Standard +0.3 This is a straightforward application of differentiation using the chain rule to find where f'(x) < 0. Students must differentiate (2x-1)^(3/2), set f'(x) = 0, and solve a simple equation. While it requires understanding that 'decreasing' means f'(x) ≤ 0 throughout the interval, the algebraic manipulation is routine for A-level, making it slightly easier than average.
4 The function f is such that \(\mathrm { f } ( x ) = ( 2 x - 1 ) ^ { \frac { 3 } { 2 } } - 6 x\) for \(\frac { 1 } { 2 } < x < k\), where \(k\) is a constant. Find the largest value of \(k\) for which f is a decreasing function.
\((2x-1)^{1/2}<2\) or \(\leqslant 2\) or \(=2\) OE
A1
Allow with \(k\) used instead of \(x\)
Largest value of \(k\) is \(\frac{5}{2}\)
A1
Allow \(k \leqslant \frac{5}{2}\) or \(k=\frac{5}{2}\). Answer must be in terms of \(k\) (not \(x\))
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x)=\left[\left(\frac{3}{2}\right)(2x-1)^{1/2}\right]\times[2]-[6]$ | B2, 1, 0 | Deduct 1 mark for each [...] incorrect |
| $f'(x)<0$ or $\leqslant 0$ or $=0$ SOI | M1 | |
| $(2x-1)^{1/2}<2$ or $\leqslant 2$ or $=2$ OE | A1 | Allow with $k$ used instead of $x$ |
| Largest value of $k$ is $\frac{5}{2}$ | A1 | Allow $k \leqslant \frac{5}{2}$ or $k=\frac{5}{2}$. Answer must be in terms of $k$ (not $x$) |
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4 The function f is such that $\mathrm { f } ( x ) = ( 2 x - 1 ) ^ { \frac { 3 } { 2 } } - 6 x$ for $\frac { 1 } { 2 } < x < k$, where $k$ is a constant. Find the largest value of $k$ for which f is a decreasing function.\\
\hfill \mbox{\textit{CAIE P1 2017 Q4 [5]}}