New test - November 03, 2015
[79 marks]
Let f(x) = e2x cos x , −1 ≤ x ≤ 2 .
′
2x
1a. Show that f (x) = e (2 cos x − sin x) .
[3 marks]
Markscheme
correctly finding the derivative of e2x , i.e. 2e2x
A1
correctly finding the derivative of cos x , i.e. − sin x
evidence of using the product rule, seen anywhere
A1
M1
e.g. f ′ (x) = 2e2x cos x − e2x sin x
f ′ (x) = 2e2x (2 cos x − sin x)
AG
N0
[3 marks]
Examiners report
A good number of candidates demonstrated the ability to apply the product and chain rules to obtain the given derivative.
1b. Let the line L be the normal to the curve of f at x = 0 .
[5 marks]
Find the equation of L .
Markscheme
evidence of finding f(0) = 1 , seen anywhere
(M1)
attempt to find the gradient of f
e.g. substituting x = 0 into
f ′ (x)
value of the gradient of f
A1
e.g.
f ′ (0)
A1
= 2 , equation of tangent is y = 2x + 1
gradient of normal = − 12
(A1)
y − 1 = − 12 x (y = − 12 x + 1)
A1
N3
[5 marks]
Examiners report
Where candidates recognized that the gradient of the tangent is the derivative, many went on to correctly find the equation of the
normal.
1c. The graph of f and the line L intersect at the point (0, 1) and at a second point P.
(i)
Find the x-coordinate of P.
(ii) Find the area of the region enclosed by the graph of f and the line L .
[6 marks]
Markscheme
(i) evidence of equating correct functions
e.g.
e2x
cos x =
x = 1.56
− 12 x + 1
A1
M1
, sketch showing intersection of graphs
N1
(ii) evidence of approach involving subtraction of integrals/areas
(M1)
e.g. ∫ [f(x) − g(x)]dx , ∫ f(x)dx − area under trapezium
fully correct integral expression
e.g.
1.56
∫0
A2
[e2x cos x − (− 12 x + 1)]dx
area = 3.28
A1
1.56 2x
e
, ∫0
cos xdx − 0.951 …
N2
[6 marks]
Examiners report
Few candidates showed the setup of the equation in part (c) before writing their answer from the GDC. Although a good number of
candidates correctly expressed the integral to find the area between the curves, surprisingly few found a correct answer. Although this
is a GDC paper, some candidates attempted to integrate this function analytically.
Let f(x) = ex (1 − x2 ) .
2a.
Show that f ′ (x) = ex (1 − 2x − x2 ) .
[3 marks]
Markscheme
M1
evidence of using the product rule
f ′ (x) = ex (1 − x2 ) + ex (−2x)
A1A1
Note: Award A1 for ex (1 − x2 ) , A1 for ex (−2x) .
f ′ (x) = ex (1 − 2x − x2 )
AG
N0
[3 marks]
Examiners report
Many candidates clearly applied the product rule to correctly show the given derivative. Some candidates missed the multiplicative
nature of the function and attempted to apply a chain rule instead.
y = f(x)
Part of the graph of y = f(x), for −6 ≤ x ≤ 2 , is shown below. The x-coordinates of the local minimum and maximum points are r
and s respectively.
2b. Write down the equation of the horizontal asymptote.
[1 mark]
Markscheme
y=0
A1
N1
[1 mark]
Examiners report
For part (b), the equation of the horizontal asymptote was commonly written as x = 0 .
2c. Write down the value of r and of s.
[4 marks]
Markscheme
at the local maximum or minimum point
f ′ (x) = 0 (ex (1 − 2x − x2 ) = 0)
⇒
1 − 2x − x2
=0
r = −2.41 s = 0.414
(M1)
(M1)
A1A1
N2N2
[4 marks]
Examiners report
Although part (c) was a “write down” question where no working is required, a good number of candidates used an algebraic method
of solving for r and s which sometimes returned incorrect answers. Those who used their GDC usually found correct values, although
not always to three significant figures.
2d. Let L be the normal to the curve of f at P(0, 1) . Show that L has equation x + y = 1 .
[4 marks]
Markscheme
f ′ (0) = 1
A1
gradient of the normal = −1
A1
evidence of substituting into an equation for a straight line
correct substitution
(M1)
A1
e.g. y − 1 = −1(x − 0) , y − 1 = −x , y = −x + 1
x+y = 1
AG
N0
[4 marks]
Examiners report
In part (d), many candidates showed some skill showing the equation of a normal, although some tried to work with the gradient of
the tangent.
2e.
Let R be the region enclosed by the curve y = f(x) and the line L.
(i)
[5 marks]
Find an expression for the area of R.
(ii) Calculate the area of R.
Markscheme
(i) intersection points at x = 0 and x = 1 (may be seen as the limits)
approach involving subtraction and integrals
fully correct expression
A2
(A1)
(M1)
N4
e.g. ∫01 (ex (1 − x2 ) − (1 − x))dx , ∫01 f(x)dx − ∫01 (1 − x)dx
(ii) area R = 0.5
A1
N1
[5 marks]
Examiners report
Surprisingly few candidates set up a completely correct expression for the area between curves that considered both integration and
the correct subtraction of functions. Using limits of −6 and 2 was a common error, as was integrating on f(x) alone. Where
candidates did write a correct expression, many attempted to perform analytic techniques to calculate the area instead of using their
GDC.
Let f(x) = e6x .
3a.
Write down f ′ (x) .
[1 mark]
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.
P(0, b)
3b.
The tangent to the graph of f at the point P(0, b) has gradient m .
(i)
[4 marks]
Show that m = 6 .
(ii) Find b .
