New test - November 09, 2017

[139 marks]

[2 marks]
1a. The values of the functions f and g and their derivatives for x = 1 and x = 8 are shown in the following table.

Let h(x) = f(x)g(x).

Find h(1).

Markscheme
expressing h(1) as a product of f(1) and g(1) (A1)

eg f(1) × g(1), 2(9)

h(1) = 18 A1 N2

[2 marks]

Examiners report
[N/A]

1b. Find h′ (8). [3 marks]

 Markscheme
attempt to use product rule (do not accept h′ = f ′ × g ′ ) (M1)

eg h′ = fg ′ + gf ′ , h′ (8) = f ′ (8)g(8) + g ′ (8)f(8)

correct substitution of values into product rule (A1)

eg h′ (8) = 4(5) + 2(−3), − 6 + 20

h′ (8) = 14 A1 N2

[3 marks]

Examiners report
[N/A]

2a. Let f(x) = cos x. [4 marks]

(i) Find the first four derivatives of f(x).

(ii) Find f (19) (x).

Markscheme
(i) f ′ (x) = − sin x, f ′′ (x) = − cos x, f (3) (x) = sin x, f (4) (x) = cos x A2 N2

(ii) valid approach (M1)

eg recognizing that 19 is one less than a multiple of 4, f (19) (x) = f (3) (x)

f (19) (x) = sin x A1 N2

[4 marks]

Examiners report
[N/A]
2b. Let g(x) = xk , where k ∈ Z+ . [5 marks]

(i) Find the first three derivatives of g(x).

k!
(ii) Given that g (19) (x) = (k−p)! (xk−19 ), find p.

Markscheme
(i) g ′ (x) = kxk−1
g ′′ (x) = k(k − 1)xk−2 , g (3) (x) = k(k − 1)(k − 2)xk−3 A1A1 N2

(ii) METHOD 1

correct working that leads to the correct answer, involving the correct expression for the 19th derivative A2
(k−19)!
eg k(k − 1)(k − 2) … (k − 18) × (k−19)! , k P19

k!
p = 19 (accept (k−19)!
xk−19 ) A1 N1

METHOD 2

correct working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general
rule for 19th coefficient A2

k k!
eg g ′′ = 2! ( ) , k(k − 1)(k − 2) = , g (3) (x)=k P3 (xk−3 )
2 (k−3)!

k k!
g (19) (x) = 19! ( ) , 19! × , k P19
19 (k−19)!×19!

k!
p = 19 (accept (k−19)!
xk−19 ) A1 N1

[5 marks]

Examiners report
[N/A]
 [7 marks]
2c. Let k = 21 and h(x) = (f (19) (x) × g (19) (x)).

(i) Find h′ (x).

−21!
(ii) Hence, show that h′ (π) = 2 π2.
Markscheme
(i) valid approach using product rule (M1)

eg uv′ + vu′ , f (19) g (20) + f (20) g (19)

correct 20th derivatives (must be seen in product rule) (A1)(A1)

21!
eg g (20) (x) = (21−20)! x, f (20) (x) = cos x

h′ (x) = sin x(21!x) + cos x ( 21! 2 21! 21! 2

2 x ) (accept sin x ( 1! x) + cos x ( 2! x )) A1 N3

(ii) substituting x = π (seen anywhere) (A1)

21! 21!
eg f (19) (π)g (20) (π) + f (20) (π)g (19) (π), sin π 1! π + cos π 2! π 2

evidence of one correct value for sin π or cos π (seen anywhere) (A1)

eg sin π = 0, cos π = −1

evidence of correct values substituted into h′ (π) A1

π π 21!
eg 21!(π) (0 − 2! ) , 21!(π) (− 2 ) , 0 + (−1) 2 π 2

Note: If candidates write only the first line followed by the answer, award A1A0A0.

−21!
2 π2 AG N0

[7 marks]

Examiners report
[N/A]
3a. Let f(x) = √4x + 5, for x ⩾ −1.25. [4 marks]

Find f ′ (1).

