Calculus

Pranit_Calculus

[387 marks]

−− − x Let f(x) = √ 1−x , 0 < x < 1. 1

3

− 1 − ′ 1a. Show that f (x) = 2 x 2 (1 − x) 2 and deduce that f is an increasing function.

[5 marks]

Markscheme EITHER derivative of f ′ (x) = −1 2

x 1−x

is

(1−x)−x(−1)

1

M1A1

(1−x) 2

− 1 ( x ) 2 1 2 2 1−x (1−x) −3

M1A1

= 12 x (1 − x) 2 AG f ′ (x) > 0 (for all 0 < x < 1) so the function is increasing

R1

OR 1

x2

f(x) =

1

(1−x) 2 1

f ′ (x)

=

−1 2

(1−x) 2 ( 1 x 2

1

1

1

3

1 1 −2 x 2

−3

1

)− 1 x 2 (1−x) 2

1−x 1

−1 2

(−1) 3

= 12 x− 2 (1 − x)− 2 + 12 x 2 (1 − x)− 2 = 12 x− 2 (1 − x)− 2 [1 − x + x]

M1A1 A1

M1

= (1 − x) 2 AG f ′ (x) > 0 (for all 0 < x < 1) so the function is increasing [5 marks]

R1

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.

1b. Show that the curve y = f(x) has one point of inflexion, and find its coordinates.

Markscheme 1

3

f ′ (x) = 12 x− 2 (1 − x)− 2 3

3

1

5

⇒ f ′′ (x) = − 14 x− 2 (1 − x)− 2 + 34 x− 2 (1 − x)− 2 −3 2

= − 14 x

−5 2

(1 − x)

f ′′ (x) = 0 ⇒ x =

[1 − 4x]

1 4

M1A1

f ′′ (x) changes sign at x = x=

1 4

⇒y=

1 √3

1 4

A1

the coordinates are ( 14 , [6 marks]

M1A1

1 ) √3

hence there is a point of inflexion

R1

[6 marks]


−−−−−

2 2 √ 1c. Use the substitution x = sin θ to show that ∫ f(x)dx = arcsin √x − x − x + c .

[11 marks]

Markscheme x = sin2 θ ⇒

dx dθ

= 2 sin θ cos θ M1A1 −−−2−− −− − x ∫ √ 1−x dx = ∫ √ sin θ2 2 sin θ cos θdθ 1− sin θ

= ∫ 2sin2 θdθ

A1

= ∫ 1 − cos 2θdθ

M1A1

= θ − 12 sin 2θ + c

A1

θ = arcsin √x

M1A1

A1

−−−−− −−−−− = sin θ cos θ = √x√1 − x = √x − x2 −− − −−−−− x hence ∫ √ 1−x dx = arcsin √x − √x − x2 + c 1 sin 2θ 2

M1A1 AG

[11 marks]


2.

Given that the graph of y = x3 − 6x2 + kx − 4 has exactly one point at which the gradient is zero, find the value of k .

[5 marks]

Markscheme dy dx

= 3x2 − 12x + k

M1A1

For use of discriminant b2 − 4ac = 0 or completing the square 3(x − 2)2 + k − 12 144 − 12k = 0

(M1)

(A1)

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

k = 12

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.

The function f(x) = 3 sin x + 4 cos x is defined for 0 < x < 2π . 3a. Write down the coordinates of the minimum point on the graph of f .

[1 mark]

3a.

Markscheme (3.79,−5)

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.

3b.

The points P(p, 3) and Q(q, 3), q > p, lie on the graph of y = f(x) .

[2 marks]

Find p and q .

Markscheme p = 1.57 or π2 , q = 6.00

A1A1

[2 marks]


3c. Find the coordinates of the point, on y = f(x) , where the gradient of the graph is 3.

[4 marks]

Markscheme f ′ (x) = 3 cos x − 4 sin x

(M1)(A1)

3 cos x − 4 sin x = 3 ⇒ x = 4.43... (y = −4)

(A1)

A1

Coordinates are (4.43,−4) [4 marks]


3d.

Find the coordinates of the point of intersection of the normals to the graph at the points P and Q.

[7 marks]

3d.

Markscheme 1 m tangent

mnormal =

(M1)

gradient at P is −4 so gradient of normal at P is gradient at Q is 4 so gradient of normal at Q is

1 4

− 14

(A1) (A1)

equation of normal at P is y − 3 = 14 (x − 1.570...) (or y = 0.25x + 2.60...)

(M1)

equation of normal at Q is y − 3 = 14 (x − 5.999...) (or y = −0.25x + 4.499...) 

(M1)

Note: Award the previous two M1 even if the gradients are incorrect in y − b = m(x − a) where (a,b) are coordinates of P and Q (or in y = mx + c with c determined using coordinates of P and Q.

intersect at (3.79, 3.55)

A1A1

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


4a. Using the definition of a derivative as f ′ (x) = lim ( h→0

f(x+h)−f(x) h

) , show that the derivative of

f(x+h)−f(x) h

)

1 2x+1

is

−2 . (2x+1) 2

[4 marks]

Markscheme let f(x) =

1 2x+1

f ′ (x) = lim (

and using the result f ′ (x) = lim ( h→0

1 − 1 2x+1 2(x+h)+1

h→0

h

)

M1A1

⇒ f ′ (x) = lim ( h[2(x+h)+1][2x+1] ) [2x+1]−[2(x+h)+1]

h→0

⇒ f ′ (x) = ⇒ f ′ (x) =

−2 lim ( h→0 [2(x+h)+1][2x+1] −2 AG (2x+1) 2

)

A1 A1

[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 n = k and then show that this leads to it being true for n = k + 1. Many candidates just write ‘Let n = k’ which is of course meaningless. The conclusion is often of the form ‘True for n = 1, n = k and n = k + 1 therefore true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for n = k ⇒ true for n = k + 1’.

