Challenging +1.2 This requires applying the quotient rule to a composite exponential function, finding where dy/dx = 0, then solving the resulting equation and substituting back. While it involves multiple techniques (quotient rule, chain rule, exponential differentiation), the algebraic manipulation is relatively straightforward once the derivative is set up, and finding stationary points is a standard procedure. The exact coordinates requirement adds modest difficulty but this remains a typical exam question testing core differentiation skills.
Need attempt at both derivatives; condone errors in chain rule. In quotient rule allow BOD in formula if \(\pm 2x\) seen unless clear that incorrect formula has been used. If omit denominator or forget to square or complete reversal of signs then M0 A0 M1 A1 A1 A1.
Obtain correct derivative in any form, e.g. \(\frac{6x(1-x^2)e^{3x^2-1}+2xe^{3x^2-1}}{(1-x^2)^2}\)
A1
If \(6x(1-x^2)e^{3x^2-1}+2xe^{3x^2-1}=0\) from the start, with no wrong formula seen, award M1A1.
Equate derivative (or its numerator) to zero and solve for \(x\)
M1
\(6x-6x^3+2x=0\) and solve. Allow for just one \(x\) value. Allow if from solution of 3 term quadratic equation, but if they get \(x=0\) the \(x\) must factorise out.
Obtain the point \((0, e^{-1})\) or exact equivalent
A1
Or for all three \(x\) coordinates found \(0\), \(\pm\frac{2\sqrt{3}}{3}\) oe and no extras but if this is the case then one pair of correct coordinates A1 and both other pairs of correct coordinates A1. Accept, e.g. \(x=0\), \(y=e^{-1}\). ISW for last 3 marks.
Obtain the point \(\left(\frac{2\sqrt{3}}{3}, -3e^3\right)\) or exact equivalent
A1
Allow \(\sqrt{(4/3)}\).
Obtain the point \(\left(-\frac{2\sqrt{3}}{3}, -3e^3\right)\) or exact equivalent
A1
Total
6
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct product or quotient rule | M1 | Need attempt at both derivatives; condone errors in chain rule. In quotient rule allow BOD in formula if $\pm 2x$ seen unless clear that incorrect formula has been used. If omit denominator or forget to square or complete reversal of signs then M0 A0 M1 A1 A1 A1. |
| Obtain correct derivative in any form, e.g. $\frac{6x(1-x^2)e^{3x^2-1}+2xe^{3x^2-1}}{(1-x^2)^2}$ | A1 | If $6x(1-x^2)e^{3x^2-1}+2xe^{3x^2-1}=0$ from the start, with no wrong formula seen, award M1A1. |
| Equate derivative (or its numerator) to zero and solve for $x$ | M1 | $6x-6x^3+2x=0$ and solve. Allow for just one $x$ value. Allow if from solution of 3 term quadratic equation, but if they get $x=0$ the $x$ must factorise out. |
| Obtain the point $(0, e^{-1})$ or exact equivalent | A1 | Or for all three $x$ coordinates found $0$, $\pm\frac{2\sqrt{3}}{3}$ oe and no extras but if this is the case then one pair of correct coordinates A1 and both other pairs of correct coordinates A1. Accept, e.g. $x=0$, $y=e^{-1}$. ISW for last 3 marks. |
| Obtain the point $\left(\frac{2\sqrt{3}}{3}, -3e^3\right)$ or exact equivalent | A1 | Allow $\sqrt{(4/3)}$. |
| Obtain the point $\left(-\frac{2\sqrt{3}}{3}, -3e^3\right)$ or exact equivalent | A1 | |
| **Total** | **6** | |
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