Markscheme
choosing chain rule (M1)
dy dy du
eg
dx
= du
× dx
, u = 4x + 5, u′ = 4

correct derivative of f A2
1
1 −2 2
eg 2 (4x + 5) × 4, f ′ (x) = √4x+5

2
f ′ (1) = 3 A1 N2

[4 marks]

Examiners report
Part a) was relatively well answered – the obvious errors seen were not using the chain rule correctly and simple fraction calculations
being wrong.
3b. Consider another function g. Let R be a point on the graph of g. The x-coordinate of R is 1. The equation of the tangent to the [2 marks]

graph at R is y = 3x + 6.

Write down g ′ (1).

Markscheme
recognize that g ′ (x) is the gradient of the tangent (M1)

eg g ′ (x) = m

g ′ (1) = 3 A1 N2

[2 marks]

Examiners report
In parts b) and c) it seemed that the students did not have a good conceptual understanding of what was actually happening in this
question. There was lack of understanding of tangents, gradients and their relationship to the original function, g. A working sketch may
have been beneficial but few were seen and many did a lot more work than required.

[2 marks]
3c. Find g(1).

Markscheme
recognize that R is on the tangent (M1)

eg g(1) = 3 × 1 + 6, sketch

g(1) = 9 A1 N2

[2 marks]

Examiners report
In parts b) and c) it seemed that the students did not have a good conceptual understanding of what was actually happening in this
question. There was lack of understanding of tangents, gradients and their relationship to the original function, g. A working sketch may
have been beneficial but few were seen and many did a lot more work than required.
 [7 marks]
3d. Let h(x) = f(x) × g(x). Find the equation of the tangent to the graph of h at the point where x = 1.
 Markscheme
f(1) = √4 + 5 (= 3) (seen anywhere) A1

h(1) = 3 × 9 (= 27) (seen anywhere) A1

choosing product rule to find h′ (x) (M1)

′ ′
eg uv + u v

correct substitution to find h′ (1) (A1)

eg f(1) × g ′ (1) + f ′ (1) × g(1)

2
h′ (1) = 3 × 3 + 3 × 9 (= 15) A1

EITHER

attempt to substitute coordinates (in any order) into the equation of a straight line (M1)

eg y − 27 = h′ (1)(x − 1), y − 1 = 15(x − 27)

y − 27 = 15(x − 1) A1 N2

OR

attempt to substitute coordinates (in any order) to find the y-intercept (M1)

eg 27 = 15 × 1 + b, 1 = 15 × 27 + b

y = 15x + 12 A1 N2

[7 marks]

Examiners report
In part d) although candidates recognized h(x) as a product and may have correctly found h(1), they did not necessarily use the
product rule to find h′ (x), instead incorrectly using h′ (x) = f ′ (x) × g ′ (x). It was rare for a candidate to get as far as finding the equation
of a straight line but those who did usually gained full marks.

[2 marks]
4a. A function f has its derivative given by f ′ (x) = 3x2 − 2kx − 9, where k is a constant.

Find f ′′ (x).

Markscheme
f ′′ (x) = 6x − 2k A1A1 N2

[2 marks]
 Examiners report
Well answered and candidates coped well with k in the expression.

[3 marks]
4b. The graph of f has a point of inflexion when x = 1.

Show that k = 3.

Markscheme
substituting x = 1 into f ′′ (M1)

eg f ′′ (1), 6(1) − 2k

recognizing f ′′ (x) = 0 (seen anywhere) M1

correct equation A1

eg 6 − 2k = 0

k=3 AG N0

[3 marks]

Examiners report
Mostly answered well with the common error being to substitute into f ′ instead of f ′′ .

[2 marks]
4c. Find f ′ (−2).
 Markscheme
correct substitution into f ′ (x) (A1)

eg 3(−2)2 − 6(−2) − 9

f ′ (−2) = 15 A1 N2

[2 marks]

Examiners report
A straightforward question that was typically answered correctly.

4d. Find the equation of the tangent to the curve of f at (−2, 1), giving your answer in the form y = ax + b. [4 marks]

Markscheme
recognizing gradient value (may be seen in equation) M1

eg a = 15, y = 15x + b

attempt to substitute (−2, 1) into equation of a straight line M1

eg 1 = 15(−2) + b, (y − 1) = m(x + 2), (y + 2) = 15(x − 1)

correct working (A1)

eg 31 = b, y = 15x + 30 + 1

y = 15x + 31 A1 N2

[4 marks]

Examiners report
Some candidates recalculated the gradient, not realising this had already been found in part c). Many understood they were finding a
linear equation but were hampered by arithmetic errors.
 [3 marks]
4e. Given that f ′ (−1) = 0, explain why the graph of f has a local maximum when x = −1.