Markscheme
(i) evidence of valid approach
e.g.
f ′ (0)
,
(M1)
6e6×0
correct manipulation
A1
e.g. 6e0 , 6 × 1
m=6
AG
N0
(ii) evidence of finding f(0)
e.g. y =
e6(0)
b=1
A1
(M1)
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.
3c.
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.
4.
Let f(x) = kx4 . The point P(1, k) lies on the curve of f . At P, the normal to the curve is parallel to y = − 18 x . Find the value [6 marks]
of k.
Markscheme
gradient of tangent = 8 (seen anywhere)
f ′ (x) = 4kx3 (seen anywhere)
(A1)
A1
recognizing the gradient of the tangent is the derivative
setting the derivative equal to 8
(A1)
e.g. 4kx3 = 8 , kx3 = 2
substituting x = 1 (seen anywhere)
k=2
A1
[6 marks]
N4
(M1)
(M1)
Examiners report
Candidates‟ success with this question was mixed. Those who understood the relationship between the derivative and the gradient of
the normal line were not bothered by the lack of structure in the question, solving clearly with only a few steps, earning full marks.
Those who were unclear often either gained a few marks for finding the derivative and substituting x = 1 , or no marks for working
that did not employ the derivative. Misunderstandings included simply finding the equation of the tangent or normal line, setting the
derivative equal to the gradient of the normal, and equating the function with the normal or tangent line equation. Among the
candidates who demonstrated greater understanding, more used the gradient of the normal (the equation − 14 k = − 18 ) than the
gradient of the tangent (4k = 8 ) ; this led to more algebraic errors in obtaining the final answer of k = 2 . A number of unsuccessful
candidates wrote down a lot of irrelevant mathematics with no plan in mind and earned no marks.
5.
Consider the curve with equation f(x) = px2 + qx , where p and q are constants. The point A(1, 3) lies on the curve. The
tangent to the curve at A has gradient 8. Find the value of p and of q .
[7 marks]
Markscheme
substituting x = 1 , y = 3 into f(x)
3 = p+q
(M1)
A1
(M1)
finding derivative
f ′ (x) = 2px + q
A1
correct substitution, 2p + q = 8
p = 5 , q = −2
A1A1
A1
N2N2
[7 marks]
Examiners report
A good number of candidates were able to obtain an equation by substituting the point 1, 3 into the function’s equation. Not as many
knew how to find the other equation by using the derivative. Some candidates thought they needed to find the equation of the tangent
line rather than recognising that the information about the tangent provided the gradient of the function at the point. While they were
usually able to find this equation correctly, it was irrelevant to the question asked.
6.
Let f(x) = ex cos x . Find the gradient of the normal to the curve of f at x = π .
[6 marks]
Markscheme
evidence of choosing the product rule
f ′ (x)
=
ex
× (− sin x) + cos x × ex
substituting π
e.g.
f ′ (π)
=
(= ex cos x − ex sin x)
A1A1
(M1)
eπ cos π − eπ sin π
taking negative reciprocal
e.g. −
(M1)
, eπ (−1 − 0) , −eπ
(M1)
1
f ′ (π)
gradient is
1
eπ
A1
N3
[6 marks]
Examiners report
Candidates familiar with the product rule easily found the correct derivative function. Many substituted π to find the tangent gradient,
but surprisingly few candidates correctly considered that the gradient of the normal is the negative reciprocal of this answer.
y = ln(3x − 1)
Consider the curve y = ln(3x − 1) . Let P be the point on the curve where x = 2 .
7a. Write down the gradient of the curve at P.
[2 marks]
Markscheme
gradient is 0.6
A2
N2
[2 marks]
Examiners report
Although the command term "write down" was used in part (a), many candidates still opted for an analytic method for finding the
derivative value. Although this value was often incorrect, many candidates knew how to find the equation of the normal and earned
follow through marks in part (b).
7b.
The normal to the curve at P cuts the x-axis at R. Find the coordinates of R.
[5 marks]
Markscheme
at R, y = 0 (seen anywhere)
A1
at x = 2 , y = ln 5 (= 1.609 …)
(A1)
gradient of normal = −1.6666 …
(A1)
evidence of finding correct equation of normal
A1
e.g. y = ln 5 = − 53 (x − 2) , y = −1.67x + c
x = 2.97 (accept 2.96)
A1
coordinates of R are (2.97,0)
N3
[5 marks]
Examiners report
Although the command term "write down" was used in part (a), many candidates still opted for an analytic method for finding the
derivative value. Although this value was often incorrect, many candidates knew how to find the equation of the normal and earned
follow through marks in part (b).
Let f(x) = sin x + 12 x2 − 2x , for 0 ≤ x ≤ π .
8a.
Find f ′ (x) .
[3 marks]
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.
g
g(0) = 5
x=2
g
g(0) = 5 . The line
g
8b.
g
Find g(4) .
[3 marks]
Markscheme
recognizing g(0) = 5 gives the point (0, 5)
recognize symmetry
(R1)
(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.
The function g can be expressed in the form g(x) = a(x − h)2 + 3 .
8c.
(i)
Write down the value of h .
(ii)
Find the value of a .
[4 marks]
Markscheme
(i)
h=2
(ii)
substituting into g(x) = a(x − 2)2 + 3 (not the vertex)
A1 N1
eg 5 = a(0 − 2)2 + 3 , 5 = a(4 − 2)2 + 3
working towards solution
eg 5 = 4a + 3 , 4a = 2
a=
1
2
A1
[4 marks]
N2
(A1)
(M1)
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.
8d.
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
g(x) = 12 (x − 2)2 + 3 = 12 x2 − 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
x=
π
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).
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