Standard +0.8 This is a straightforward application of l'Hôpital's rule requiring verification of indeterminate form, differentiation of trigonometric functions, and evaluation at the limit. While it's Further Maths content, the execution is mechanical with no conceptual subtlety—students simply differentiate numerator and denominator once and substitute. The 5 marks reflect standard working rather than genuine difficulty.
Question 13:
13 | Explains that numerator and
denominator are both zero at
x=π | 2.4 | E1 | Let
f(x)= xsin2x
g(x)=cos (x)
2
Then
f(π)=0and g(π)=0
So, by l’Hôpital’s rule,
f(x) f '(x)
lim =lim
x→πg(x) x→πg'(x)
f '(x)=2xcos2x+sin2x
g'(x)=−1sin (x)
2 2
f '(π)=2πcos2π+sin2π=2π
g'(π)=−1sin (π)=−1
2 2 2
f(x) f '(x) 2π
lim =lim = =−4π
x→πg(x) x→πg'(x) −1
2
Obtains derivatives of
numerator and denominator | 1.1a | M1
Correctly evaluates both
correct derivatives of
numerator and denominator
at x=π, may be
unsimplified | 1.1b | A1
f '(π)
Forms their
g'(π) | 1.1a | M1
Completes a reasoned
argument using l’Hôpital’s
rule to obtain the required
result, including a clear
demonstration of the limiting
process.
Can score E0 M1A1M1R1 | 2.1 | R1
Total | 5
Q | Marking instructions | AO | Marks | Typical solution
Use l'Hôpital's rule to prove that
$$\lim_{x \to \pi} \frac{x \sin 2x}{\cos\left(\frac{x}{2}\right)} = -4\pi$$
[5 marks]
\hfill \mbox{\textit{AQA Further Paper 1 2023 Q13 [5]}}