Markscheme
METHOD 1 (2nd derivative)

recognizing f ′′ < 0 (seen anywhere) R1

′′
substituting x = −1 into f (M1)

eg f ′′ (−1), 6(−1) − 6

f ′′ (−1) = −12 A1

therefore the graph of f has a local maximum when x = −1 AG N0

METHOD 2 (1st derivative)

recognizing change of sign of f ′ (x) (seen anywhere) R1

eg sign chart

correct value of f ′ for −1 < x < 3 A1

′
eg f (0) = −9

correct value of f ′ for x value to the left of −1 A1

eg f ′ (−2) = 15

therefore the graph of f has a local maximum when x = −1 AG N0

[3 marks]

Total [14 marks]

Examiners report
Using change of sign of the first derivative was the most common approach used with a sign chart or written explanation. However, few
candidates then supported their approach by calculating suitable values for f ′ (x). This was necessary because the question already
identified a local maximum, hence candidates needed to explain why this was so. Some candidates did not mention the ‘first derivative’
just that ‘it’ was increasing/decreasing. Few candidates used the more efficient second derivative test.
 [5 marks]
5a. The following diagram shows the graph of a function f . There is a local minimum point at A, where x > 0.

The derivative of f is given by f ′ (x) = 3x2 − 8x − 3.

Find the x-coordinate of A.

Markscheme
recognizing that the local minimum occurs when f ′ (x) = 0 (M1)
2
valid attempt to solve 3x − 8x − 3 = 0 (M1)

eg factorization, formula

correct working A1
8±√64+36
(3x + 1)(x − 3), x = 6

x=3 A2 N3

−1
Note: Award A1 if both values x = 3 , x = 3 are given.

[5 marks]

Examiners report
The majority of candidates approached part (a) correctly, and most recognized that only one solution was possible within the given
domain.
5b. The y-intercept of the graph is at (0, 6). Find an expression for f(x). [6 marks]

m
The graph of a function g is obtained by reflecting the graph of f in the y-axis, followed by a translation of ( ).
n

Markscheme
valid approach (M1)

f(x) = ∫ f ′ (x)dx
f(x) = x3 − 4x2 − 3x + c (do not penalize for missing “+c”) A1A1A1

c=6 (A1)

f(x) = x3 − 4x2 − 3x + 6 A1 N6

[6 marks]

Examiners report
Nearly all candidates answered part (b) correctly, earning all the available marks for integrating the polynomial and solving for C .

6a. Let f(x) = px3 + px2 + qx. [2 marks]

Find f ′ (x).

Markscheme
f ′ (x) = 3px2 + 2px + q A2 N2

Note: Award A1 if only 1 error.

[2 marks]

Examiners report
[N/A]

[5 marks]
6b. Given that f ′ (x) ⩾ 0, show that p2 ⩽ 3pq.
 Markscheme
evidence of discriminant (must be seen explicitly, not in quadratic formula) (M1)

eg b2 − 4ac
correct substitution into discriminant (may be seen in inequality) A1
2 2
eg (2p) − 4 × 3p × q, 4p − 12pq
f ′ (x) ⩾ 0 then f ′ has two equal roots or no roots (R1)
recognizing discriminant less or equal than zero R1
2
eg Δ ⩽ 0, 4p − 12pq ⩽ 0
correct working that clearly leads to the required answer A1
2 2
eg p − 3pq ⩽ 0, 4p ⩽ 12pq
p2 ⩽ 3pq AG N0
[5 marks]

Examiners report

7a. The following diagram shows part of the graph of y = f(x). [4 marks]

The graph has a local maximum at A, where x = −2, and a local minimum at B, where x = 6.

On the following axes, sketch the graph of y = f ′ (x).

 Markscheme

A1A1A1A1 N4

Note: Award A1 for x-intercept in circle at −2, A1 for x-intercept in circle at 6.

