Moderate -0.8 This is a scaffolded error-correction question on differentiation from first principles at a specific point. Students must identify algebraic mistakes in given working and complete two steps with heavy guidance. The context (sin x at π/2) is standard, and the question structure provides significant support, making it easier than average despite being a 4-mark question.
Jodie is attempting to use differentiation from first principles to prove that the gradient of \(y = \sin x\) is zero when \(x = \frac{\pi}{2}\)
Jodie's teacher tells her that she has made mistakes starting in Step 4 of her working. Her working is shown below.
\includegraphics{figure_11}
Complete Steps 4 and 5 of Jodie's working below, to correct her proof.
[4 marks]
Step 4 \quad For gradient of curve at A,
Step 5 \quad Hence the gradient of the curve at A is given by
Question 11:
11 | Replaces h=0with h→0or
better seen anywhere | 2.3 | M1 | For gradient of curve at A,
let h→0 then
( )−1 ( )
cos h sin h
→0and →1
h h
Hence the gradient of the curve at
A is given by
π π
sin ×0+cos ×1=0
2 2
Uses limit notation fully correctly
( )
sin h
Accept →0here
h
Accept full limit notation here | 2.5 | A1
( )
sin h
=1 seen
h
OE eg sin(h)=h | 2.3 | B1
Writes last line explicitly as
π π
sin ×0+cos ×1=0
2 2
Accept 1×0+0×1=0 | 2.2a | B1
Total | 4
Q | Marking instructions | AO | Mark | Typical solution
Jodie is attempting to use differentiation from first principles to prove that the gradient of $y = \sin x$ is zero when $x = \frac{\pi}{2}$
Jodie's teacher tells her that she has made mistakes starting in Step 4 of her working. Her working is shown below.
\includegraphics{figure_11}
Complete Steps 4 and 5 of Jodie's working below, to correct her proof.
[4 marks]
Step 4 \quad For gradient of curve at A,
Step 5 \quad Hence the gradient of the curve at A is given by
\hfill \mbox{\textit{AQA Paper 1 2019 Q11 [4]}}