2n n!

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

[9 marks]

Markscheme let y =

1 2x+1

we want to prove that let n = 1 ⇒ ⇒

dy dx

=

dy dx

−2 (2x+1) 2

dn y dxn

= (−1)1

= (−1)n

21 1! (2x+1) 1+1

2n n! (2x+1) n+1

M1

which is the same result as part (a)

hence the result is true for n = 1

R1

assume the result is true for n = k :

dk y

d

k+1

y

dxk+1

⇒ ⇒ ⇒ ⇒

=

dk+1 y dxk+1 dk+1 y dxk+1 dk+1 y dxk+1 dk+1 y dxk+1

k d [(−1) k 2 k!k+1 dx (2x+1)

]

dxk

= (−1)k

2k k! (2x+1) k+1

M1

M1

=

d [ (−1) k 2k k!(2x + 1) −k−1 ] dx

=

(−1)k 2k k!(−k − 1)(2x + 1)−k−2

=

(−1)k+1 2k+1 (k + 1)!(2x + 1)−k−2

=

2k+1 (k+1)! (−1)k+1 (2x+1) k+2

(A1) ×2

A1 (A1)

A1

hence if the result is true for n = k , it is true for n = k + 1 since the result is true for n = 1 , the result is proved by mathematical induction Note: Only award final R1 if all the M marks have been gained.

R1

[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 n = k and then show that this leads to it being true for n = k + 1. Many candidates just write ‘Let n = k’ which is of course meaningless. The conclusion is often of the form ‘True for n = 1, n = k and n = k + 1 therefore true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for n = k ⇒ true for n = k + 1’.

5a. Sketch the curve y =

minimum points.

cos x √x2 +1

, − 4 ⩽ x ⩽ 4 showing clearly the coordinates of the x-intercepts, any maximum points and any

[4 marks]

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 [1 mark]

A1


5c.

Find the equation of the normal to the curve at x = 1 .

[3 marks]

5c.

Markscheme −1 gradient of normal is −0.786 (= 1.272...) (A1) when x = 1, y = 0.3820... (A1) Equation of normal is y – 0.382 = 1.27(x – 1) A1 (⇒ y = 1.27x − 0.890) [3 marks]


6.

A normal to the graph of y = arctan(x − 1) , for x > 0, has equation y = −2x + c , where x ∈ R .

[6 marks]

Find the value of c.

Markscheme d (arctan(x − 1)) dx

=

1 1+(x−1) 2

mN = −2 and so mT =

1 2

1 1+(x−1) 2

x = 2 (as x > 0)

A1

c = 4 + π4

A1

A1

(R1)

Attempting to solve

Substituting x = 2 and y =

(or equivalent)

π 4

=

1 2

(or equivalent) for x

to find c

M1

M1

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 −2x + c rather than the correct numerical value. A few candidates either didn’t simplify arctan(1) to π4 , or stated it to be 45 or π2 .

7.

Find the equation of the normal to the curve 5xy2 − 2x2 = 18 at the point (1, 2) .

[7 marks]

Markscheme dy

5y2 + 10xy dx − 4x = 0

A1A1A1

Note: Award A1A1 for correct differentiation of 5xy2 . A1 for correct differentiation of −2x2 and 18. dy

At the point (1, 2), 20 + 20 dx − 4 = 0 ⇒

dy dx

= − 45

(A1)

Gradient of normal =

5 4

A1

Equation of normal y − 2 = 54 (x − 1) y= y=

5 x− 4 5 x+ 4

5 4 3 4

+

M1

8 4

(4y = 5x + 3)

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.

The curve C has equation y = 18 (9 + 8x2 − x4 ) . 8a. Find the coordinates of the points on C at which

dy dx

=0.

[4 marks]

Markscheme dy dx

= 2x − 12 x3

x (2 −

1 2 x ) 2

A1

=0

x = 0, ± 2 dy dx

= 0 at (0, 98 ) , (−2, 258 ) , (2, 258 )

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.

8b. The tangent to C at the point P(1, 2) cuts the x-axis at the point T. Determine the coordinates of T.

[4 marks]

Markscheme at x =1, gradient of tangent =

3 2

(A1)

Note: In the following, allow FT on incorrect gradient. equation of tangent is y − 2 = 32 (x − 1) (y = 32 x + 12 ) meets x-axis when y = 0 , −2 = x=

3 (x − 1) 2

(A1)

(M1)

− 13

coordinates of T are (− 13 ,0)

A1

[4 marks]


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

[7 marks]

Markscheme gradient of normal = − 23

(A1)

equation of normal is y − 2 = − 23 (x − 1) (y = − 23 x + 83 ) at x = 0 , y =

8 3

(M1)

A1

Note: In the following, allow FT on incorrect coordinates of T and N. −− −− lengths of PN = √ 139 , PT = √ 529 A1A1 − − − − area of triangle PTN = 12 × √ 139 × √ 529 M1 =

13 9

(or equivalent e.g.

√676 ) 18

A1

[7 marks]


9.

1

1

Consider the functions f and g defined by f(x) = 2 x and g(x) = 4 − 2 x , x ≠ 0 . (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.

[7 marks]

Markscheme 1

(a)

1

2x = 4 −2x

attempt to solve the equation x=1

A1

so P is (1, 2) , as f(1) = 2

(b)

M1

1

f ′ (x) = − 12 2 x ln 2 x

A1

N1

A1

attempt to substitute x-value found in part (a) into their f ′ (x)

M1

f ′ (1) = −2 ln 2 y − 2 = −2 ln 2(x − 1) (or equivalent)

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.

