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Date May 2016 Marks available 7 Reference code 16M.1.sl.TZ1.10
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 10 Adapted from N/A

Question

Let \(f(x) = \sqrt {4x + 5} \), for \(x \geqslant  - 1.25\).

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 graph at R is \(y = 3x + 6\).

Find \(f'(1)\).

[4]
a.

Write down \(g'(1)\).

[2]
b.

Find \(g(1)\).

[2]
c.

Let \(h(x) = f(x) \times g(x)\). Find the equation of the tangent to the graph of \(h\) at the point where \(x = 1\).

[7]
d.

Markscheme

choosing chain rule     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}u = 4x + 5,{\text{ }}u' = 4\)

correct derivative of \(f\)     A2

eg\(\,\,\,\,\,\)\(\frac{1}{2}{(4x + 5)^{ - \frac{1}{2}}} \times 4,{\text{ }}f'(x) = \frac{2}{{\sqrt {4x + 5} }}\)

\(f'(1) = \frac{2}{3}\)    A1     N2

[4 marks]

a.

recognize that \(g'(x)\) is the gradient of the tangent     (M1)

eg\(\,\,\,\,\,\)\(g'(x) = m\)

\(g'(1) = 3\)    A1     N2

[2 marks]

b.

recognize that R is on the tangent     (M1)

eg\(\,\,\,\,\,\)\(g(1) = 3 \times 1 + 6\), sketch

\(g(1) = 9\)    A1     N2

[2 marks]

c.

\(f(1) = \sqrt {4 + 5} {\text{ }}( = 3)\) (seen anywhere)     A1

\(h(1) = 3 \times 9{\text{ }}( = 27)\) (seen anywhere)     A1

choosing product rule to find \(h'(x)\)     (M1)

eg\(\,\,\,\,\,\)\(uv' + u'v\)

correct substitution to find \(h'(1)\)     (A1)

eg\(\,\,\,\,\,\)\(f(1) \times g'(1) + f'(1) \times g(1)\)

\(h'(1) = 3 \times 3 + \frac{2}{3} \times 9{\text{ }}( = 15)\)     A1

EITHER

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

eg\(\,\,\,\,\,\)\(y - 27 = h'(1)(x - 1),{\text{ }}y - 1 = 15(x - 27)\)

\(y - 27 = 15(x - 1)\)     A1     N2

OR

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

eg\(\,\,\,\,\,\)\(27 = 15 \times 1 + b,{\text{ }}1 = 15 \times 27 + b\)

\(y = 15x + 12\)     A1     N2

[7 marks]

d.

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.

a.

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.

b.

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.

c.

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 \(h'(x) = f'(x) \times g'(x)\). 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.

d.

Syllabus sections

Topic 6 - Calculus » 6.2 » The chain rule for composite functions.
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