Standard +0.3 This requires differentiating using the chain rule, setting f'(x) ≤ 0, and solving an inequality involving a surd. It's a straightforward application of calculus to determine where a function is decreasing, slightly above average difficulty due to the fractional power and algebraic manipulation required.
2 The function f is defined by \(\mathrm { f } ( x ) = \frac { 1 } { 3 } ( 2 x - 1 ) ^ { \frac { 3 } { 2 } } - 2 x\) for \(\frac { 1 } { 2 } < x < a\). It is given that f is a decreasing function.
Find the maximum possible value of the constant \(a\).
SOI. Rearranging and then squaring, must have power of \(\frac{1}{2}\) not present. Allow '\(=0\)' at this stage but do not allow '\(\geq 0\)' or '\(> 0\)'. If '\(-2\)' missed then must see \(\leqslant\) or \(<\) for the M1
Value \([of\ a]\) is \(2\tfrac{1}{2}\) or \(a = 2\tfrac{1}{2}\)
A1
WWW, OE e.g. \(\frac{5}{2}\), 2.5. Do not allow from '\(=0\)' unless some reference to negative gradient.
4
**Question 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[f^{-1}(x) =]\ \left((2x-1)^{1/2}\right) \times \left(\frac{1}{3} \times 2 \times \frac{3}{2}\right)(-2)$ | **B2, 1, 0** | Expect $(2x-1)^{1/2} - 2$ |
| $(2x-1)^{1/2} - 2 \leqslant 0\ \rightarrow\ 2x - 1 \leqslant 4$ or $2x - 1 < 4$ | **M1** | SOI. Rearranging and then squaring, must have power of $\frac{1}{2}$ not present. Allow '$=0$' at this stage but do not allow '$\geq 0$' or '$> 0$'. If '$-2$' missed then must see $\leqslant$ or $<$ for the M1 |
| Value $[of\ a]$ is $2\tfrac{1}{2}$ or $a = 2\tfrac{1}{2}$ | **A1** | WWW, OE e.g. $\frac{5}{2}$, 2.5. Do not allow from '$=0$' unless some reference to negative gradient. |
| | **4** | |
2 The function f is defined by $\mathrm { f } ( x ) = \frac { 1 } { 3 } ( 2 x - 1 ) ^ { \frac { 3 } { 2 } } - 2 x$ for $\frac { 1 } { 2 } < x < a$. It is given that f is a decreasing function.
Find the maximum possible value of the constant $a$.\\
\hfill \mbox{\textit{CAIE P1 2021 Q2 [4]}}