Show that \(A = 10\) .

Given that \(f(15) = 3.49\) (correct to 3 significant figures), find the value of k.

(i)     Using your value of k , find \(f'(x)\) .

(ii)    Hence, explain why f is a decreasing function.

(iii)   Write down the equation of the horizontal asymptote of the graph f .

Let \(g(x) = - {x^2} + 12x - 24\) .

Find the area enclosed by the graphs of f and g .

substituting (0, 13) into function     M1 

e.g. \(13 = A{{\rm{e}}^0} + 3\)

\(13 = A + 3\)     A1

\(A = 10\)     AG     N0

[2 marks]

substituting into \(f(15) = 3.49\)     A1

e.g. \(3.49 = 10{{\rm{e}}^{15k}} + 3\) , \(0.049 = {{\rm{e}}^{15k}}\)

evidence of solving equation     (M1)

e.g. sketch, using \(\ln \)

\(k = - 0.201\) (accept \(\frac{{\ln 0.049}}{{15}}\) )     A1     N2

[3 marks]

(i) \(f(x) = 10{{\rm{e}}^{ - 0.201x}} + 3\)

\(f(x) = 10{{\rm{e}}^{ - 0.201x}} \times - 0.201\) \(( = - 2.01{{\rm{e}}^{ - 0.201x}})\)     A1A1A1     N3

Note: Award A1 for \(10{{\rm{e}}^{ - 0.201x}}\) , A1 for \( \times - 0.201\) , A1 for the derivative of 3 is zero.

(ii) valid reason with reference to derivative     R1     N1

e.g. \(f'(x) < 0\) , derivative always negative

(iii) \(y = 3\)     A1     N1

[5 marks]

finding limits \(3.8953 \ldots \), \(8.6940 \ldots \) (seen anywhere)     A1A1

evidence of integrating and subtracting functions     (M1)

correct expression     A1

e.g. \(\int_{3.90}^{8.69} {g(x) - f(x){\rm{d}}x} \) , \(\int_{3.90}^{8.69} {\left[ {\left( { - {x^2} + 12x - 24} \right) - \left( {10{{\rm{e}}^{ - 0.201x}} + 3} \right)} \right]} {\rm{d}}x\)

area \(= 19.5\)     A2     N4

[6 marks]

This question was quite well done by a great number of candidates indicating that calculus is a topic that is covered well by most centres. Parts (a) and (b) proved very accessible to many candidates.

This question was quite well done by a great number of candidates indicating that calculus is a topic that is covered well by most centres. Parts (a) and (b) proved very accessible to many candidates.

The chain rule in part (c) was also carried out well. Few however, recognized the command term “hence” and that \(f'(x) < 0\) guarantees a decreasing function. A common answer for the equation of the asymptote was to give \(y = 0\) or \(x = 3\) .

In part (d), it was again surprising and somewhat disappointing to see how few candidates were able to use their GDC effectively to find the area between curves, often not finding correct limits, and often trying to evaluate the definite integral without the GDC, which led nowhere.