1.57 or , q = 6.00π₂

[4 marks]

3c. Find the coordinates of the point, on , where the gradient of the graph is 3.

Markscheme

(M1)(A1) (A1) A1 Coordinates are [4 marks]

Examiners report

Candidates answered parts (a) and (b) of this question well and, although many were also successful in part (c), just a few candidates gave answers to the required level of accuracy. Part d) was rather challenging for many candidates. The most common errors among the candidates who attempted this question were the confusion between tangents and normals and incorrect final answers due to premature rounding. y= f(x) (x) = 3 cosx − 4 sin x f′ 3 cosx − 4 sin x = 3 ⇒ x = 4.43... (y = −4) (4.43,−4)

(4)

3d.

Markscheme

(M1)

gradient at P is so gradient of normal at P is (A1) gradient at Q is 4 so gradient of normal at Q is (A1)

equation of normal at P is (M1)

equation of normal at Q is (M1)

Note: Award the previous two M1 even if the gradients are incorrect in where are coordinates of P and Q (or in with c determined using coordinates of P and Q.

intersect at A1A1

Note: Award N2 for 3.79 without other working. [7 marks]

Examiners report

Candidates answered parts (a) and (b) of this question well and, although many were also successful in part (c), just a few candidates gave answers to the required level of accuracy. Part d) was rather challenging for many candidates. The most common errors among the candidates who attempted this question were the confusion between tangents and normals and incorrect final answers due to premature rounding. = mnormal mtangent1 −4 1 4 −1 4 y− 3 = (x − 1.570...) (or y = 0.25x + 2.60...)1 4 y− 3 = (x − 5.999...) (or y = −0.25x +1₄ 4.499...__{ }_ _) y− b = m(x − a) (a,b) y= mx + c (3.79, 3.55) [4 marks]

4a. Using the definition of a derivative as , show that the derivative of .

Markscheme

let and using the result M1A1 A1 A1 AG [4 marks]

Examiners report

Even though the definition of the derivative was given in the question, solutions to (a) were often disappointing with algebraic errors fairly common, usually due to brackets being omitted or manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates fail to understand that they have to assume that the result is true for and then show that this leads to it being true for . Many candidates just write ‘Let ’ which is of course meaningless. The conclusion is often of the form ‘True for therefore true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for true for ’.

(x) = ( ) f′ lim h→0 f(x+h)−f(x) h is 1 2x+1 (2x+1)−2 2 f(x) = 1 2x+1 f′(x) =lim_h_→0( ) f(x+h)−f(x) h (x) = ( ) f′ lim h→0 − 1 2(x+h)+1 1 2x+1 h ⇒ (x) =f′ lim( ) h→0 [2x+1]−[2(x+h)+1] h[2(x+h)+1][2x+1] ⇒ (x) =f′ lim( ) h→0 −2 [2(x+h)+1][2x+1] ⇒ (x) =f′ −2 (2x+1)2 n= k n= k + 1 n= k n= 1, n = k and n = k + 1 n= k ⇒ n= k + 1 [9 marks]

4b. Prove by induction that the nth derivative of (2x + 1)−1 is (−1)n 2n! .

n

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let

we want to prove that

let M1

which is the same result as part (a) hence the result is true for R1

assume the result is true for M1 M1

(A1) A1 (A1) A1

hence if the result is true for , it is true for

since the result is true for , the result is proved by mathematical induction R1 Note: Only award final R1 if all the M marks have been gained.

[9 marks]

Examiners report

Even though the definition of the derivative was given in the question, solutions to (a) were often disappointing with algebraic errors fairly common, usually due to brackets being omitted or manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates fail to understand that they have to assume that the result is true for and then show that this leads to it being true for . Many candidates just write ‘Let ’ which is of course meaningless. The conclusion is often of the form ‘True for therefore true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for true for ’.

y=_2x+11 = (−1 y dn dxn )n n! 2n (2x+1)n+1 n= 1 ⇒dy_dx= (−1)1 211! (2x+1)1+1 ⇒dy_dx = −2 (2x+1)2 n= 1 n= k : dky= (−1 dxk ) k 2kk! (2x+1)k+1 = [ ] y dk+1 dxk+1 d dx (−1) k 2kk! (2x+1)k+1 ⇒_dxdk+1k+1y = [ k! ] d dx (−1) k₂k _{(2x + 1)}−k−1 ⇒dk+1y = (−1 k!(−k − 1)(2x + 1 × 2 dxk+1 )k2 k ₎−k−2 ⇒dk+1y = (−1 (k + 1)!(2x + 1 dxk+1 )k+12 k+1 ₎−k−2 ⇒_dxdk+1k+1y = (−1)k+1 (k+1)! 2k+1 (2x+1)k+2 n= k n= k + 1 n= 1 n= k n= k + 1 n= k n= 1, n = k and n = k + 1 n= k ⇒ n= k + 1

5a. Sketch the curve showing clearly the coordinates of the x-intercepts, any maximum points and any [4 marks] minimum points.

y= cos x , − 4⩽ x ⩽ 4

+1

x2 √

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Markscheme

A1A1A1A1

Note: Award A1 for correct shape. Do not penalise if too large a domain is used, A1 for correct x-intercepts,

A1 for correct coordinates of two minimum points, A1 for correct coordinates of maximum point.

Accept answers which correctly indicate the position of the intercepts, maximum point and minimum points. [4 marks]

Examiners report

Most candidates were able to make a meaningful start to this question, but many made errors along the way and hence only a relatively small number of candidates gained full marks for the question. Common errors included trying to use degrees, rather than radians, trying to use algebraic methods to find the gradient in part (b) and trying to find the equation of the tangent rather than the equation of the normal in part (c).

5b. Write down the gradient of the curve at x = 1 . [1 mark]

Markscheme

gradient at x = 1 is –0.786 A1 [1 mark]

Examiners report

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Markscheme

gradient of normal is (A1) when x = 1, y = 0.3820... (A1)

Equation of normal is y – 0.382 = 1.27(x – 1) A1 [3 marks]

Examiners report

(= 1.272...)

−1 −0.786

(⇒ y = 1.27x − 0.890)

6. _{A normal to the graph of} _{, for} _{, has equation} _{, where} _. [6 marks] Find the value of c.