10. A tangent to the graph of y = ln x passes through the origin.

(a)

Sketch the graphs of y = ln x and the tangent on the same set of axes, and hence find the equation of the tangent.

(b)

Use your sketch to explain why ln x ⩽

(c)

Show that xe ⩽ ex for x > 0 .

(d)

Determine which is larger, π e or eπ .

x e

for x > 0 .

[17 marks]

Markscheme (a)

A3 Note: Award A1 for each graph A1 for the point of tangency.

(M1)

point on curve and line is (a, ln a) y = ln(x) dy dx

=

1 x

⇒

dy dx

= a1 (when x = a)

(M1)A1

EITHER gradient of line, m, through (0, 0) and (a, ln a) is ⇒

ln a a

=

1 a

⇒ ln a = 1 ⇒ a = e ⇒ m =

1 e

ln a a

(M1)A1

M1A1

OR y − ln a = a1 (x − a)

(M1)A1

passes through 0 if ln a − 1 = 0

M1

a=e⇒m=

1 e

A1

THEN ∴ y = 1e x

A1

[11 marks]

(b)

the graph of ln x never goes above the graph of y = 1e x , hence ln x ⩽

[1 mark]

(c)

ln x ⩽

x e

⇒ e ln x ⩽ x ⇒ ln xe ⩽ x

exponentiate both sides of

ln xe

⩽x⇒

[3 marks]

(d)

equality holds when x = e

letting x = π ⇒ π e < eπ

A1

R1 N0

xe

M1A1 ⩽ ex

R1AG

x e

R1

[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 8y ln x − 2x2 + 4y2 = 7. Find the equation of the tangent to the curve at the point where x =

[7 marks]

1 and y > 0.

Markscheme dy

dy

8y × x1 + 8 dx ln x − 4x + 8y dx = 0

M1A1A1

Note: M1 for attempt at implicit differentiation. A1 for differentiating 8y ln x, A1 for differentiating the rest.

when x = 1, 8y × 0 − 2 × 1 + 4y2 = 7 y2 =

9 4

at (1,

3 dy ) 2 dx

⇒y= =

3 2

(as y > 0)

− 23

(M1)

A1

A1

y − 32 = − 23 (x − 1) or y = − 23 x + 136

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 dy equation to find y. Others wasted a lot of time rearranging the derivative to make dx the subject, rather than simply putting in the particular values for x and y.

12. The normal to the curve xe−y + ey = 1 + x, at the point (c, ln c), has a y-intercept c2 + 1.

Determine the value of c.

[7 marks]

Markscheme EITHER differentiating implicitly: dy

dy

1 × e−y − xe−y dx + ey dx = 1

M1A1

at the point (c, ln c) 1 c

dy

dy

− c × 1c dx + c dx = 1

dy dx

=

1 c

M1

(A1)

(c ≠ 1)

OR reasonable attempt to make expression explicit

(M1)

xe−y + ey = 1 + x x + e2y = ey (1 + x) e2y − ey (1 + x) + x = 0 (A1)

(ey − 1)(ey − x) = 0 ey = 1, ey = x y = 0, y = ln x

A1

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

=

1 x

gradient of tangent =

1 c

A1

Note: If candidate starts with y = ln x with no justification, award (M0)(A0)A1A1.

THEN the equation of the normal is y − ln c = −c(x − c) x = 0, y =

c2

M1

+1

c2 + 1 − ln c = c2

(A1)

ln c = 1 c=e

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.

13a.

A particle P moves in a straight line with displacement relative to origin given by

[10 marks]

s = 2 sin(πt) + sin(2πt), t ⩾ 0, 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 0 ⩽ t ⩽ 4.

Markscheme (i)

the period is 2

(ii)

v=

a=

dv dt

(iii)

=

ds dt

A1

= 2π cos(πt) + 2π cos(2πt)

−2π 2

sin(πt) − 4π 2

(M1)A1 (M1)A1

sin(2πt)

v=0

2π (cos(πt) + cos(2πt)) = 0 EITHER cos(πt) + 2cos2 (πt) − 1 = 0

M1

(2 cos(πt) − 1) (cos(πt) + 1) = 0 cos(πt) =

1 2

or cos(πt) = −1

t = 13 , t = 1 t=

5 , 3

t=

(A1)

A1

A1

7 , 3

t=

11 , 3

t=3

A1

OR 2 cos( πt2 ) cos( 3πt )=0 2 cos( πt2 )

= 0 or

t = 13 , 1

A1

t=

5 7 , , 3 3

3,

11 3

cos( 3πt ) 2

M1 =0

A1A1

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 cos(πt) but then failed to find all the possible values of t.

[8 marks]

13b. Consider the function

f(x) = A sin(ax) + B sin(bx), A, a, B, b, x ∈ R. Use mathematical induction to prove that the(2n)th derivative of f is given by (f (2n) (x) = (−1)n (Aa2n sin(ax) + Bb2n sin(bx)), for all n ∈ Z+ .

Markscheme P(n) : f (2n) (x) = (−1)n (Aa2n sin(ax) + Bb2n sin(bx)) P(1) : f ′′ (x) = (Aa cos(ax) + Bb cos(bx))

′

M1

= −Aa2 sin(ax) − Bb2 sin(bx) = −1 (Aa2 sin(ax) + Bb2 sin(bx))

A1

∴ P(1) true assume that P(k) : f (2k) (x) = (−1)k (Aa2k sin(ax) + Bb2k sin(bx)) is true

M1

consider P(k + 1) f (2k+1) (x) = (−1)k (Aa2k+1 cos(ax) + Bb2k+1 cos(bx)) f (2k+2) (x) = (−1)k (−Aa2k+2 sin(ax) − Bb2k+2 sin(bx)) = (−1)k+1 (Aa2k+2 sin(ax) + Bb2k+2 sin(bx))

M1A1 A1

A1

P(k) true implies P(k + 1) true, P(1) true so P(n) true ∀n ∈ Z+

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.

