C3 June 2017 Q7

Notes

M1: Attempts the product rule to differentiate \(2x(x^2 - 1)^5\) to a form \(A\left(x^2 - 1\right)^5 + Bx^n\left(x^2 - 1\right)^4\) where \(n = 1\) or 2. and \(A, B > 0\) If the rule is stated it must be correct, and not with a "\(-\)" sign.

A1: Any unsimplified but correct form \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = 2\left(x^2 - 1\right)^5 + 20x^2\left(x^2 - 1\right)^4\)

M1: For taking a common factor of \(\left(x^2 - 1\right)^4\) out of a suitable expression
Look for \(A\left(x^2 - 1\right)^5 \pm Bx^n\left(x^2 - 1\right)^4 = \left(x^2 - 1\right)^4\left\{A\left(x^2 - 1\right) \pm Bx^n\right\}\) but you may condone missing brackets
It can be scored from a \(vu'\)-\(uv'\) or similar.

A1: \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = \left(x^2 - 1\right)^4\left(22x^2 - 2\right)\) Expect \(\mathrm{g}(x)\) to be simplified but accept \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \left(x^2 - 1\right)^4 2\left(11x^2 - 1\right)\)
There is no need to state g(\(x\)) and remember to isw after a correct answer. This must be in part (a).

Notes

M1: Sets their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} \geqslant 0, > 0\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and proceeds to find one of the critical values for their \(\mathrm{g}(x)\) or their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) rearranged and \(\div\left(x^2 - 1\right)^4\) if g(\(x\)) not found. \(\mathrm{g}(x)\) should be at least a 2TQ with real roots. If g(x) is factorised, the usual rules apply. The M cannot be awarded from work just on \(\left(x^2 - 1\right)^4 \geqslant 0\) ie \(x = \pm 1\)
You may see and accept decimals for the M.

A1: cao \(x \geqslant \dfrac{1}{\sqrt{11}} \quad x \leqslant -\dfrac{1}{\sqrt{11}}\) or exact equivalent only. Condone \(x \geqslant \dfrac{1}{\sqrt{11}} \quad x \leqslant -\dfrac{1}{\sqrt{11}}\), with \(x \geqslant 1, x \leqslant -1\)
Accept exact equivalents such as \(x \geqslant \dfrac{\sqrt{11}}{11} \quad x \leqslant -\dfrac{\sqrt{11}}{11}\); \(|x| \geqslant \dfrac{1}{\sqrt{11}}\); \(\left\{\left(-\infty, -\dfrac{\sqrt{11}}{11}\right] \cup \left[\dfrac{\sqrt{11}}{11}, \infty\right)\right\}\)
Condone the word "and" appearing between the two sets of values.
Withhold the final mark if \(x \geqslant \dfrac{1}{\sqrt{11}} \quad x \leqslant -\dfrac{1}{\sqrt{11}}\), appears with values not in this region eg \(x \leqslant 1, x \geqslant -1\)

Alt 1 (ii)

Alt 2 (ii)

Notes

B1: Differentiates and achieves a correct line involving \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\)
Accept \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{1}{\sec 2y} \times 2\sec 2y\tan 2y\), \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = -\dfrac{1}{\cos 2y} \times -2\sin 2y\)   \(2\sec 2y\tan 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^x\)

M1: For inverting their expression for \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) to achieve an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
The variables (on the rhs) must be consistent, you may condone slips on the coefficients but not the terms.
In the alternative method it is for correctly changing the subject

M1: Scored for using \(\tan^2 2y = \pm 1 \pm \sec^2 2y\) and \(\sec 2y = \mathrm{e}^x\) to achieve \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) in terms of \(x\)
Alternatively they could use \(\sin^2 2y + \cos^2 2y = 1\) with \(\cos 2y = \mathrm{e}^{-x}\) to achieve \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) in terms of \(x\)
For the M mark you may condone \(\sec^2 2y = \left(\mathrm{e}^x\right)^2\) appearing as \(\mathrm{e}^{x^2}\)

A1: cso \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2\sqrt{\mathrm{e}^{2x} - 1}}\)   Final answer, do not allow if students then simplify this to eg. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2\mathrm{e}^x - 1}\)
Condone \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm\dfrac{1}{2\sqrt{\mathrm{e}^{2x} - 1}}\) but do not allow \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{2\sqrt{\mathrm{e}^{2x} - 1}}\)
Allow a misread on \(x = \ln(\sec y)\) for the two method marks only