Markscheme

(or equivalent) A1 (R1)

Attempting to solve (or equivalent) for x M1 A1

Substituting and to find c M1 A1 N1

[6 marks]

Examiners report

There was a disappointing response to this question from a fair number of candidates. The differentiation was generally correctly performed, but it was then often equated to rather than the correct numerical value. A few candidates either didn’t simplify arctan(1) to , or stated it to be 45 or . y= arctan(x − 1) x> 0 y= −2x + c x∈R (arctan(x − 1)) = d dx _1+(x−1)1 2 = −2 and so = mN mT 1₂ = 1 1+(x−1)2 1 2 x= 2 (as x > 0) x= 2 y=π₄ c= 4 +π 4 −2x + c π 4 π 2 [7 marks]

(8)

Markscheme

A1A1A1

Note: Award A1A1 for correct differentiation of . A1 for correct differentiation of and 18. At the point (1, 2),

(A1) Gradient of normal

A1 Equation of normal

M1

A1 [7 marks]

Examiners report

It was pleasing to see that a significant number of candidates understood that implicit differentiation was required and that they were able to make a reasonable attempt at this. A small number of candidates tried to make the equation explicit. This method will work, but most candidates who attempted this made either arithmetic or algebraic errors, which stopped them from gaining the correct answer. 5 + 10xyy2 dy_dx− 4x = 0 5xy2 −2x2 20 + 20dy_dx− 4 = 0 ⇒dy_dx = −4 5 =5 4 y− 2 = (x − 1)5₄ y= x − +5₄ 5₄ 8₄ y= x +5₄ 3₄ (4y = 5x + 3) [4 marks] 8a.

The curve C has equation . Find the coordinates of the points on C at which .

Markscheme

A1

at A1A1A1

Note: Award A2 for all three x-values correct with errors/omissions in y-values. [4 marks]

Examiners report

The whole of this question seemed to prove accessible to a high proportion of candidates.

(a) was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at 0. (b) was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c).

The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded.

y= (9 + 81₈ x2− )x4 = 0 dy dx = 2x − dy dx 12x3 x(2 −1 ) = 0 2x2 x= 0, ± 2 = 0 dy dx (0, ), (−2, ), (2, ) 9 8 25 8 25 8 [4 marks]

(9)

at x =1, gradient of tangent (A1)

Note: In the following, allow FT on incorrect gradient. equation of tangent is (A1) meets x-axis when y = 0 , (M1)

coordinates of T are A1 [4 marks]

Examiners report

=3₂ y− 2 = (x − 1) (y = x + )3 2 3 2 1 2 −2 = (x − 1)3 2 x= −1₃ (− ,0)1 3 [7 marks]

8c. The normal to C at the point P cuts the y-axis at the point N. Find the area of triangle PTN.

Markscheme

gradient of normal (A1)

equation of normal is (M1) at x = 0 , A1

Note: In the following, allow FT on incorrect coordinates of T and N. lengths of , A1A1

area of triangle M1 (or equivalent e.g. ) A1

[7 marks]

Examiners report

= −2 3 y− 2 = − (x − 1) (y = − x + )2 3 23 8 3 y=8₃ PN = 13 9 −− √ PT = 52 9 −− √ PTN = ×1 × 2 13 9 −− √ 52 9 −− √ =13 9 676 √ 18

9. _{Consider the functions f and g defined by} _and _, _. [7 marks] (a) Find the coordinates of P, the point of intersection of the graphs of f and g .

(b) Find the equation of the tangent to the graph of f at the point P.

f(x) = 21x g(x) = 4 − 2

1

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Markscheme

(a)

attempt to solve the equation M1 x = 1 A1

so P is (1, 2) , as A1 N1 (b) A1

attempt to substitute x-value found in part (a) into their M1 M1A1 N0

[7 marks]

Examiners report

Most candidates answered part (a) correctly although some candidates showed difficulty solving the equation using valid methods. Part (b) was less successful with many candidates failing to apply chain rule to obtain the derivative of the exponential function.

= 4 − 21x 2 1 x f(1) = 2 (x) = − ln2 f′ 1 x22 1 x (x) f′ (1) = −2 ln2 f′ y− 2 = −2 ln2(x − 1) (or equivalent)

10. _{A tangent to the graph of} _{passes through the origin.} [17 marks] (a) Sketch the graphs of and the tangent on the same set of axes, and hence find the equation of the tangent.

(b) Use your sketch to explain why for . (c) Show that for .

(d) Determine which is larger, or .

y= lnx y= lnx lnx⩽x e x> 0 ⩽ xe ex _x_{> 0} πe _eπ

(11)

(a)

A3

Note: Award A1 for each graph A1 for the point of tangency.

point on curve and line is (M1) (M1)A1 EITHER

gradient of line, m, through (0, 0) and is (M1)A1 M1A1 OR (M1)A1 passes through 0 if M1 A1 THEN A1 [11 marks]

(b) the graph of never goes above the graph of , hence R1 [1 mark]

(c) M1A1

exponentiate both sides of R1AG [3 marks]

(d) equality holds when R1 letting A1 N0 (a, lna) y= ln(x) = ⇒ = (when x = a) dy dx x1 dy dx a1 (a, lna) ln a a ⇒ln a = ⇒ lna = 1 ⇒ a = e ⇒ m = a 1 a 1 e y− lna = (x − a)1_a lna − 1 = 0 a= e ⇒ m =1 e ∴ y = x1 e lnx y= x1 e lnx⩽ x e lnx⩽ ⇒ elnx ⩽ x ⇒ ln ⩽ xx e xe lnxe⩽ x ⇒xe⩽ex x= e x= π ⇒πe<eπ

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[2 marks] Total [17 marks]

Examiners report

This was the least accessible question in the entire paper, with very few candidates achieving high marks. Sketches were generally done poorly, and candidates failed to label the point of intersection. A ‘dummy’ variable was seldom used in part (a), hence in most cases it was not possible to get more than 3 marks. There was a lot of good guesswork as to the coordinates of the point of intersection, but no reasoning showed. Many candidates started with the conclusion in part (c). In part (d) most candidates did not distinguish between the inequality and strict inequality.

11. A curve is defined by the equation . Find the equation of the tangent to the curve at the point where x =[7 marks] 1 and .