14. Consider the curve y = xex and the line y = kx, k ∈ R .

(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)

(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 0 < k < 1, show that A < 1.

Find the values of k for which the curve y = xex and the line y = kx meet in two distinct points.

[23 marks]

Markscheme (a)

(i)

A1

xex = 0 ⇒ x = 0

so, they intersect only once at (0, 0)

(ii)

M1A1

y′ = ex + xex = (1 + x)ex

y′ (0) = 1

A1

θ = arctan1 =

π 4

(θ = 45∘ )

A1

[5 marks]

(b)

when k = 1, y = x M1

xex = x ⇒ x(ex − 1) = 0 A1

⇒x=0

y′ (0) = 1 which equals the gradient of the line y = x

R1

AG

so, the line is tangent to the curve at origin

Note: Award full credit to candidates who note that the equation x(ex − 1) = 0 has a double root x = 0 so y = x is a tangent.

[3 marks]

(c)

(i)

xex = kx ⇒ x(ex − k) = 0

⇒ x = 0 or x = ln k k > 0 and k ≠ 1

M1

A1

A1

A1A1

(ii)

(0, 0) and (ln k, k ln k)

(iii)

A = ∣∣∫0ln k kx − xex dx∣∣

M1A1

Note: Do not penalize the omission of absolute value.

(iv)

attempt at integration by parts to find ∫ xex dx

∫ xex dx = xex − ∫ ex dx = ex (x − 1) as 0 < k < 1 ⇒ ln k < 0

A1

R1

A = ∫ln k kx − xex dx = [ k2 x2 − (x − 1)ex ] 0

= 1 − ( k2 (ln k) 2 − (ln k − 1)k)

A1

= 1 − k2 ( (ln k) 2 − 2 ln k + 2) = 1 − k2 ( (ln k − 1) 2 + 1)

M1A1

since k2 ( (ln k − 1) 2 + 1) > 0 A<1

AG


R1

0 ln k

A1

M1

Examiners report Many candidates solved (a) and (b) correctly but in (c), many failed to realise that the equation xex = kx 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).

15. Find the equation of the normal to the curve x3 y3 − xy = 0 at the point (1, 1).

[7 marks]

Markscheme x3 y3 − xy = 0 3x2 y3 + 3x3 y2 y′ − y − xy′ = 0 Note: Award A1 for correctly differentiating each term.

x = 1, y = 1

3 + 3y ′ − 1 − y ′ = 0

2y′ = −2 y′ = −1

(M1)A1

gradient of normal = 1

(A1)

equation of the normal y − 1 = x − 1

A1

N2

y=x 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.

y = m(x − m)

(1 − x)y = 1

Markscheme EITHER y=

1 1−x

⇒ y′ =

1 (1−x) 2

M1A1

solve simultaneously

M1

1 1−x

= m(x − m) and

1 (1−x) 2

1 1−x

=

1 (x − 1 2 ) (1−x) 2 (1−x)

=m A1

Note: Accept equivalent forms.

(1 − x)3 − x(1 − x)2 + 1 = 0, x ≠ 1 x = 1.65729 … ⇒ y =

1 1−1.65729…

= −1.521379 …

tangency point (1.66, –1.52)

A1A1

(−1.52137 …)2

A1

m=

= 2.31

OR (1 − x)y = 1 m(1 − x)(x − m) = 1

M1

m(x − x2 − m + mx) = 1 mx2 − x(m + m2 ) + (m2 + 1) = 0 (M1)

b2 − 4ac = 0 (m + m2 )2

A1

− 4m(m2 + 1) = 0 A1

m = 2.31

substituting m = 2.31 … into mx2 − x(m + m2 ) + (m2 + 1) = 0 A1

x = 1.66 y=

(M1)

1 1−1.65729

= −1.52

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.

The curve C is given by y =

17a.

Show that

dy dx

=

cos 2 x− x2 sin x (x+cos x) 2

x cos x , x+cos x

, x ⩾ 0.

for x ⩾ 0. [4 marks]

Markscheme dy dx

=

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

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”.

=

x cos x+ cos 2 x− x2 sin x−x cos x sin x−x cos x+x cos x sin x

=

cos 2 x− x2 sin x

A1

(x+cos x) 2 (x+cos x) 2

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 − sin x rather than 1 − sin x. A disappointing number of candidates failed to calculate the correct gradient at the specified point.

[3 marks]

17b. Find the equation of the tangent to C at the point ( π ,0). 2

Markscheme the derivative has value –1

(A1)

the equation of the tangent line is (y − 0) = (−1) (x − π2 ) (y =

π 2

M1A1

− x)

[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 − sin x rather than 1 − sin x. A disappointing number of candidates failed to calculate the correct gradient at the specified point.

18.

Given that y =

1 , 1−x

use mathematical induction to prove that

dn y dxn

=

n! (1−x)

n+1

, n ∈ Z+ .

