Standard +0.3 This is a straightforward stationary points question requiring differentiation (using chain rule for the second term), solving a quadratic equation, and using the second derivative test. All steps are standard AS-level techniques with no conceptual challenges, making it slightly easier than average.
5 A curve has equation \(y = 8 x + ( 2 x - 1 ) ^ { - 1 }\). Find the values of \(x\) at which the curve has a stationary point and determine the nature of each stationary point, justifying your answers.
When \(x=\frac{1}{4}\), \(\frac{d^2y}{dx^2}(=-64)\) and/or \(<0\) MAX
DB1
Alt. methods for last 3 marks (values either side of \(\frac{1}{4}\) & \(\frac{3}{4}\)) must indicate which \(x\)-values
When \(x=\frac{3}{4}\), \(\frac{d^2y}{dx^2}(=64)\) and/or \(>0\) MIN
DB1 [7]
cannot use \(x=\frac{1}{2}\). (M1A1A1)
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx}=[8]+[-2][(2x-1)^{-2}]$ | B2,1,0 | |
| $=0\rightarrow 4(2x-1)^2=1$ oe eg $16x^2-16x+3=0$ | M1 | Set to zero, simplify and attempt to solve |
| $x=\frac{1}{4}$ and $\frac{3}{4}$ | A1 | Needs both $x$ values. Ignore $y$ values |
| $\frac{d^2y}{dx^2}=8(2x-1)^{-3}$ | B1$\checkmark$* | ft to $k(2x-1)^{-3}$ where $k>0$ |
| When $x=\frac{1}{4}$, $\frac{d^2y}{dx^2}(=-64)$ and/or $<0$ MAX | DB1 | Alt. methods for last 3 marks (values either side of $\frac{1}{4}$ & $\frac{3}{4}$) must indicate which $x$-values |
| When $x=\frac{3}{4}$, $\frac{d^2y}{dx^2}(=64)$ and/or $>0$ MIN | DB1 [7] | cannot use $x=\frac{1}{2}$. (M1A1A1) |
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5 A curve has equation $y = 8 x + ( 2 x - 1 ) ^ { - 1 }$. Find the values of $x$ at which the curve has a stationary point and determine the nature of each stationary point, justifying your answers.
\hfill \mbox{\textit{CAIE P1 2016 Q5 [7]}}