Markscheme

M1A1A1

Note: M1 for attempt at implicit differentiation. A1 for differentiating , A1 for differentiating the rest.

when (M1) A1 at A1 or A1 [7 marks]

Examiners report

The implicit differentiation was generally well done. Some candidates did not realise that they needed to substitute into the original equation to find . Others wasted a lot of time rearranging the derivative to make the subject, rather than simply putting in the particular values for and .

8y lnx − 2x2+ 4y2= 7 y> 0 8y × + 8 lnx − 4x + 8y1 = 0 x dy dx dy dx 8y lnx x= 1, 8y × 0 − 2 × 1 + 4y2= 7 = ⇒ y = (as y > 0) y2 9₄ 3₂ (1, )3 = − 2 dy dx 23 y− = − (x − 1)3 2 2 3 y= − x + 2 3 13 6 y dy_dx x y

12. _{The normal to the curve} _{, at the point (c,} _{), has a y-intercept} _. [7 marks] Determine the value of c.

(13)

EITHER differentiating implicitly: M1A1 at the point (c, ) M1 (A1) OR

reasonable attempt to make expression explicit (M1)

(A1) A1

Note: Do not penalize if y = 0 not stated.

gradient of tangent A1

Note: If candidate starts with with no justification, award (M0)(A0)A1A1. THEN

the equation of the normal is M1 (A1) A1

[7 marks]

Examiners report

This was the first question to cause the majority of candidates a problem and only the better candidates gained full marks. Weaker candidates made errors in the implicit differentiation and those who were able to do this often were unable to simplify the expression they gained for the gradient of the normal in terms of c; a significant number of candidates did not know how to simplify the logarithms appropriately. 1 ×e−y− x_e−y dy+ = 1 dx ey dydx lnc − c × + c = 1 1 c 1 c dy dx dy dx = (c ≠ 1) dy dx 1c xe−y+_ey= 1 + x x+e2y= (1 + x)_ey − (1 + x) + x = 0 e2y _ey ( − 1)( − x) = 0ey _ey = 1, = x ey _ey y= 0, y = lnx = dy dx x1 =1 c y= lnx y− lnc = −c(x − c) x= 0, y =c2+ 1 + 1 − lnc = c2 c2 lnc = 1 c= e

(14)

13a._{A particle P moves in a straight line with displacement relative to origin given by} [10 marks]

where t is the time in seconds and the displacement is measured in centimetres. (i) Write down the period of the function s.

(ii) Find expressions for the velocity, v, and the acceleration, a, of P. (iii) Determine all the solutions of the equation v = 0 for .

Markscheme

(i) the period is 2 A1

(ii) (M1)A1 (M1)A1 (iii) EITHER M1 (A1) A1 A1 A1 OR M1 A1A1 A1 A1 [10 marks]

Examiners report

In (a), only a few candidates gave the correct period but the expressions for velocity and acceleration were correctly obtained by most candidates. In (a)(iii), many candidates manipulated the equation v = 0 correctly to give the two possible values for but then failed to find all the possible values of t.

s= 2 sin(πt) + sin(2πt), t⩾ 0, 0⩽ t ⩽ 4 v=ds_dt= 2π cos(πt) + 2π cos(2πt) a=dv= −2 sin(πt) − 4 sin(2πt) dt π2 π2 v= 0 2π (cos(πt) + cos(2πt)) = 0 cos(πt) + 2cos2(πt) − 1 = 0 (2 cos(πt) − 1)(cos(πt) + 1) = 0 cos(πt) = or cos(πt) = −11 2 t= , t = 11₃ t= , t = , t = , t = 35 3 7 3 113 2 cos( )cos( ) = 0πt 2 3πt 2 cos( ) = 0 or cos( ) = 0πt 2 3πt 2 t= , 11 3 t= , , 3, 5₃ 7₃ 11₃ cos(πt)

13b._{Consider the function} [8 marks]

Use mathematical induction to prove that the derivative of f is given by , for all .

f(x) = A sin(ax) + B sin(bx), A, a, B, b, x ∈R.

(2n)th ₍_f(2n)_{(x) = (−1 (A}₎n _a2n_{sin(ax) + B}_b2n_sin(bx))

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M1 A1 true assume that is true M1 consider M1A1 A1 A1

true implies true, true so true R1

Note: Award the final R1 only if the previous three M marks have been awarded.

[8 marks]

Examiners report

Solutions to (b) were disappointing in general with few candidates giving a correct solution.

P(n) :f(2n)(x) = (−1 (A)n _a2nsin(ax) + B_b2nsin(bx))

P(1) :f′′(x) =(Aacos(ax) + Bb cos(bx)) ′ = −A sin(ax) − B sin(bx)a2 b2

= −1 (A sin(ax) + B sin(bx))a2 b2 ∴ P(1) P(k) :f(2k)(x) = (−1 (A)k _a2ksin(ax) + B_b2ksin(bx)) P(k + 1) (x) = (−1 (A cos(ax) + B cos(bx)) f(2k+1) )k _a2k+1 _b2k+1

(x) = (−1 (−A sin(ax) − B sin(bx))

f(2k+2) )k _a2k+2 _b2k+2

= (−1)k+1(A_a2k+2sin(ax) + B_b2k+2sin(bx))

P(k) P(k + 1) P(1) P(n) ∀n ∈Z+

14. _{Consider the curve} _{and the line} _. [23 marks]

(a) Let k = 0.

(i) Show that the curve and the line intersect once.

(ii) Find the angle between the tangent to the curve and the line at the point of intersection. (b) Let k =1. Show that the line is a tangent to the curve.

(c) (i) Find the values of k for which the curve and the line meet in two distinct points. (ii) Write down the coordinates of the points of intersection.

(iii) Write down an integral representing the area of the region A enclosed by the curve and the line. (iv) Hence, given that , show that .

y= xex _y_{= kx, k ∈}R

y= xex _y_{= kx}

(16)

Markscheme

(a) (i) A1 so, they intersect only once at (0, 0)

(ii) M1A1 A1 A1 [5 marks] (b) when M1 A1

which equals the gradient of the line R1 so, the line is tangent to the curve at origin AG

Note: Award full credit to candidates who note that the equation has a double root x = 0 so y = x is a tangent.

[3 marks]

(c) (i) M1 A1

A1

(ii) (0, 0) and A1A1

(iii) M1A1

Note: Do not penalize the omission of absolute value.

