The mean value theorem states that if \(f\) is a continuous function on \([a,{\text{ }}b]\) and differentiable on \(]a,{\text{ }}b[\) then \(f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\) for some \(c \in ]a,{\text{ }}b[\).

(i)     Find the two possible values of \(c\) for the function defined by \(f(x) = {x^3} + 3{x^2} - 2\) on the interval \([ - 3,{\text{ }}1]\).

(ii)     Illustrate this result graphically.

(i)     The function \(f\) is continuous on \([a,{\text{ }}b]\), differentiable on \(]a,{\text{ }}b[\) and \(f'(x) = 0\) for all \(x \in ]a,{\text{ }}b[\). Show that \(f(x)\) is constant on \([a,{\text{ }}b]\).

(ii)     Hence, prove that for \(x \in [0,{\text{ }}1],{\text{ }}2\arccos x + \arccos (1 - 2{x^2}) = \pi \).

(i)     \(f'(x) = 3{x^2} + 6x\)     A1

gradient of chord \( = 1\)     A1

\(3{c^2} + 6c = 1\)

\(c = \frac{{ - 3 \pm 2\sqrt 3 }}{3}{\text{ }}( =  - 2.15,{\text{ }}0.155)\)     A1A1

Note:     Accept any answers that round to the correct 2sf answers \(( - 2.2,{\text{ }}0.15)\).

(ii)     

award A1 for correct shape and clear indication of correct domain, A1 for chord (from \(x =  - 3\) to \(x = 1\)) and A1 for two tangents drawn at their values of \(c\)     A1A1A1

[7 marks]

(i)     METHOD 1

(if a theorem is true for the interval \([a,{\text{ }}b]\), it is also true for any interval \([{x_1},{\text{ }}{x_2}]\) which belongs to \([a,{\text{ }}b]\))

suppose \({x_1},{\text{ }}{x_2} \in [a,{\text{ }}b]\)     M1

by the \(MVT\), there exists \(c\) such that \(f'(c) = \frac{{f({x_2}) - f({x_1})}}{{{x_2} - {x_1}}} = 0\)     M1A1

hence \(f({x_1}) = f({x_2})\)     R1

as \({x_1},{\text{ }}{x_2}\) are arbitrarily chosen, \(f(x)\) is constant on \([a,{\text{ }}b]\)

Note:     If the above is expressed in terms of \(a\) and \(b\) award M0M1A0R0.

METHOD 2

(if a theorem is true for the interval \([a,{\text{ }}b]\), it is also true for any interval \([{x_1},{\text{ }}{x_2}]\) which belongs to \([a,{\text{ }}b]\))

suppose \(x \in [a,{\text{ }}b]\)     M1

by the \(MVT\), there exists \(c\) such that \(f'(c) = \frac{{f(x) - f(a)}}{{x - a}} = 0\)     M1A1

hence \(f(x) = f(a) = \) constant     R1

(ii)     attempt to differentiate \((x) = 2\arccos x + \arccos (1 - 2{x^2})\)     M1

\( - 2\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{{ - 4x}}{{\sqrt {1 - {{(1 - 2{x^2})}^2}} }}\)     A1A1

\( =  - 2\frac{1}{{\sqrt {1 - {x^2}} }} + \frac{{4x}}{{\sqrt {4{x^2} - 4{x^4}} }} = 0\)     A1

Note:     Only award A1 for \(0\) if a correct attempt to simplify the denominator is also seen.

\(f(x) = f(0) = 2 \times \frac{\pi }{2} + 0 = \pi \)     A1AG

Note:     This A1 is not dependent on previous marks.

Note:     Allow any value of \(x \in [0,{\text{ }}1]\).

[9 marks]

Total [16 marks]

(i)     This was well done by most candidates.

(ii)     This was generally poorly done, with many candidates failing to draw the curve correctly as they did not appreciate the importance of the given domain. Another common error was to draw the graph of the derivative rather than the function.

(i)     This was very poorly done. A lot of the arguments seemed to be stating what was being required to be proved, eg ‘because the derivative is equal to 0 the line is flat’. Most candidates did not realise the importance of testing a point inside the interval, so the most common solutions seen involved the Mean Value Theorem applied to the end points. In addition there was some confusion between the Mean Value Theorem and Rolle’s Theorem.

(ii)     It was pleasing that so many candidates spotted the link with the previous part of the question. The most common error after this point was to differentiate incorrectly. Candidates should be aware this is a ‘prove’ question, and so it was not sufficient simply to state, for example, \(f(0) = \pi \).