AQA Further Paper 2 2020 June — Question 3 1 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2020
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeDerivative of inverse trig function
DifficultyModerate -0.5 This is a straightforward application of the standard derivative formula d/dx(sin⁻¹x) = 1/√(1-x²), requiring only substitution of x=1/5 and simplification. Being multiple choice with 1 mark makes it easier than average, though it's from Further Maths so involves a less common derivative that students must recall correctly.
Spec1.07l Derivative of ln(x): and related functions4.08g Derivatives: inverse trig and hyperbolic functions

Find the gradient of the tangent to the curve $$y = \sin^{-1} x$$ at the point where \(x = \frac{1}{5}\) Circle your answer. [1 mark] \(\frac{5\sqrt{6}}{12}\) \quad \(\frac{2\sqrt{6}}{5}\) \quad \(\frac{4\sqrt{3}}{25}\) \quad \(\frac{25}{24}\)

Question 3:
AnswerMarks Guidance
3Circles
5√61.1b B1
12 Total1 12
QMarking Instructions AO
Question 3:
3 | Circles
5√6 | 1.1b | B1 | 5√6
12 Total | 1 | 12
Q | Marking Instructions | AO | Marks | Typical solution
Find the gradient of the tangent to the curve
$$y = \sin^{-1} x$$
at the point where $x = \frac{1}{5}$

Circle your answer.
[1 mark]

$\frac{5\sqrt{6}}{12}$ \quad $\frac{2\sqrt{6}}{5}$ \quad $\frac{4\sqrt{3}}{25}$ \quad $\frac{25}{24}$

\hfill \mbox{\textit{AQA Further Paper 2 2020 Q3 [1]}}