(iv) attempt at integration by parts to find M1 A1 as R1 A1 A1 M1A1 since R1 AG [15 marks] Total [23 marks] xex= 0 ⇒ x = 0 = + x = (1 + x) y′ ex _ex _ex (0) = 1 y′ θ= arctan1 = (θ =π ) 4 45 ∘ k= 1, y = x xex= x ⇒ x( − 1) = 0_ex ⇒ x = 0 (0) = 1 y′ y= x x( − 1) = 0ex xex= kx ⇒ x( − k) = 0_ex ⇒ x = 0 or x = lnk k> 0 and k ≠ 1 (lnk, klnk) A=∣∣∫₀ln kkx− x dxex ∣∣ ∫ x dxex ∫ x dx = x − ∫ dx = (x − 1)ex _ex _ex _ex 0 < k < 1 ⇒ lnk < 0 A=∫_{ln k}0 kx− x dx =ex [k _{− (x − 1)} ] 2x2 ex 0 ln k = 1 −(k − (lnk − 1)k) 2(lnk) 2 = 1 −k( − 2 lnk + 2) 2 (lnk) 2 = 1 −k( + 1) 2 (lnk − 1) 2 ( + 1) > 0 k 2 (lnk − 1) 2 A< 1

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Many candidates solved (a) and (b) correctly but in (c), many failed to realise that the equation has two roots under certain conditions and that the point of the question was to identify those conditions. Most candidates made a reasonable attempt to write down the appropriate integral in (c)(iii) with the modulus signs and limits often omitted but no correct solution has yet been seen to (c)(iv).

xex= kx

[7 marks]

15. _{Find the equation of the normal to the curve} _{at the point (1, 1).}

Markscheme

Note: Award A1 for correctly differentiating each term.

(M1)A1 gradient of normal = 1 (A1)

equation of the normal A1 N2

Note: Award A2R5 for correct answer and correct justification.

[7 marks]

Examiners report

This implicit differentiation question was well answered by most candidates with many achieving full marks. Some candidates made algebraic errors which prevented them from scoring well in this question.

Other candidates realised that the equation of the curve could be simplified although the simplification was seldom justified. − xy = 0 x3y3 − xy = 0 x3y3 3x2y3+ 3x3y2y′− y − x = 0y′ x= 1, y = 1 3 + 3 − 1 −y′ y′= 0 2 = −2y′ = −1 y′ y− 1 = x − 1 y= x y= m(x − m) (1 − x)y = 1

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Markscheme

EITHER

M1A1 solve simultaneously M1

A1 Note: Accept equivalent forms.

tangency point (1.66, –1.52) A1A1 A1 OR M1 A1 (M1) A1 substituting (M1) A1 A1 tangency point (1.66, –1.52) [7 marks]

Examiners report

Very few candidates answered this question well but among those a variety of nice approaches were seen. This question required some organized thinking and good understanding of the concepts involved and therefore just strong candidates were able to go beyond the first steps. Sadly a few good answers were spoiled due to early rounding.

y= 1 ⇒ = 1−x y′ 1 (1−x)2 = m(x − m) and = m 1 1−x (1−x)1 2 = (x − ) 1 1−x 1 (1−x)2 1 (1−x)2 (1 − x − x(1 − x + 1 = 0, x ≠ 1)3 ₎2 x= 1.65729… ⇒ y = 1 = −1.521379… 1−1.65729… m= (−1.52137… = 2.31)2 (1 − x)y = 1 m(1 − x)(x − m) = 1 m(x −x2− m + mx) = 1 mx2− x(m +_m2) + (_m2+ 1) = 0 − 4ac = 0 b2 (m +m2)2− 4m(m2+ 1) = 0 m= 2.31 m= 2.31… into mx2− x(m +m2) + (m2+ 1) = 0 x= 1.66 y= 1 = −1.52 1−1.65729 17a. [4 marks]

The curve C is given by , for .

Show that . y=_xx_{+cos x}cos x x⩾ 0 = , x⩾ 0 dy dx x− sin x cos2 _x2 (x+cos x)2

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M1A1A1

Note: Award M1 for attempt at differentiation of a quotient and a product condoning sign errors in the quotient formula and the trig differentiations, A1 for correct derivative of “u”, A1 for correct derivative of “v”.

A1 AG

[4 marks]

Examiners report

The majority of candidates earned significant marks on this question. The product rule and the quotient rule were usually correctly applied, but a few candidates made an error in differentiating the denominator, obtaining rather than . A

disappointing number of candidates failed to calculate the correct gradient at the specified point. =

dy dx

(x+cos x)(cos x−x sin x)−x cos x(1−sin x) (x+cos x)2

=xcos x+cos2x−_x2sin x−x cos x sin x−x cos x+x cos x sin x (x+cos x)2

=cos2x−_x2sin x (x+cos x)2

−sin x 1 − sin x

[3 marks]

17b._{Find the equation of the tangent to C at the point} _.

Markscheme

the derivative has value –1 (A1)

the equation of the tangent line is M1A1 [3 marks]

Examiners report

The majority of candidates earned significant marks on this question. The product rule and the quotient rule were usually correctly applied, but a few candidates made an error in differentiating the denominator, obtaining rather than . A

disappointing number of candidates failed to calculate the correct gradient at the specified point. ( ,0)π 2 (y − 0) = (−1)(x − )(y = − x)π 2 π 2 −sin x 1 − sin x

18. _{Given that}_y₌ 1 _{, use mathematical induction to prove that} _. [7 marks]

1−x = , n ∈ y dn dxn n! (1−x)n+1 Z+

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Markscheme

proposition is true for n = 1 since M1 A1

Note: Must see the 1! for the A1.

assume true for n = k, , i.e. M1 consider (M1)

A1 A1

hence, is true whenever is true, and is true, and therefore the proposition is true for all positive integers R1 Note: The final R1 is only available if at least 4 of the previous marks have been awarded.

[7 marks]

Examiners report

Most candidates were awarded good marks for this question. A disappointing minority thought that the th derivative was the th derivative multiplied by the first derivative. Providing an acceptable final statement remains a perennial issue.

