Write down \(f(0)\) .

Solve \(f(x) = 95\) .

Find the range of \(f\) .

Show that \(f'(x) = \frac{{1000{{\rm{e}}^{ - 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ - 0.2x}})}^2}}}\) .

Find the maximum rate of change of \(f\) .

\(f(0) = \frac{{100}}{{51}}\) (exact), \(1.96\)     A1     N1

[1 mark]

setting up equation     (M1)

eg   \(95 = \frac{{100}}{{1 + 50{{\rm{e}}^{ - 0.2x}}}}\) , sketch of graph with horizontal line at \(y = 95\)

\(x = 34.3\)     A1     N2

[2 marks]

upper bound of \(y\) is \(100\)     (A1)

lower bound of \(y\) is \(0\)     (A1)

range is \(0 < y < 100\)     A1     N3

[3 marks]

METHOD 1

setting function ready to apply the chain rule     (M1)

eg   \(100{(1 + 50{{\rm{e}}^{ - 0.2x}})^{ - 1}}\) 

evidence of correct differentiation (must be substituted into chain rule)     (A1)(A1)

eg   \(u' = - 100{(1 + 50{{\rm{e}}^{ - 0.2x}})^{ - 2}}\) , \(v' = (50{{\rm{e}}^{ - 0.2x}})( - 0.2)\) 

correct chain rule derivative     A1

eg   \(f'(x) = - 100{(1 + 50{{\rm{e}}^{ - 0.2x}})^{ - 2}}(50{{\rm{e}}^{ - 0.2x}})( - 0.2)\) 

correct working clearly leading to the required answer     A1

eg   \(f'(x) = 1000{{\rm{e}}^{ - 0.2x}}{(1 + 50{{\rm{e}}^{ - 0.2x}})^{ - 2}}\) 

\(f'(x) = \frac{{1000{{\rm{e}}^{ - 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ - 0.2x}})}^2}}}\)     AG     N0

METHOD 2

attempt to apply the quotient rule (accept reversed numerator terms)     (M1)

eg   \(\frac{{vu' - uv'}}{{{v^2}}}\) , \(\frac{{uv' - vu'}}{{{v^2}}}\)

evidence of correct differentiation inside the quotient rule     (A1)(A1)

eg   \(f'(x) = \frac{{(1 + 50{{\rm{e}}^{ - 0.2x}})(0) - 100(50{{\rm{e}}^{ - 0.2x}} \times  - 0.2)}}{{{{(1 + 50{{\rm{e}}^{ - 0.2x}})}^2}}}\) , \(\frac{{100( - 10){{\rm{e}}^{ - 0.2x}} - 0}}{{{{(1 + 50{{\rm{e}}^{ - 0.2x}})}^2}}}\)

any correct expression for derivative (\(0\) may not be explicitly seen)     (A1)

eg   \(\frac{{ - 100(50{{\rm{e}}^{ - 0.2x}} \times  - 0.2)}}{{{{(1 + 50{{\rm{e}}^{ - 0.2x}})}^2}}}\)

correct working clearly leading to the required answer     A1

eg   \(f'(x) = \frac{{0 - 100( - 10){{\rm{e}}^{ - 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ - 0.2x}})}^2}}}\) , \(\frac{{ - 100( - 10){{\rm{e}}^{ - 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ - 0.2x}})}^2}}}\)

\(f'(x) = \frac{{{\rm{1000}}{{\rm{e}}^{ - 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ - 0.2x}})}^2}}}\)     AG     N0

[5 marks]

METHOD 1

sketch of \(f'(x)\)     (A1)

eg

recognizing maximum on \(f'(x)\)     (M1)

eg dot on max of sketch

finding maximum on graph of \(f'(x)\)     A1

eg   (\(19.6\), \(5\)) , \(x = 19.560 \ldots \)

maximum rate of increase is \(5\)     A1 N2

METHOD 2

recognizing \(f''(x) = 0\)     (M1)

finding any correct expression for  \(f''(x) = 0\)     (A1)

eg   \(\frac{{{{(1 + 50{{\rm{e}}^{ - 0.2x}})}^2}( - 200{{\rm{e}}^{ - 0.2x}}) - (1000{{\rm{e}}^{ - 0.2x}})(2(1 + 50{{\rm{e}}^{ - 0.2x}})( - 10{{\rm{e}}^{ - 0.2x}}))}}{{{{(1 + 50{{\rm{e}}^{ - 0.2x}})}^4}}}\)

finding \(x = 19.560 \ldots \)     A1

maximum rate of increase is \(5\)     A1     N2

[4 marks]

Candidates had little difficulty with parts (a), (b) and (c).

Candidates had little difficulty with parts (a), (b) and (c). Successful analytical approaches were often used in part (b) but again, this was not the most efficient or expected method.

Candidates had little difficulty with parts (a), (b) and (c). In part (c), candidates gained marks by correctly identifying upper and lower bounds but often did not express them properly using an appropriate notation.

In part (d), the majority of candidates opted to use the quotient rule and did so with some degree of competency, but failed to recognize the command term “show that” and consequently did not show enough to gain full marks. Approaches involving the chain rule were also successful but with the same point regarding sufficiency of work.

Part (e) was poorly done as most were unable to interpret what was required. There were a few responses involving the use of the “trace” feature of the GDC which often led to inaccurate answers and a number of candidates incorrectly reported \(x = 19.6\) as their final answer. Some found the maximum value of \(f\) rather than \(f'\).