State the mean value theorem for a function that is continuous on the closed interval \([a,{\text{ }}b]\) and differentiable on the open interval \(]a,{\text{ }}b[\).

(i)     Find \(g(0)\).

(ii)     Find \(g(h)\).

(iii)     Apply the mean value theorem to the function \(g(x)\) on the closed interval \([0,{\text{ }}h]\) to show that there exists \(c\) in the open interval \(]0,{\text{ }}h[\) such that \(g'(c) = 0\).

(iv)     Find \(g'(x)\).

(v)     Hence show that \( - (h - c)f''(c) + \frac{{2(h - c)}}{{{h^2}}}\left( {f(h) - f(0) - hf'(0)} \right) = 0\).

(vi)     Deduce that \(f(h) = f(0) + hf'(0) + \frac{{{h^2}}}{2}{\text{ }}f''(c)\).

Hence show that, for \(h > 0\)

\(1 - \cos (h) \leqslant \frac{{{h^2}}}{2}\).

there exists \(c\) in the open interval \(]a,{\text{ }}b[\) such that     A1

\(\frac{{f(b) - f(a)}}{{b - a}} = f'(c)\)    A1

Note: Open interval is required for the A1.

[2 marks]

(i)     \(g(0) = f(h) - f(0) - hf'(0) - \frac{{{h^2}}}{{{h^2}}}\left( {{\text{ }}f(h) - f(0) - hf'(0)} \right)\)

\( = 0\)    A1

(ii)     \(g(h) = f(h) - f(h) - 0 - 0\)

\( = 0\)    A1

(iii)     (\(g(x)\) is a differentiable function since it is a combination of other differentiable functions \(f\), \({f'}\) and polynomials.)

there exists \(c\) in the open interval \(]0,{\text{ }}h[\) such that

\(\frac{{g(h) - g(0)}}{h} = g'(c)\)    A1

\(\frac{{g(h) - g(0)}}{h} = 0\)    A1

hence \(g'(c) = 0\)     AG

(iv)     \(g'(x) =  - f'(x) + f'(x) - (h - x)f''(x) + \frac{{2(h - x)}}{{{h^2}}}\left( {f(h) - f(0) - hf'(0)} \right)\)     A1A1

Note: A1 for the second and third terms and A1 for the other terms (all terms must be seen).

\( =  - (h - x)f''(x) + \frac{{2(h - x)}}{{{h^2}}}\left( {f(h) - f(0) - hf'(0)} \right)\)

(v)     putting \(x = c\) and equating to zero     M1

\( - (h - c)f''(c) + \frac{{2(h - c)}}{{{h^2}}}\left( {f(h) - f(0) - hf'(0)} \right) = g'(c) = 0\)    AG

(vi)     \( - f''(c) + \frac{2}{{{h^2}}}\left( {f(h) - f(0) - hf'(0)} \right) = 0\)     A1

since \(h - c \ne 0\)     R1

\(\frac{{{h^2}}}{2}f''(c) = f(h) - f(0) - hf'(0)\)

\(f(h) = f(0) + hf'(0) + \frac{{{h^2}}}{2}f''(c)\)    AG

[9 marks]

letting \(f(x) = \cos (x)\)     M1

\(f'(x) =  - \sin (x)\)    \(f''(x) =  - \cos (x)\)     A1

\(\cos (h) = 1 + 0 - \frac{{{h^2}}}{2}\cos (c)\)     A1

\(1 - \cos (h) = \frac{{{h^2}}}{2}\cos (c)\)    (A1)

since \(\cos (c) \leqslant 1\)     R1

\(1 - \cos (h) \leqslant \frac{{{h^2}}}{2}\)    AG

Note: Allow \(f(x) = a \pm b\cos x\).

[5 marks]