= dy dx _(1−x)1 2 = 1! (1−x)2 k∈Z+ d_dxkyk= k! (1−x)k+1 = y dk+1 dxk+1 d(dky) dxk dx = (k + 1)k!(1 − x)−(k+1)−1 = (k+1)! (1−x)k+2 Pk+1 Pk P1 (k + 1) (k) [3 marks] 19a.

The curve C with equation satisfies the differential equation

and y = e when x = 2.

Find the equation of the tangent to C at the point (2, e).

Markscheme

A1

at (2, e) the tangent line is M1 hence A1

[3 marks]

Examiners report

Nearly always correctly answered.

y= f(x) = (x + 2), y > 1, dy dx y lny = (2 + 2) = 4e dy dx e ln e y− e = 4e(x − 2) y= 4ex − 7e 19b._Find_f_(x)_. [11 marks]

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M1

using substitution (M1)(A1) (A1) A1A1 at (2, e), M1 A1 M1A1 since y > 1, R1

Note:M1 for attempt to make y the subject.

[11 marks]

Examiners report

Most candidates separated the variables and attempted the integrals. Very few candidates made use of the condition y > 1, so losing 2 marks. = (x + 2) ⇒ dy = (x + 2)dx dy dx y ln y ln y y ∫ ln yy dy = ∫ (x + 2)dx u= lny; du = dy1_y ⇒ ∫ln yy dy = ∫ udu = 1 2u2 ⇒(ln y)₂ 2 =x2+ 2x + c 2 = 6 + c (ln e)2 2 ⇒ c = −11 2 ⇒(ln y)₂ 2 =x2+ 2x − ⇒ (lny = + 4x − 11 2 112 )2 x2 lny = ±_√x−−−−−−−−−2+ 4x − 11−⇒ y =_e±_√x2+4x−11 f(x) = e_√x2+4x−11

19c._{Determine the largest possible domain of f.} [6 marks]

Markscheme

EITHER

A1

using the quadratic formula M1

critical values are A1 using a sign diagram or algebraic solution M1

A1A1 OR

A1

by methods of completing the square M1 A1

(M1) A1A1 [6 marks]

Examiners report

Part (c) was often well answered, sometimes with follow through. + 4x − 11 > 0 x2 (= −2 ± ) −4± 60√ 2 15 −− √ x< −2 −√−−15; x > −2 +√−−15 + 4x − 11 > 0 x2 (x + 2 > 15)2 ⇒ x + 2 < −√−−15 or x + 2 >√−−15 x< −2 −√−−15; x > −2 +√−−15

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19d._{Show that the equation} _{has no solution.} [4 marks]

Markscheme

M1 A1 A1 R1AG [4 marks]

Examiners report

Only the best candidates were successful on part (d).

f(x) = (x)f′ f(x) = (x) ⇒ f(x) =f′ _{ln f(x)}f(x) (x + 2) ⇒ ln(f(x)) = x + 2 (⇒ x + 2 =_√−x−−−−−−−−2+ 4x − 11−) ⇒ (x + 2 =)2 _x2+ 4x − 11 ⇒_x2+ 4x + 4 =_x2+ 4x − 11 ⇒ 4 = −11, hence f(x) ≠ (x)f′ 20. [19 marks]

Let be a function defined by , . The diagram below shows a region bound by the graph of and the line .

A and C are the points of intersection of the line and the graph of , and B is the minimum point of . (a) If A, B and C have x-coordinates , and , where , , ,find the values of , and . (b) Find the range of .

(c) Find the equation of the normal to the graph of f at the point C, giving your answer in the form . (d) The region is rotated through about the x-axis to generate a solid.

(i) Write down an integral that represents the volume of this solid. (ii) Show that .

f f(x) = x + 2 cosx x ∈ [0, 2π] S f y= x y= x f f aπ₂ bπ₆ cπ₂ a b c∈N a b c f y= px + q S 2π V V = 6π2

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(a) METHOD 1 using GDC , , A1A2A1 METHOD 2 , ... M1 , A1 M1 or A1

Note: Final M1A1 is independent of previous work. [4 marks]

(b) (or ) (M1) (or ) (M1)

the range is (or [ , ]) A1 [3 marks]

(c) (M1) A1

gradient of normal (M1)

equation of the normal is (M1) (or equivalent decimal values) A1 N4 [5 marks]

(d) (i) (or equivalent) A1A1 Note: Award A1 for limits and A1 for and integrand.

(ii)

using integration by parts M1

and the identity , M1

A1A1 Note: Award A1 for and A1 for sin .

A1 AG N0

Note: Do not accept numerical answers. [7 marks] a= 1 b = 5 c = 3 x= x + 2 cosx ⇒ cosx = 0 ⇒ x =π 2 3π 2 a= 1 c = 3 1 − 2 sin x = 0 ⇒ sin x = ⇒ x =1 2 π 6 5π 6 b= 5 f( ) =5π₆ 5π₆ −√3 0.886 f(2π) = 2π + 2 8.28 [5π₆ −√3, 2π + 2] 0.886 8.28 (x) = 1 − 2 sin x f′ ( ) = 3 f′ 3π₆ = −1 3 y−3π₂ = − (x − )1₃ 3π₂ y= − x + 2π1 3 V = π∫3π2 ( − )dx π 2 x2 (x + 2 cosx)2 π V = π∫3π2 ( − )dx π 2 x2 _{(x + 2 cosx)}2 = −π∫3π2 (4x cosx + 4 x)dx π 2 cos2 4cos2x= 2 cos2x + 2

V= −π[(4x sin x + 4 cosx) + (sin 2x + 2x)]

3π 2

π

2

4x sin x + 4 cosx 2x + 2x

= −π [(6πsin3π+ 4 cos + sin 3π + 3π) − (2πsin + 4 cos + sin π + π)]

2 3π 2 π 2 π 2 = −π (−6π + 3π − π) = 6π2

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Total [19 marks]

Examiners report

Generally there were many good attempts to this, more difficult, question. A number of students found to be equal to 1, rather than 5. In the final part few students could successfully work through the entire integral successfully.

b

21a.Consider the functions and . [5 marks]

(i) Find . (ii) Find .

(iii) Hence, show that is increasing on .