[7 marks]

Markscheme proposition is true for n = 1 since 1! (1−x) 2

=

dy dx

=

1 (1−x) 2

M1

A1

Note: Must see the 1! for the A1.

assume true for n = k, k ∈ Z+ , i.e. consider

dk+1 y dxk+1

=

dk y

d(

dxk

)

=

(1−x) k+2

dxk

=

k! (1−x) k+1

M1

(M1)

dx

= (k + 1)k!(1 − x)−(k+1)−1 (k+1)!

dk y

A1

A1

hence, Pk+1 is true whenever Pk is true, and P1 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 (k + 1)th derivative was the (k)th derivative multiplied by the first derivative. Providing an acceptable final statement remains a perennial issue.

The curve C with equation y = f(x) satisfies the differential equation dy y = (x + 2), y > 1, dx ln y and y = e when x = 2. 19a. Find the equation of the tangent to C at the point (2, e).

[3 marks]

Markscheme dy dx

=

e (2 + 2) ln e

= 4e

A1

at (2, e) the tangent line is y − e = 4e(x − 2) hence y = 4ex − 7e

M1

A1

[3 marks]

Examiners report Nearly always correctly answered.

19b. Find f(x).

[11 marks]

Markscheme dy dx

∫

=

y (x + 2) ln y

ln y dy y

⇒

ln y dy y

= ∫ (x + 2)dx

using substitution u = ln y; du = 1y dy ⇒∫ ⇒

ln y dy y

(ln y) 2 2

at (2, e),

=

= ∫ udu = 12 u2 x2 2

(ln e) 2

2

⇒ c = − 112 ⇒

(ln y) 2 2

M1

= (x + 2)dx

=

+ 2x + c = 6+c

(M1)(A1)

(A1)

A1A1 M1

A1 x2 2

+ 2x − 112 ⇒ (ln y)2 = x2 + 4x − 11

−−−−−−−−−− ln y = ±√x2 + 4x − 11 ⇒ y = e± √x2 +4x−11 since y > 1, f(x) = e√x2 +4x−11

M1A1

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.

19c.

[6 marks]

Determine the largest possible domain of f.

Markscheme EITHER x2 + 4x − 11 > 0

A1

using the quadratic formula

M1

−− (= −2 ± √15 )

A1

using a sign diagram or algebraic solution −− −− x < −2 − √15 ; x > −2 + √15 A1A1

M1

critical values are

−4± √60 2

OR x2 + 4x − 11 > 0

A1

by methods of completing the square

M1

A1 −− −− ⇒ x + 2 < −√15 or x + 2 > √15 (M1) −− −− x < −2 − √15 ; x > −2 + √15 A1A1 (x + 2)2 > 15

[6 marks]

Examiners report Part (c) was often well answered, sometimes with follow through.

19d.

[4 marks]

Show that the equation f(x) = f ′ (x) has no solution.

Markscheme f(x) = f ′ (x) ⇒ f(x) =

f(x) ln f(x)

(x + 2)

M1

−−−−−−−−−− ⇒ ln(f(x)) = x + 2 (⇒ x + 2 = √x2 + 4x − 11 )

A1

⇒ (x + 2)2 = x2 + 4x − 11 ⇒ x2 + 4x + 4 = x2 + 4x − 11 ⇒ 4 = −11, hence f(x) ≠ f ′ (x)

A1

R1AG

[4 marks]

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

Let f be a function defined by f(x) = x + 2 cos x , x ∈ [0, 2π] . The diagram below shows a region S bound by the graph of f and the line y = x .

A and C are the points of intersection of the line y = x and the graph of f , and B is the minimum point of f . 20. (a)

If A, B and C have x-coordinates a π2 , b π6 and c π2 , where a , b, c ∈ N , find the values of a , b and c .

(b)

Find the range of f .

(c)

Find the equation of the normal to the graph of f at the point C, giving your answer in the form y = px + q .

(d)

The region S is rotated through 2π about the x-axis to generate a solid. (i)

Write down an integral that represents the volume V of this solid.

(ii)

Show that V = 6π 2 .

[19 marks]

Markscheme (a)

METHOD 1

using GDC A1A2A1

a = 1, b = 5, c = 3 METHOD 2

x = x + 2 cos x ⇒ cos x = 0 ⇒ x = π2 ,

3π 2

M1

... A1

a = 1, c = 3

M1

1 − 2 sin x = 0 ⇒ sin x =

1 2

⇒x=

π 6

or

5π 6

A1

b=5

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

(b)

f ( 5π6 ) =

5π 6

− √3 (or 0.886)

f(2π) = 2π + 2 (or 8.28)

(M1)

(M1)

the range is [ 5π6 − √3, 2π + 2] (or [0.886, 8.28])

A1

[3 marks]

(c)

f ′ (x) = 1 − 2 sin x

f ′ ( 3π6 ) = 3

(M1)

A1

gradient of normal = − 13

(M1)

equation of the normal is y − 3π2 = − 13 (x − 3π2 )

(M1)

y = − 13 x + 2π (or equivalent decimal values)

A1

N4

[5 marks]

(d)

3π

V = π ∫ π 2 ( x2 − (x + 2 cos x) 2 )dx (or equivalent)

(i)

A1A1

2

Note: Award A1 for limits and A1 for π and integrand.

(ii)

3π

V = π ∫ π 2 ( x2 − (x + 2 cos x) 2 )dx 2

= −π ∫

3π 2 π 2

(4x cos x + 4cos2 x)dx

using integration by parts and the identity

4cos2 x

M1

= 2 cos 2x + 2 ,

M1

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

3π 2 π 2

A1A1

Note: Award A1 for 4x sin x + 4 cos x and A1 for sin 2x + 2x . = −π [(6π sin 3π2 + 4 cos 3π2 + sin 3π + 3π) − (2πsin π2 + 4 cos π2 + sin π + π)] = −π (−6π + 3π − π) = 6π 2

AG

N0

Note: Do not accept numerical answers. [7 marks]

A1

Total [19 marks]

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

Consider the functions f(x) = (ln x)2 , x > 1 and g(x) = ln(f(x)), x > 1. (i) Find f ′ (x). (ii) Find g′ (x). (iii) Hence, show that g(x) is increasing on ]1, ∞[.

