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]