Markscheme

(i) attempt at chain rule (M1) A1

(ii) attempt at chain rule (M1) A1 (iii) is positive on A1 so is increasing on AG [5 marks]

Examiners report

[N/A] f(x) = (lnx , x > 1)2 g(x) = ln(f(x)), x > 1 (x) f′ (x) g′ g(x) ]1, ∞[ (x) = f′ 2 ln x_x (x) = g′ _x_{ln x}2 (x) g′ ]1, ∞[ g(x) ]1, ∞[

21b.Consider the differential equation [12 marks]

(i) Find the general solution of the differential equation in the form .

(ii) Show that the particular solution passing through the point with coordinates is given by .

(iii) Sketch the graph of your solution for , clearly indicating any asymptotes and any maximum or minimum points. (lnx)dy+ y = , x > 1. dx 2 x 2x − 1 (lnx) y= h(x) (e, )e2 _y₌x2−x+e (ln x)2 x> 1

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(i) rearrange in standard form: (A1) integrating factor:

(M1) (A1)

multiply by integrating factor (M1)

M1 attempt to integrate: M1

A1

(ii) attempt to use the point to determine c: M1

eg, or or

A1 AG (iii)

graph with correct shape A1

minimum at (accept answers to a minimum of 2 s.f) A1 asymptote shown at A1

Note: y-coordinate of minimum not required for A1;

Equation of asymptote not required for A1 if VA appears on the sketch. Award A0 for asymptotes if more than one asymptote are shown [12 marks]

Examiners report

[N/A] + y= , x > 1 dy dx 2 xln x 2x−1 (ln x)2 e∫ 2 dx xln x = eln((ln x)2) = (lnx)2 (lnx)2 dy + y= 2x − 1 dx 2 ln x x (y ) = 2x − 1 (or y = ∫ 2x − 1dx) d dx (lnx) 2 _(lnx)2 (lnx y =)2 _x2− x + c y=x2−x+c (ln x)2 (e, )e2 (lne)2_e2=_e2− e + c _e2₌e2−e+c (ln e)2 e2=e2− e + c c = e y=x2−x+e (ln x)2 x= 3.1 x= 1

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[2 marks]

22a.

Consider the function .

The sketch below shows the graph of and its tangent at a point A.

Show that .

Markscheme

M1A1 AG [2 marks]

Examiners report

[N/A] f(x) =ln x_x , x > 0 y= f(x) (x) = f′ 1−ln x_x₂ (x) = f′ x× −ln x1x x2 =1−ln x x2 [3 marks]

22b.Find the coordinates of B, at which the curve reaches its maximum value.

Markscheme

has solution M1A1 A1

hence maximum at the point [3 marks]

Examiners report

[N/A] = 0 1−ln x x2 x= e y=1 e (e, )1 e [5 marks]

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M1A1

Note: The M1A1 should be awarded if the correct working appears in part (b). point of inflexion where M1

so A1A1 C has coordinates [5 marks]

Examiners report

[N/A] (x) = f′′ x(− )−2x(1−ln x) 2 1 x x4 =2 ln x−3 x3 (x) = 0 f′′ x=e32, y = 3 2e −3 2 ( , e32 3 ) 2e −3 2

22d.The graph of crosses the -axis at the point A. [4 marks]

Find the equation of the tangent to the graph of at the point A.

Markscheme

A1 (A1) (M1) through (1, 0) equation is A1 [4 marks]

Examiners report

[N/A] y= f(x) x f f(1) = 0 (1) = 1 f′ y= x + c y= x − 1

22e.The graph of crosses the -axis at the point A. [7 marks]

Find the area enclosed by the curve , the tangent at A, and the line .

y= f(x) x

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Markscheme

METHOD 1

area M1A1A1

Note: Award M1 for integration of difference between line and curve, A1 for correct limits, A1 for correct expressions in either order.

(M1)A1 A1

A1 METHOD 2

area = area of triangle M1A1

Note: A1 is for correct integral with limits and is dependent on the M1. (M1)A1

area of triangle M1A1 A1 [7 marks]

Examiners report

[N/A] =∫₁ex− 1 −ln xdx x ∫ ln xdx = (+c) x (ln x)2 2 ∫ (x − 1)dx =x2− x(+c) 2 =[1 − x − ] 2x2 1 2(lnx) 2 e 1 = (1 − e − ) − ( − 1) 2e2 12 12 =1 − e 2e2 −∫e dx 1 ln xx ∫ ln xdx = (+c) x (ln x)2 2 = (e − 1)(e − 1)1 2 (e − 1)(e − 1) − ( ) = − e 1 2 1 2 1 2e2 [4 marks] 23a. Let .

Find the equations of the horizontal and vertical asymptotes of the curve .

Markscheme

so is an asymptote (M1)A1 so is an asymptote (M1)A1 [4 marks]

Examiners report

[N/A] f(x) =e_e2xx₋₂+1 y= f(x) x→ −∞ ⇒ y → −1₂ y= −1₂ − 2 = 0 ⇒ x = ln2 ex _x_{= ln2 (= 0.693)}

23b.(i) Find . [8 marks]

(ii) Show that the curve has exactly one point where its tangent is horizontal. (iii) Find the coordinates of this point.

(x)

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(i) M1A1

(ii) when M1

A1A1

Note: Award A1 for zero, A1 for other two solutions.

Accept any answers which show a zero, a negative and a positive. as exactly one solution R1

Note: Do not award marks for purely graphical solution. (iii) (1.44, 8.47) A1A1 [8 marks]

Examiners report

[N/A] (x) = f′ 2( −2)e −( +1) x _e2x _e2x _ex ( −2)ex 2 =e3x−4_e2x−_ex ( −2)ex 2 (x) = 0 f′ e3x− 4_e2x−_ex= 0 ( − 4 − 1) = 0 ex _e2x _ex = 0, = −0.236, = 4.24 (or = 2 ± ) ex _ex _ex _ex _√₅ > 0 ex [4 marks]

23c.Find the equation of , the normal to the curve at the point where it crosses the y-axis.

Markscheme

(A1) so gradient of normal is (M1) (A1) so equation of is A1 [4 marks]

Examiners report

[N/A] L1 (0) = −4 f′ 1 4 f(0) = −2 L1 y= x − 21₄ [5 marks] 23d.

The line is parallel to and tangent to the curve . Find the equation of the line .