[5 marks]

21a.

Markscheme (i)

attempt at chain rule 2 ln x x

=

(ii)

attempt at chain rule

g′ (x) =

2 x ln x

(M1)

A1

f ′ (x)

(M1)

A1

g′ (x) is positive on ]1, ∞[

(iii)

so g(x) is increasing on ]1, ∞[

A1 AG

[5 marks]

Examiners report [N/A]

[12 marks]

21b. Consider the differential equation

(ln x)

dy 2 2x − 1 + y= , x > 1. dx x (ln x)

(i)

Find the general solution of the differential equation in the form y = h(x).

(ii)

Show that the particular solution passing through the point with coordinates (e, e2 ) is given by y =

(iii)

Sketch the graph of your solution for x > 1, clearly indicating any asymptotes and any maximum or minimum points.

x2 −x+e (ln x) 2

.

Markscheme (i)

rearrange in standard form:

dy dx

2x−1

+ x ln2 x y =

(ln x) 2

(A1)

, x>1

integrating factor: e∫

2 x ln x

dx

ln( (ln x) 2 )

(M1)

=e

(A1)

= (ln x)2

(M1)

multiply by integrating factor dy (ln x)2 dx + 2 lnx x d (y(ln x) 2 ) = dx

y = 2x − 1 2x − 1 (or y(ln x) 2 = ∫ 2x − 1dx)

attempt to integrate: (ln x)2 y y=

=

x2

x2 −x+c (ln x) 2

M1

−x+c A1

attempt to use the point (e, e2 ) to determine c:

(ii)

(ln e)2 e2

eg,

c=e y=

M1

=

e2

− e + c or

e2

=

e2−e+c (ln e) 2

or

e2

=

e2

M1 −e+c

A1

x2 −x+e (ln x) 2

AG

(iii)

graph with correct shape

A1

minimum at x = 3.1 (accept answers to a minimum of 2 s.f) asymptote shown at x = 1

A1

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]


Consider the function f(x) =

ln x , x

x > 0.

The sketch below shows the graph of y = f(x) and its tangent at a point A.

′ 22a. Show that f (x) =

1−ln x x2

[2 marks]

.

Markscheme f ′ (x) = =

x× 1x −ln x x2

1−ln x x2

M1A1

AG

[2 marks]


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

[3 marks]

Markscheme 1−ln x x2 y = 1e

= 0 has solution x = e

M1A1

A1

hence maximum at the point (e, 1e ) [3 marks]


22c. Find the coordinates of C, the point of inflexion on the curve.

[5 marks]

Markscheme x2 (− 1x )−2x(1−ln x)

f ′′ (x) = =

M1A1

x4

2 ln x−3 x3

The M1A1 should be awarded if the correct working appears in part (b).

Note:

point of inflexion where f ′′ (x) = 0 3

so x = e 2 , y = 32 e

−3 2

M1

A1A1

C has coordinates ( e 2 , 32 e 3

−3 2

)

[5 marks]


22d. The graph of y = f(x) crosses the x-axis at the point A.

[4 marks]

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

Markscheme f(1) = 0

A1

=1

(A1)

y = x+c

(M1)

f ′ (1)

through (1, 0) equation is y = x − 1

A1

[4 marks]


22e. The graph of y = f(x) crosses the x-axis at the point A.

Find the area enclosed by the curve y = f(x), the tangent at A, and the line x = e.

[7 marks]

Markscheme METHOD 1 e

area = ∫1 x − 1 − lnxx dx Note:

M1A1A1

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

order. (ln x) 2 (+c) (M1)A1 2 x2 ∫ (x − 1)dx = 2 − x(+c) A1 e = [ 12 x2 − x − 12 (ln x) 2 ] 1

∫

ln x dx x

=

= ( 12 e2 − e − 12 ) − ( 12 − 1) = 12 e2 − e

A1

METHOD 2 area = area of triangle − ∫1e Note:

ln x dx x

A1 is for correct integral with limits and is dependent on the M1.

(ln x) 2 (+c) (M1)A1 2 area of triangle = 12 (e − 1)(e − 1) 1 (e − 1)(e − 1) − ( 12 ) = 12 e2 − e 2

∫

ln x dx x

M1A1

=

M1A1 A1

[7 marks]


Let f(x) =

e2x +1 . ex −2

23a. Find the equations of the horizontal and vertical asymptotes of the curve y = f(x).

[4 marks]

Markscheme x → −∞ ⇒ y → − 12 so y = − 12 is an asymptote

(M1)A1

ex − 2 = 0 ⇒ x = ln 2 so x = ln 2 (= 0.693) is an asymptote

(M1)A1

[4 marks]


23b. (i)

(ii) (iii)

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

[8 marks]

Markscheme 2(ex −2)e2x −(e2x +1)ex

f ′ (x) =

(i)

=

(ex −2) 2

= 0 when e3x − 4e2x − ex = 0

ex (e2x ex

M1A1

e3x −4e2x − ex

f ′ (x)

(ii)

(ex −2) 2

− 4ex

= 0,

ex

M1

− 1) = 0

= −0.236, ex = 4.24 (or ex = 2 ± √5)

A1A1

Award A1 for zero, A1 for other two solutions.

Note:

Accept any answers which show a zero, a negative and a positive. as ex > 0 exactly one solution Note: (iii)

R1

Do not award marks for purely graphical solution. (1.44, 8.47)

A1A1

[8 marks]


23c. Find the equation of L1 , the normal to the curve at the point where it crosses the y-axis.

