Moderate -0.5 This is a straightforward application of differentiation to prove monotonicity. Students need to find f'(x) = 3x² - 6x + 15, complete the square or use the discriminant to show it's always positive, then conclude the function is increasing. While it requires multiple steps and understanding of the connection between derivative sign and monotonicity, it's a standard technique taught explicitly in AS-level courses with no novel problem-solving required.
Question 8:
8 | Explains clearly that f(x)
is increasing ⇔f '(x)>0
(for all values of x)
or
Explains ⇒f(x) is increasing
f '(x)>0for all values of x
This may appear at any appropriate point in
their argument | AO2.4 | E1 | For all x, f '(x)>0⇒f(x) is an
increasing function
f(x)= x3 −3x2 +15x−1
⇒f '(x)=3x2 −6x+15
⇒f '(x)=3(x−1)2 +12
∴f '(x)has a minimum value of 12
therefore f '(x)>0 for all values of x
OR
for f '(x), b2 −4ac=−144
∴f '(x)≠0 for any real x, so f '(x) is
either always positive or always
negative.
f '(0)=15
therefore f '(x)>0 for all values of x
OR
f ''(x)=6x−6, which = 0 when x = 1
so min f '(x) is f '(1)=12
therefore f '(x)>0 for all values of x
Thus, since, f '(x)>0 for all values of
x it is proven that f(x) is an increasing
function.
Differentiates – at least two correct terms | AO1.1a | M1
All terms correct | AO1.1b | A1
Attempts a correct method which could lead | AO3.1a | AO3.1a | M1 | M1
to f '(x)>0
Correctly deduces f '(x)>0 | AO2.2a | AO2.2a | A1 | A1
(for all values of x)
Writes a clear statement that links the steps | AO2.1 | AO2.1 | R1 | R1
in the argument together, the deduction
about a positive gradient for all values of x
proves that the given function is increasing
for all values of x
Total | 6
Q | Marking Instructions | AO | Marks | Typical Solution