Standard +0.3 This is a straightforward point of inflection question requiring students to find f''(x), evaluate f''(0), and check the sign change of f''(x) around x=0. While it requires understanding the definition of inflection points (not just f''=0 but sign change), the differentiation is routine and the verification is mechanical, making it slightly easier than average.
The function f is defined by
$$f(x) = x^2 + 2 \cos x \text{ for } -\pi \leq x \leq \pi$$
Determine whether the curve with equation \(y = f(x)\) has a point of inflection at the point where \(x = 0\)
Fully justify your answer.
[4 marks]
Question 10:
10 | Differentiates to obtain
2x±2sinx | 1.1a | M1 | f′( )=2x−2sinx
x
f′′( )=2−2cosx
x
f′′( )=2−2cos ( )=0
0 0
f′′(−0.1 )=9.99×10 −3
f′′( 0.1 )=9.99×10 −3
f′′( )
x does not change sign either
side of x = 0
Therefore, the curve does not have a
point of inflection at x = 0
Differentiates again to obtain
2−2cosx | 1.1b | A1
f′′( )=0
Concludes 0 and tests
f′( ) f′′( )
the sign of their x or x
either side of x = 0
Or
f′′( )≥0
Deduces that x | 2.1 | M1
Completes a reasoned
argument to conclude that
y = f(x) does not have a point of
inflection at x = 0
Or
Completes a reasoned
argument to conclude that
y = f(x) does not have a point of
inflection by consideration of the
function | 2.4 | R1
Question 10 Total | 4
Q | Marking instructions | AO | Marks | Typical solution
The function f is defined by
$$f(x) = x^2 + 2 \cos x \text{ for } -\pi \leq x \leq \pi$$
Determine whether the curve with equation $y = f(x)$ has a point of inflection at the point where $x = 0$
Fully justify your answer.
[4 marks]
\hfill \mbox{\textit{AQA Paper 2 2024 Q10 [4]}}