[4 marks]

Markscheme f ′ (0) = −4

(A1)

so gradient of normal is f(0) = −2

1 4

(M1)

(A1)

so equation of L1 is y = 14 x − 2

A1

[4 marks]


The line L2 is parallel to L1 and tangent to the curve y = f(x). [5 marks]

23d. Find the equation of the line L2 .

Markscheme f ′ (x) =

1 4

so x = 1.46

M1 (M1)A1

f(1.46) = 8.47

(A1)

equation of L2 is y − 8.47 = 14 (x − 1.46) (or y =

1 x + 8.11) 4

[5 marks]


A1

The function f is defined by f(x) = {

1 − 2x, x ≤ 2 − 3, x > 2

3 (x − 2)2 4

[2 marks]

24a. Determine whether or not fis continuous.

Markscheme 1 − 2(2) = −3 and 34 (2 − 2)2 − 3 = −3

A1

both answers are the same, hence f is continuous (at x = 2) Note:

R1

R1 may be awarded for justification using a graph or referring to limits. Do not award A0R1.

[2 marks]


24b. The graph of the function g is obtained by applying the following transformations to the graph of f:

2 a reflection in the y–axis followed by a translation by the vector ( ). 0

Find g(x).

Markscheme reflection in the y-axis 1 + 2x, x ≥ −2 f(−x) = { 3 (x + 2)2 − 3, x < −2 4 Note:

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

2 translation ( ) 0 2x − 3, x ≥ 0 g(x) = { 3 2 x − 3, x < 0 4 Note:

(M1)

(M1)A1A1

Award (M1) for attempting to substitute (x − 2) 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]


2

(x2 + y2 ) = 4xy2 25a.

dy dx

[4 marks]

dx

Markscheme METHOD 1 expanding the brackets first: x4 + 2x2 y2 + y4 = 4xy2 dy

M1 dy

dy

M1A1A1

4x3 + 4xy2 + 4x2 y dx + 4y3 dx = 4y2 + 8xy dx

Award M1 for an attempt at implicit differentiation.

Note:

Award A1 for each side correct. dy dx

=

− x3 −xy 2 + y 2 xy 2 −2xy+ y 3

or equivalent

A1

METHOD 2

2 (x2 + y2 ) (2x + 2y dx ) = 4y2 + 8xy dx dy

dy

M1A1A1

Award M1 for an attempt at implicit differentiation.

Note:

Award A1 for each side correct. (x2 + y2 ) (x + y dx ) = y2 + 2xy dx dy

dy

dy

dy

dy

x3 + x2 y dx + y2 x + y3 dx = y2 + 2xy dx dy dx

=

− x3 −xy 2 + y 2 yx2 −2xy+ y 3

or equivalent

M1

A1

[5 marks]


[3 marks]

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

Markscheme METHOD 1 at (1, 1),

dy dx

y=1

A1

is undefined

M1A1

METHOD 2 gradient of normal = − dy1 = − dx

at (1, 1) gradient = 0 y=1

(yx2 −2xy+ y 3 ) (− x3 −xy 2 + y 2 )

M1

A1

A1

[3 marks]


A function f is defined by f(x) = 12 (ex + e−x ) , x ∈ R . 26a. (i) Explain why the inverse function f −1 does not exist.

(ii) (iii)

[14 marks]

Show that the equation of the normal to the curve at the point P where x = ln 3 is given by 9x + 12y − 9 ln 3 − 20 = 0. Find the x-coordinates of the points Q and R on the curve such that the tangents at Q and R pass through (0, 0).

Markscheme (i)

either counterexample or sketch or

recognising that y = k (k > 1) intersects the graph of y = f(x) twice function is not 1 − 1 (does not obey horizontal line test) AG

so f −1 does not exist

(ii) f ′ (x) = 12 (ex − e−x ) f ′ (ln 3) = m=

4 3

− 34

M1

R1

(A1)

(A1)

(= 1.33)

M1

f(ln 3) =

5 3

(= 1.67)

A1

EITHER y− 5 3

x−ln 3

4y −

= − 34 20 3

M1 A1

= −3x + 3 ln 3

OR 5 3

= − 34 ln 3 + c

M1

5 3

c= y=

+ 34 ln 3 − 34 x + 53 + 34 ln 3

A1

12y = −9x + 20 + 9 ln 3 THEN AG

9x + 12y − 9 ln 3 − 20 = 0 (iii)

The tangent at (a, f(a)) has equation y − f(a) = f ′ (a)(x − a).

f ′ (a) =

f(a) a

ea − e−a =

(or equivalent) ea + e−a a

(or equivalent)

attempting to solve for a a = ±1.20

(M1)

(A1) A1

(M1)

A1A1

[14 marks]

Examiners report In part (a) (i), successful candidates typically sketched the graph of y = f(x), applied the horizontal line test to the graph and concluded that the function was not 1 − 1 (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 9x + 12y − 9 ln 3 − 20 = 0. 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 y = f(x) and y = xf ′ (x) on the same set of axes and found the x-coordinates of the intersection points.

26b. The domain of f is now restricted to x ⩾ 0.

(i) (ii)

Find an expression for

[8 marks]

f −1 (x).

Find the volume generated when the region bounded by the curve y = f(x) and the lines x = 0 and y = 5 is rotated through an

angle of 2π radians about the y-axis.

