CAIE P1 2013 June — Question 1 3 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2013
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeProve or show increasing/decreasing function
DifficultyModerate -0.8 This is a straightforward application of differentiation to prove monotonicity. Students need to find f'(x) using the chain rule, then show f'(x) > 0 for all x. The algebra is simple: f'(x) = 6(2x-5)² + 1, which is clearly always positive since it's a square term plus 1. This is easier than average as it requires only one standard technique with no problem-solving insight.
Spec1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation

1 It is given that \(\mathrm { f } ( x ) = ( 2 x - 5 ) ^ { 3 } + x\), for \(x \in \mathbb { R }\). Show that f is an increasing function.

AnswerMarks Guidance
\(f'(x) = (2x-5)^2 \times 2 + 1\) or \(24\left(x - \frac{5}{2}\right)^2 + 1\)B1B1 B1 for \(3(2x-5)^2\), B1 for \((×2+1)\) SC B1 for \(24x^2 - 120x + 151\)
\(> 0\) (allow \(\geq\))B1√ Dep on \(k(2x-5)^2 + c\) (\(k > 0\), \(c \geq 0\)) Subst of particular values is B0
$f'(x) = (2x-5)^2 \times 2 + 1$ or $24\left(x - \frac{5}{2}\right)^2 + 1$ | B1B1 | B1 for $3(2x-5)^2$, B1 for $(×2+1)$ SC B1 for $24x^2 - 120x + 151$
$> 0$ (allow $\geq$) | B1√ | Dep on $k(2x-5)^2 + c$ ($k > 0$, $c \geq 0$) Subst of particular values is B0
1 It is given that $\mathrm { f } ( x ) = ( 2 x - 5 ) ^ { 3 } + x$, for $x \in \mathbb { R }$. Show that f is an increasing function.

\hfill \mbox{\textit{CAIE P1 2013 Q1 [3]}}