AQA Paper 1 Specimen — Question 17 6 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiation from First Principles
TypeFirst principles: trigonometric functions
DifficultyChallenging +1.2 This question requires applying the first principles definition of the derivative to a trigonometric function, using the compound angle formula for sin(A+h), and evaluating standard limits (sin h/h → 1, (cos h - 1)/h → 0). While it's a multi-step proof requiring careful algebraic manipulation and knowledge of key limits, it's a standard textbook exercise in differentiation from first principles with no novel insight required. The 6 marks reflect the length rather than exceptional difficulty.
Spec1.07g Differentiation from first principles: for small positive integer powers of x
\(f(x) = \sin x\) Using differentiation from first principles find the exact value of \(f'\left(\frac{\pi}{6}\right)\) Fully justify your answer. [6 marks]
Question 17:
AnswerMarks
17
Translatesf '  into
6
  
sin h  sin 
6  6
lim
AnswerMarks Guidance
h0 hAO1.1a M1
 sin h  sin 
 6  6
f ' lim  
 
6 h0  h 
 
 
   
 sin coshcos sinhsin 
6 6 6
lim 
h0  h 
 
 
1 3 1
 cosh sinh 
2 2 2
lim 
h0  h 
 
 
1cosh1 3 sinh
lim    
h0 2 h  2 h 
  h 
2sin2
    
1 2 3 sinh
lim   
h0 2 2h  2 h 
   
  2  
 h h
 sin  sin 
2 2 3 sinh
lim lim  lim
 h0 2  h0 h  2 h0 h
    
  2 
3
01 1
2
3
2
Uses sin (A + B) identity to
 
replace sin h , to
6 
commence argument
(at least two lines of argument
AnswerMarks Guidance
seen)AO2.1 M1
Obtains correct two term
expression involving cos h and
AnswerMarks Guidance
sin hAO1.1b A1
Deduce what happens as h0,
for one part of ‘their’ expression
sinh
using the limit of
h
OR by using small angles
AnswerMarks Guidance
approximationsAO2.2a R1
Deduce what happens as h0 ,
for the second part of ‘their’
expression using the limit of
(cos h – 1)h
OR by using small angle
AnswerMarks Guidance
approximationsAO2.2a R1
Completes a rigorous argument
leading to the correct exact
value, with all the steps in the
AnswerMarks Guidance
method clearly shown.AO2.1 R1
Total6
TOTAL100 
Question 17:
17 | 
Translatesf '  into
6
  
sin h  sin 
6  6
lim
h0 h | AO1.1a | M1 |    
 sin h  sin 
 6  6
f ' lim  
 
6 h0  h 
 
 
   
 sin coshcos sinhsin 
6 6 6
lim 
h0  h 
 
 
1 3 1
 cosh sinh 
2 2 2
lim 
h0  h 
 
 
1cosh1 3 sinh
lim    
h0 2 h  2 h 
  h 
2sin2
    
1 2 3 sinh
lim   
h0 2 2h  2 h 
   
  2  
 h h
 sin  sin 
2 2 3 sinh
lim lim  lim
 h0 2  h0 h  2 h0 h
    
  2 
3
01 1
2
3

2
Uses sin (A + B) identity to
 
replace sin h , to
6 
commence argument
(at least two lines of argument
seen) | AO2.1 | M1
Obtains correct two term
expression involving cos h and
sin h | AO1.1b | A1
Deduce what happens as h0,
for one part of ‘their’ expression
sinh
using the limit of
h
OR by using small angles
approximations | AO2.2a | R1
Deduce what happens as h0 ,
for the second part of ‘their’
expression using the limit of
(cos h – 1)h
OR by using small angle
approximations | AO2.2a | R1
Completes a rigorous argument
leading to the correct exact
value, with all the steps in the
method clearly shown. | AO2.1 | R1
Total | 6
TOTAL | 100
$f(x) = \sin x$

Using differentiation from first principles find the exact value of $f'\left(\frac{\pi}{6}\right)$

Fully justify your answer.
[6 marks]

\hfill \mbox{\textit{AQA Paper 1  Q17 [6]}}
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