Markscheme

M1 so (M1)A1 (A1) equation of is A1 (or ) [5 marks]

Examiners report

[N/A] L2 L1 y= f(x) L2 (x) = f′ 1₄ x= 1.46 f(1.46) = 8.47 L2 y− 8.47 = (x − 1.46)1₄ y= x + 8.111₄

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[2 marks]

24a.

The function f is defined by

Determine whether or not is continuous.

Markscheme

and A1

both answers are the same, hence f is continuous (at ) R1

Note: R1 may be awarded for justification using a graph or referring to limits. Do not award A0R1. [2 marks]

Examiners report

[N/A] f(x) ={ 3_{(x − 2 − 3,}1 − 2x, 4 )2 x≤ 2 x> 2 f 1 − 2(2) = −3 3(2 − 2 − 3 = −3 4 )2 x= 2

24b.The graph of the function is obtained by applying the following transformations to the graph of : [4 marks] a reflection in the –axis followed by a translation by the vector .

Find .

Markscheme

reflection in the y-axis

(M1)

Note: Award M1 for evidence of reflecting a graph in y-axis. translation

(M1)A1A1

Note: Award (M1) for attempting to substitute for

x

, or translating a graph along positive x-axis. Award A1 for the correct domains (this mark can be awarded independent of the M1).

Award A1 for the correct expressions. [4 marks]

Examiners report

[N/A] g f y ( )2 0 g(x) f(−x) ={ 1 + 2x, (x + 2 − 3, 3 4 )2 x≥ −2 x< −2 ( )2 0 g(x) ={ 2x − 3, − 3, 3 4x2 x≥ 0 x< 0 (x − 2) 25a. = 4x ( + )x2 y2 2 y2 dy dx

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Markscheme

METHOD 1

expanding the brackets first: M1

M1A1A1 Note: Award M1 for an attempt at implicit differentiation. Award A1 for each side correct.

or equivalent A1 METHOD 2

M1A1A1 Note: Award M1 for an attempt at implicit differentiation. Award A1 for each side correct.

M1 or equivalent A1 [5 marks]

Examiners report

[N/A] + 2 + = 4x x4 x2y2 y4 y2 4x3+ 4x + 4 yy2 x2 dy_dx+ 4y3 dy_dx= 4 + 8xyy2 dy_dx = dy dx − −x +x3 y2 y2 xy2−2xy+_y3 2 ( + )(2x + 2y ) = 4 + 8xyx2 y2 dy_dx y2 dy_dx ( + )(x + y ) = + 2xyx2 y2 dy_dx y2 dy_dx + y + x + = + 2xy x3 x2 dy_dx y2 y3 dy_dx y2 dy_dx = dy dx − −x +x3 y2 y2 yx2−2xy+_y3 [3 marks]

25b.Find the equation of the normal to the curve at the point (1, 1).

Markscheme

METHOD 1 at (1, 1), is undefined M1A1 A1 METHOD 2 gradient of normal M1 at (1, 1) gradient A1 A1 [3 marks]

Examiners report

[N/A] dy dx y= 1 = − 1 = − dy dx (y −2xy+ )x2 _y3 (− −x + )x3 _y2 _y2 = 0 y= 1 26a. [14 marks] A function is defined by . (i) Explain why the inverse function does not exist.

(ii) Show that the equation of the normal to the curve at the point P where is given by . (iii) Find the x-coordinates of the points Q and R on the curve such that the tangents at Q and R pass through .

f f(x) = ( +1 ), x ∈R

2 ex e−x

f−1

x= ln3 9x + 12y − 9 ln3 − 20 = 0

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Markscheme

(i) either counterexample or sketch or

recognising that intersects the graph of twice M1 function is not (does not obey horizontal line test) R1

so does not exist AG (ii) (A1) (A1) M1 A1 EITHER M1 A1 OR M1 A1 THEN AG

(iii) The tangent at has equation . (M1) (or equivalent) (A1)

(or equivalent) A1 attempting to solve for a (M1)

A1A1 [14 marks]

Examiners report

In part (a) (i), successful candidates typically sketched the graph of , applied the horizontal line test to the graph and concluded that the function was not (it did not obey the horizontal line test).

In part (a) (ii), a large number of candidates were able to show that the equation of the normal at point P was

. A few candidates used the gradient of the tangent rather than using it to find the gradient of the normal. Part (a) (iii) challenged most candidates. Most successful candidates graphed and on the same set of axes and found the x-coordinates of the intersection points.

y= k (k > 1) y= f(x) 1 − 1 f−1 (x) = ( − ) f′ 1₂ ex _e−x (ln3) = (= 1.33) f′ 4₃ m= −3 4 f(ln3) = (= 1.67)5 3 = − y−5 3 x−ln 3 3 4 4y −20 = −3x + 3 ln3 3 = − ln3 + c 5 3 3 4 c= + ln35 3 3 4 y= − x + + ln33 4 5 3 3 4 12y = −9x + 20 + 9 ln3 9x + 12y − 9 ln3 − 20 = 0

(a, f(a)) y− f(a) = (a)(x − a)f′

(a) = f′ f(a)_a − = ea _e−a ea+_e−a a a= ±1.20 y= f(x) 1 − 1 9x + 12y − 9 ln3 − 20 = 0 y= f(x) y= x (x)f′

26b.The domain of is now restricted to . [8 marks]

(i) Find an expression for .

(ii) Find the volume generated when the region bounded by the curve and the lines and is rotated through an angle of radians about the y-axis.

f x⩾ 0

(x)

f−1

y= f(x) x= 0 y= 5

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(i)

M1A1

Note: Award M1 for either attempting to rearrange or interchanging x and y. A1

A1 A1

Note: Award A1 for correct notation and for stating the positive “branch”.

(ii) (M1)(A1)

Note: Award M1 for attempting to use . A1

[8 marks]

Examiners report

Part (b) (i) challenged most candidates. While a large number of candidates seemed to understand how to find an inverse function, poor algebra skills (e.g. erroneously taking the natural logarithm of both sides) meant that very few candidates were able to form a quadratic in either or . 2y =ex+_e−x − 2y + 1 = 0 e2x _ex = ex 2y± 4 −4√y2 2 = y ± ex _√−_y−−−2_{− 1}− x= ln(y ±√−y−−−2− 1−) (x) = ln(x + ) f−1 _√−x−−−2− 1− V = π∫₁5(ln(y +_√y−−−−2− 1−))2dy V = π∫d dy c x2 = 37.1 (unit )s3 ex _ey

27. A family of cubic functions is defined as . [13 marks]

(a) Express in terms of k

(i) ;

(ii) the coordinates of the points of inflexion on the graphs of . (b) Show that all lie on a straight line and state its equation.