Award A1 for approximately correct shape.
Only if this A1 is awarded, award A1 for a negative y-intercept.

[4 marks]

Examiners report
[N/A]

7b. Write down the following in order from least to greatest: f(0), f ′ (6), f ′′ (−2). [2 marks]

Markscheme
f ′′ (−2), f ′ (6), f(0) A2 N2
[2 marks]

Examiners report
[N/A]

[6 marks]
8. Let f(x) = e2x . The line L is the tangent to the curve of f at (1, e2 ).
Find the equation of L in the form y = ax + b.
 Markscheme
recognising need to differentiate (seen anywhere) R1
eg f ′ , 2e2x
attempt to find the gradient when x = 1 (M1)
eg f ′ (1)
f ′ (1) = 2e2 (A1)
attempt to substitute coordinates (in any order) into equation of a straight line (M1)
eg y − e2 = 2e2 (x − 1), e2 = 2e2 (1) + b
correct working (A1)
eg y − e2 = 2e2 x − 2e2 , b = −e2
y = 2e2 x − e2 A1 N3
[6 marks]

Examiners report
[N/A]

[2 marks]
9a. Consider f(x) = ln(x4 + 1) .

Find the value of f(0) .

Markscheme
substitute 0 into f (M1)
eg ln(0 + 1) , ln 1
f(0) = 0 A1 N2
[2 marks]

Examiners report
Many candidates left their answer to part (a) as ln 1. While this shows an understanding for substituting a value into a function, it
leaves an unfinished answer that should be expressed as an integer.

[5 marks]
9b. Find the set of values of x for which f is increasing.
 Markscheme
1
f ′ (x) = x4 +1
× 4x3 (seen anywhere) A1A1
1
Note: Award A1 for x4 +1
and A1 for 4x3 .

recognizing f increasing where f ′ (x) > 0 (seen anywhere) R1

′
eg f (x) > 0 , diagram of signs
attempt to solve f ′ (x) > 0 (M1)
eg 4x3 = 0 , x3 > 0
f increasing for x > 0 (accept x ≥ 0 ) A1 N1
[5 marks]

Examiners report
Candidates who attempted to consider where f is increasing generally understood the derivative is needed. However, a number of
candidates did not apply the chain rule, which commonly led to answers such as “increasing for all x”. Many set their derivative equal to
zero, while neglecting to indicate in their working that f ′ (x) > 0 for an increasing function. Some created a diagram of signs, which
provides appropriate evidence as long as it is clear that the signs represent f ′ .

4x2 (3−x4 ) [5 marks]

9c. The second derivative is given by f ′′ (x) = 2
.
(x4 +1)

The equation f ′′ (x) = 0 has only three solutions, when x = 0 , ±√4 3 (±1.316 …) .

(i) Find f ′′ (1) .

(ii) Hence, show that there is no point of inflexion on the graph of f at x = 0 .

Markscheme
(i) substituting x = 1 into f ′′ (A1)
4(3−1) 4×2
eg , 4
(1+1)2

f ′′ (1) = 2 A1 N2

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

eg no change of sign in f ′′ (x) , no change in concavity,
f ′ increasing both sides of zero
attempt to find f ′′ (x) for x < 0 (M1)
2 4
4(−1) (3−(−1) )
eg f ′′ (−1) , 4 2
, diagram of signs
((−1) +1)

correct working leading to positive value A1

eg f ′′ (−1) = 2 , discussing signs of numerator and denominator
there is no point of inflexion at x = 0 AG N0

[5 marks]
 Examiners report
Finding f ′′ (1) proved no challenge, however, using this value to show that no point of inflexion exists proved elusive for many.
Some candidates recognized the signs must not change in the second derivative. Few candidates presented evidence in the form of
a calculation, which follows from the “hence” command of the question. In this case, a sign diagram without numerical evidence
was not sufficient.

9d. There is a point of inflexion on the graph of f at x = √4 3 (x = 1.316 …) . [3 marks]

Sketch the graph of f , for x ≥ 0 .

