Challenging +1.2 This is a Further Maths question requiring Maclaurin series manipulation and limit evaluation. Part (a) involves algebraic manipulation of standard series (given in formulae booklet) to derive a general termβsystematic but requires careful index handling. Part (b) uses the derived series to evaluate an indeterminate limit, requiring recognition that substituting the series resolves the 0/0 form. While multi-step and requiring series fluency, both parts follow standard Further Maths techniques without requiring novel insight or particularly deep problem-solving.
Starting from the series given in the formulae booklet, show that the general term of the Maclaurin series for
$$\frac{\sin x}{x} - \cos x$$
is
$$(-1)^{r+1} \frac{2r}{(2r + 1)!} x^{2r}$$
[4 marks]
Show that
$$\lim_{x \to 0} \left[ \frac{\sin x}{x} - \cos x \right] \frac{1}{1 - \cos x} = \frac{2}{3}$$
[4 marks]
Question 11:
--- 11(a) ---
11(a) | Finds the second or
third simplified term of
the series for or
sinπ₯π₯
finds the general term
π₯π₯ | 1.1a | M1 | 2 4 ππ 2ππ
sinπ₯π₯ π₯π₯ π₯π₯ (β1) π₯π₯
= 1β + ββ―+ +β―
π₯π₯ 3! 5! (2ππ+1)!
2 4 ππ 2ππ
π₯π₯ π₯π₯ (β1) π₯π₯
cosπ₯π₯ = 1β + ββ―+
2! 4! (2ππ)!
1 1 1β(2ππ+1)
β =
(2ππ+1)! (2ππ)! (2ππ+1)!
β2ππ
=
(2ππ+1)!
ππ 2ππ ππ 2ππ ππ 2ππ
(β1) π₯π₯ (β1) π₯π₯ (β1) π₯π₯ (β2ππ)
β΄ β =
(2ππ+1)! (2ππ)! (2ππ+1)!
ππ+1 2ππ 2ππ
= (β1) π₯π₯
(2ππ+1)!
Finds general term of
series for with
sinπ₯π₯
2ππ | 1.1a | M1
π₯π₯ π₯π₯
Subtracts general terms
of and
sinπ₯π₯ | 1.1a | M1
π₯π₯ cosπ₯π₯
Completes a rigorous
argument to show the
required result, including
1 1
β
(2ππ+1)! (2ππ)!
1β(2ππ+1)
=
(2ππ+1)!
β2ππ
AG
= | 2.1 | R1
--- 11(b) ---
11(b) | (2ππ+1)!
Selects a method to
determine the value of
the limit by finding the
first non-zero term of
o ne series. | 3.1a | M1 | First non-zero terms of series expansion of
are and
2 4
sinπ₯π₯ π₯π₯ π₯π₯
π₯π₯ βcosπ₯π₯ 3 β30
First non-zero terms of series expansion of
are and
2 4
π₯π₯ π₯π₯
1βcosπ₯π₯ 2 β24
2 4 2
π₯π₯ π₯π₯ 1 π₯π₯
β +β― β +β― 1 οΏ½
3 30 3 30 3
π₯π₯liβm0οΏ½ 2 4 οΏ½ =π₯π₯liβm0οΏ½ 2 οΏ½ =
π₯π₯ π₯π₯ 1 π₯π₯ 1 οΏ½
β +β― β +β― 2
2 24 2 24
2
=
3
Explains or shows that
each series has terms in
h igher powers of x | 2.4 | E1
Deduces that the
required limit can be
determined by dividing
numerator and
denominator by the
lowest power of x
or by using lβHΓ΄pitalβs
r ule. | 2.2a | M1
Completes a rigorous
argument to show the
required result
sinπ₯π₯
βcosπ₯π₯ 2
π₯π₯
π₯π₯liβm0οΏ½ οΏ½ =
1βcosπ₯π₯ 3 | 2.1 | R1
Total | 8
Q | Marking Instructions | AO | Marks | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Starting from the series given in the formulae booklet, show that the general term of the Maclaurin series for
$$\frac{\sin x}{x} - \cos x$$
is
$$(-1)^{r+1} \frac{2r}{(2r + 1)!} x^{2r}$$
[4 marks]
\item Show that
$$\lim_{x \to 0} \left[ \frac{\sin x}{x} - \cos x \right] \frac{1}{1 - \cos x} = \frac{2}{3}$$
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2020 Q11 [8]}}