(c) Show that for all values of k, the tangents to the graphs of at are parallel, and find the equation of the tangent lines. (x) = − k + x, k ∈ fk k2x3 x2 Z+ (x) and (x) f′_k f′′_k Pk fk Pk fk Pk

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Markscheme

(a) (i) A1

A1 (ii)

Setting

M1

A1

M1

A1 Hence, [6 marks]

(b)

Equation of the straight line is

A1

As this equation is independent of k, all lie on this straight line

R1 [2 marks]

(c)

Gradient of tangent at :

M1A1 As the gradient is independent of k, the tangents are parallel.

R1

(A1) The equation is

A1 [5 marks]

Total [13 marks]

Examiners report

Many candidates scored the full 6 marks for part (a). The main mistake evidenced was to treat k as a variable, and hence use the product rule to differentiate. Of the many candidates who attempted parts (b) and (c), few scored the R1 marks in either part, but did manage to get the equations of the straight lines.

(x) = 3 − 2kx + 1 f′k k2x2 (x) = 6 x − 2k f′′k k2 (x) = 0 f′′ ⇒ 6 x − 2k = 0 ⇒ x =k2 _3k1 f( ) =_3k1 k2( )_3k1 3− k( )_3k1 2+( )_3k1 = 7 27k is ( , ) Pk _3k1 _27k7 y= x7 9 Pk Pk ( ) = ( ) = 3 − 2k( ) + 1 = f′ Pk f′ _3k1 k2( )_3k1 2 1 3k 23 = × + c ⇒ c = 7 27k 23 3k1 27k1 y= x +2 3 1 27k [3 marks] 28a.

The function f is defined, for , by . Determine whether f is even, odd or neither even nor odd.

Markscheme

M1 A1 therefore f is even A1 [3 marks]

Examiners report

[N/A] −π ⩽ x ⩽ 2 π 2 f(x) = 2 cosx + x sin x f(−x) = 2 cos(−x) + (−x)sin(−x) = 2 cosx + x sin x (= f(x))

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A1 A1 so AG [2 marks]

Examiners report

[N/A]

(x) = −2 sin x + sin x + x cosx (= −sin x + x cosx)

f′

(x) = −cosx + cosx − x sin x (= −x sin x)

f′′

(0) = 0

f′′

28c._{John states that, because} _{, the graph of f has a point of inflexion at the point (0, 2) . Explain briefly whether John’s} [2 marks] statement is correct or not.

Markscheme

John’s statement is incorrect because

either; there is a stationary point at (0, 2) and since f is an even function and therefore symmetrical about the y-axis it must be a maximum or a minimum

or; is even and therefore has the same sign either side of (0, 2) R2 [2 marks]

Examiners report

[N/A] (0) = 0 f′′ (x) f′′ [6 marks] 29a.

The function f is defined by

where a , .

Given that f and its derivative, , are continuous for all values in the domain of f , find the values of a and b .

f(x) ={ 2x − 1, ax2+ bx − 5, x⩽ 2 2 < x < 3 b∈R f′

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Markscheme

f continuous M1 A1 A1 A1 solve simultaneously M1 to obtain a = –1 and b = 6 A1 [6 marks]

Examiners report

[N/A] ⇒ limf(x) = f(x) x→2− _xlim_→2÷ 4a + 2b = 8 (x) ={ f′ 2, 2ax + b, x< 2 2 < x < 3 continuous ⇒ (x) = (x) f′ lim x→2−f ′ _lim x→2÷f ′ 4a + b = 2 [3 marks]

29b._{Show that f is a one-to-one function.}

Markscheme

for A1

since for all values in the domain of f , f is increasing R1 therefore one-to-one AG [3 marks]

Examiners report

[N/A] x⩽ 2, (x) = 2 > 0f′ 2 < x < 3, (x) = −2x + 6 > 0f′ (x) > 0 f′ [5 marks]

29c._{Obtain expressions for the inverse function} _{and state their domains.}

Markscheme

M1

M1 therefore

A1A1A1

Note: Award A1 for the first line and A1A1 for the second line. [5 marks]

Examiners report

[N/A] f−1 x= 2y − 1 ⇒ y =x+1₂ x= − + 6y − 5 ⇒y2 y2− 6y + x + 5 = 0 y= 3 ± 4 − x√− −−−− (x) ={ f−1 , x+1 2 3 −√4 − x− −−−−, x⩽ 3 3 < x < 4

(37)

30a._{Obtain an expression for} _. [3 marks]

Markscheme

M1A1A1 [3 marks]

Examiners report

[N/A] (x) f′ (x) = sin(πx) + π ln(x + 1)cos(πx) f′ _x₊₁1 [4 marks]

30b._{Sketch the graphs of f and on the same axes, showing clearly all x-intercepts.}

Markscheme

A4

Note: Award A1A1 for graphs, A1A1 for intercepts. [4 marks]

Examiners report

[N/A]

f′

[2 marks]

(38)

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Examiners report

[N/A]

[3 marks]

30d._{Find the equation of the normal to the graph of f where x = 0.75 , giving your answer in the form y = mx + c .}

Markscheme

A1 so equation of normal is M1 A1 [3 marks]

Examiners report

[N/A] (0.75) = −0.839092 f′ y− 0.39570812 =_0.8390921 (x − 0.75) y= 1.19x − 0.498

30e._{Consider the points} _, _and _{where a , b and c} _{are the solutions of the equation}[6 marks] . Find the area of the triangle ABC.

Markscheme

A1

Note: Accept coordinates for B and C rounded to 3 significant figures. area (ci + dj) (ei + fj) M1A1

A1 A1 [6 marks]

Examiners report

[N/A] A (a , f(a)) B(b , f(b)) C (c , f(c)) (a < b < c) f(x) = (x)f′ A(0, 0) B(0.548…  , ) c 0.432…  d  C(1.44…  , ) e −0.881…  f  ΔABC = |1 2 × | = (de − cf)1 2 = 0.554