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 (0, 0) , A1 for increasing throughout.
Sketch need not be drawn to scale. Only essential features need to be clear.
[3 marks]

Examiners report
Few candidates created a correct graph from the information given or found in the question. This included the point (0, 0), the fact
that the function is always increasing for x > 0 , the concavity at x = 1 and the change in concavity at the given point of inflexion.
Many incorrect attempts showed a graph concave down to the right of x = 0 , changing to concave up.

10a. Consider f(x) = x2 sin x . [4 marks]

Find f ′ (x) .

Markscheme
evidence of choosing product rule (M1)
′ ′
eg uv + vu
correct derivatives (must be seen in the product rule) cos x , 2x (A1)(A1)
f ′ (x) = x2 cos x + 2x sin x A1 N4
[4 marks]
 Examiners report
Many candidates correctly applied the product rule for the derivative, although a common error was to answer f ′ (x) = 2x cos x .

10b. Find the gradient of the curve of f at x = π . [3 marks]

2

Markscheme
π
substituting 2
into their f ′ (x) (M1)
2
eg f ′ ( π2 ) , ( π2 ) cos( π2 ) + 2 ( π2 ) sin( π2 )
π π
correct values for both sin 2
and cos 2
seen in f ′ (x) (A1)

eg 0 + 2 ( π2 ) × 1

f ′ ( π2 ) = π A1 N2

[3 marks]

Examiners report
Candidates generally understood that the gradient of the curve uses the derivative, although in some cases the substitution was
made in the original function. Some candidates did not know the values of sine and cosine at π2 .

11a. Let f(x) = sin x + 1 x2 − 2x , for 0 ≤ x ≤ π . [3 marks]

2

Find f ′ (x) .

Markscheme
f ′ (x) = cos x + x − 2 A1A1A1 N3
Note: Award A1 for each term.
[3 marks]

Examiners report
In part (a), most candidates were able to correctly find the derivative of the function.

[3 marks]
11b. Let g be a quadratic function such that g(0) = 5 . The line x = 2 is the axis of symmetry of the graph of g .

Find g(4) .
 Markscheme
recognizing g(0) = 5 gives the point (0, 5) (R1)
recognize symmetry (M1)
eg vertex, sketch

g(4) = 5 A1 N3
[3 marks]

Examiners report
In part (b), many candidates did not understand the significance of the axis of symmetry and the known point (0, 5), and so were
unable to find g(4) using symmetry. A few used more complicated manipulations of the function, but many algebraic errors were
seen.

[4 marks]
11c. The function g can be expressed in the form g(x) = a(x − h)2 + 3 .

(i) Write down the value of h .

(ii) Find the value of a .

 Markscheme
(i) h=2 A1 N1

(ii) substituting into g(x) = a(x − 2)2 + 3 (not the vertex) (M1)
eg 5 = a(0 − 2)2 + 3 , 5 = a(4 − 2)2 + 3
working towards solution (A1)
eg 5 = 4a + 3 , 4a = 2
1
a= 2
A1 N2

[4 marks]

Examiners report
In part (c), a large number of candidates were able to simply write down the correct value of h, as intended by the command term
in this question. A few candidates wrote down the incorrect negative value. Most candidates attempted to substitute the x and y
values of the known point correctly into the function, but again many arithmetic and algebraic errors kept them from finding the
correct value for a.

11d. Find the value of x for which the tangent to the graph of f is parallel to the tangent to the graph of g . [6 marks]

Markscheme
1 1 2
g(x) = 2
(x − 2)2 + 3 = 2
x − 2x + 5

correct derivative of g A1A1

eg 2 × 12 (x − 2) , x − 2

evidence of equating both derivatives (M1)

′ ′
eg f = g
correct equation (A1)
eg cos x + x − 2 = x − 2
working towards a solution (A1)
eg cos x = 0 , combining like terms
π
x= 2
A1 N0

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

[6 marks]

Examiners report
Part (d) required the candidates to find the derivative of g, and to equate that to their answer from part (a). Although many
candidates were able to simplify their equation to cos x = 0, many did not know how to solve for x at this point. Candidates who
had made errors in parts (a) and/or (c) were still able to earn follow-through marks in part (d).
12a. In this question, you are given that cos π = 1 , and sin π =
√3 [3 marks]
3 2 3 2
.