Markscheme 2y = ex + e−x

(i)

e2x − 2yex + 1 = 0 Note:

M1A1

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

2y± √4y 2 −4

A1 −− − − − = y ± √y 2 − 1 −−−−− x = ln(y ± √y2 − 1 ) A1 −−−−− f −1 (x) = ln(x + √x2 − 1 ) ex =

2

ex

Note:

A1

Award A1 for correct notation and for stating the positive “branch”. −−−−− 2 5 V = π ∫1 (ln(y + √y2 − 1 )) dy

(ii) Note:

(M1)(A1)

Award M1 for attempting to use V = π ∫cd x2 dy.

= 37.1 (units3 )

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 ex or ey .

+ 27. A family of cubic functions is defined as fk (x) = k2 x3 − kx2 + x, k ∈ Z .

(a)

Express in terms of k

(i)

f ′ k (x) and f ′′ k (x) ;

(ii)

the coordinates of the points of inflexion Pk on the graphs of fk .

[13 marks]

(b)

Show that all Pk lie on a straight line and state its equation.

(c)

Show that for all values of k, the tangents to the graphs of fk at Pk are parallel, and find the equation of the tangent lines.

Markscheme (a)

f ′ k (x) = 3k2 x2 − 2kx + 1

(i)

A1

f ′′ k (x) = 6k2 x − 2k

M1

Setting f ′′ (x) = 0

(ii)

A1

⇒ 6k2 x − 2k = 0 ⇒ x = 3

1 3k

A1 2

f ( 3k1 ) = k2 ( 3k1 ) − k( 3k1 ) + ( 3k1 ) 7 27k

=

M1

A1

7 Hence, Pk is ( 3k1 , 27k )

[6 marks] Equation of the straight line is y = 79 x

(b)

A1

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

R1

[2 marks] (c)

Gradient of tangent at Pk : 2

f ′ (Pk ) = f ′ ( 3k1 ) = 3k2 ( 3k1 ) − 2k ( 3k1 ) + 1 =

2 3

M1A1

As the gradient is independent of k, the tangents are parallel. 7 27k

= 23 × 3k1 + c ⇒ c =

The equation is y =

1 27k 2 1 x + 27k 3

R1

(A1) A1


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.

The function f is defined, for − π2 ⩽ x ⩽

π 2

, by f(x) = 2 cos x + x sin x .

28a. Determine whether f is even, odd or neither even nor odd.

[3 marks]

Markscheme f(−x) = 2 cos(−x) + (−x) sin(−x) = 2 cos x + x sin x (= f(x)) therefore f is even

M1

A1

A1

[3 marks]


28b. Show that f ′′ (0) = 0 .

[2 marks]

Markscheme f ′ (x) = −2 sin x + sin x + x cos x (= − sin x + x cos x) f ′′ (x) = − cos x + cos x − x sin x (= −x sin x) so f ′′ (0) = 0

A1

A1

AG

[2 marks]


28c. John states that, because f ′′ (0) = 0 , 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; f ′′ (x) is even and therefore has the same sign either side of (0, 2)

R2

[2 marks]


The function f is defined by f(x) = {

2x − 1, ax2 + bx − 5,

x⩽2 2
where a , b ∈ R . 29a. Given that f and its derivative, f ′ , are continuous for all values in the domain of f , find the values of a and b .

[6 marks]

Markscheme f continuous ⇒ lim− f(x) = lim÷ f(x) x→2

x→2

M1

A1

4a + 2b = 8

2, f ′ (x) = { 2ax + b,

x<2 2
A1

f ′ continuous ⇒ lim− f ′ (x) = lim f ′ (x) x→2÷

x→2

4a + b = 2

A1 M1

solve simultaneously

A1

to obtain a = –1 and b = 6 [6 marks]


[3 marks]

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

Markscheme for x ⩽ 2, f ′ (x) = 2 > 0 for 2 < x < 3,

f ′ (x)

A1

= −2x + 6 > 0

A1

since f ′ (x) > 0 for all values in the domain of f , f is increasing

R1

AG

therefore one-to-one [3 marks]


29c. Obtain expressions for the inverse function f −1 and state their domains.

Markscheme x = 2y − 1 ⇒ y = x=

M1

x+1 2

−y2

+ 6y − 5 ⇒ −−−−− y = 3 ± √4 − x

y2

− 6y + x + 5 = 0

M1

therefore f −1 (x) = {

x+1 , 2

−−−−− 3 − √4 − x ,

x⩽3 3
A1A1A1

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


[5 marks]

The function f is defined on the domain [0, 2] by f(x) = ln(x + 1) sin(πx) . [3 marks]

30a. Obtain an expression for f ′ (x) .

Markscheme f ′ (x) =

1 sin(πx) + π ln(x + 1) cos(πx) x+1

M1A1A1

[3 marks]


[4 marks]

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

Markscheme

A4

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


30c. Find the x-coordinates of the two points of inflexion on the graph of f .

[2 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 .

[3 marks]

Markscheme f ′ (0.75) = −0.839092

A1

so equation of normal is y − 0.39570812 = y = 1.19x − 0.498

1 0.839092

(x − 0.75)

M1

A1

[3 marks]


30e. Consider the points A (a , f(a)) , B (b , f(b)) and C (c , f(c)) where a , b and c (a < b < c) are the solutions of the equation [6 marks]

f(x) = f ′ (x) . Find the area of the triangle ABC.

Markscheme A(0, 0)

  B( 0.548 … , 0.432 … ) c

d

  C(1.44 … , −0.881 … )

A1

f

e

A1

Note: Accept coordinates for B and C rounded to 3 significant figures.

area ΔABC = 12 |(ci + dj) × (ei + fj)| =

1 (de − cf) 2

= 0.554

M1A1

A1

A1

[6 marks]


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Printed for Fountainhead School

Calculus

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