The displacement of an object from a fixed point, O is given by s(t) = t − sin 2t for 0 ≤ t ≤ π .

Find s′ (t) .

Markscheme
s′ (t) = 1 − 2 cos 2t A1A2 N3
Note: Award A1 for 1, A2 for −2 cos 2t .
[3 marks]

Examiners report
The derivative in part (a) was reasonably well done, but errors here often caused trouble in later parts. Candidates occasionally
attempted to use the double angle identity for sin 2t before differentiating, but they rarely were successful in then applying the
product rule.

12b. In this interval, there are only two values of t for which the object is not moving. One value is t = π . [4 marks]
6

Find the other value.

Markscheme
evidence of valid approach (M1)
′
e.g. setting s (t) = 0
correct working A1
1
e.g. 2 cos 2t = 1 , cos 2t = 2
π 5π
2t = 3
, 3
,… (A1)
5π
t= 6
A1 N3

[4 marks]

Examiners report
In part (b), most candidates understood that they needed to set their derivative equal to zero, but fewer were able to take the next
step to solve the resulting double angle equation. Again, some candidates over-complicated the equation by using the double angle
identity. Few ended up with the correct answer 5π6
.

12c. Show that s′ (t) > 0 between these two values of t . [3 marks]
 Markscheme
evidence of valid approach (M1)
π 5π
e.g. choosing a value in the interval 6
<t< 6

correct substitution A1

e.g. s′ ( π2 ) = 1 − 2 cos π

s′ ( π2 ) = 3 A1

s′ (t) > 0 AG N0
[3 marks]

Examiners report
In part (c), many candidates knew they needed to test a value between π/6 and their value from part (b), but fewer were able to
successfully complete that calculation. Some candidates simply tested their boundary values while others unsuccessfully attempted
to make use of the second derivative.

[5 marks]
12d. Find the distance travelled between these two values of t .

Markscheme
evidence of approach using s or integral of s′ (M1)
5π

e.g. ∫ s′ (t)dt ; s ( 5π
6
) , s ( π6 ) ; [t − sin 2t] π
6

substituting values and subtracting (M1)

√3 √3
e.g. s ( 5π
6
) − s ( π6 ) , ( π6 − 2
) − ( 5π
6
− (− 2
))

correct substitution A1
5π 5π √3 √3
e.g. 6
− sin 3
− [ π6 − sin π3 ] , ( 5π
6
− (− 2
)) − ( π6 − 2
)
2π
distance is 3
+ √3 A1A1 N3
2π
Note: Award A1 for 3
, A1 for √3 .

[5 marks]

Examiners report
Although many candidates did not attempt part (d), those who did often demonstrated a good understanding of how to use the
displacement function s or the integral of their derivative from part (a). Candidates who had made an error in part (b) often could
not finish, as sin(2t) could not be evaluated at their value without a calculator. Of those who had successfully found the other
boundary of 5π/6 , a common error was giving the incorrect sign of the value of sin(5π/3) . Again, this part was a good
discriminator between the grade 6 and 7 candidates.

[1 mark]
13a. Let f(x) = e6x .

Write down f ′ (x) .

 Markscheme
f ′ (x) = 6e6x A1 N1
[1 mark]

Examiners report
On the whole, candidates handled this question quite well with most candidates correctly applying the chain rule to an exponential
function and successfully finding the equation of the tangent line.

[4 marks]
13b. The tangent to the graph of f at the point P(0, b) has gradient m .

(i) Show that m = 6 .

(ii) Find b .

Markscheme
(i) evidence of valid approach (M1)
e.g. f ′ (0) , 6e6×0
correct manipulation A1
e.g. 6e0 , 6 × 1
m = 6 AG N0
(ii) evidence of finding f(0) (M1)
e.g. y = e6(0)
b=1 A1 N2
[4 marks]

Examiners report
On the whole, candidates handled this question quite well with most candidates correctly applying the chain rule to an exponential
function and successfully finding the equation of the tangent line. Some candidates lost a mark in (b)(i) for not showing sufficient
working leading to the given answer.

13c. Hence, write down the equation of this tangent. [1 mark]

Markscheme
y = 6x + 1 A1 N1
[1 mark]

Examiners report
On the whole, candidates handled this question quite well.
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