C3 June 2016 Q2
2.\[y = \frac{4x}{x^2 + 5}\]
| Scheme | Marks |
|---|---|
| \(y = \dfrac{4x}{\left(x^2 + 5\right)} \Rightarrow \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = \dfrac{4\left(x^2 + 5\right) - 4x \times 2x}{\left(x^2 + 5\right)^2}\) | M1A1 |
| \(\Rightarrow \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = \dfrac{20 - 4x^2}{\left(x^2 + 5\right)^2}\) | M1A1 |
| (4) |
Notes
M1: Attempt to use the quotient rule \(\dfrac{vu' - uv'}{v^2}\) with \(u = 4x\) and \(v = x^2 + 5\). If the rule is quoted it must be correct. It may be implied by their \(u = 4x, u' = A,\ v = x^2 + 5, v' = Bx\) followed by their \(\dfrac{vu' - uv'}{v^2}\)
If the rule is neither quoted nor implied only accept expressions of the form \(\dfrac{A\left(x^2 + 5\right) - 4x \times Bx}{\left(x^2 + 5\right)^2}, A, B > 0\) You may condone missing (invisible) brackets
Alternatively uses the product rule with \(u(/v) = 4x\) and \(v(/u) = \left(x^2 + 5\right)^{-1}\). If the rule is quoted it must be correct. It may be implied by their \(u = 4x, u' = A,\ v = x^2 + 5, v' = Bx\left(x^2 + 5\right)^{-2}\) followed by their \(vu' + uv'\). If the rule is neither quoted nor implied only accept expressions of the form \(A\left(x^2 + 5\right)^{-1} \pm 4x \times Bx\left(x^2 + 5\right)^{-2}\)
A1: \(\mathrm{f}'(x)\) correct (unsimplified). For the product rule look for versions of \(4\left(x^2 + 5\right)^{-1} - 4x \times 2x\left(x^2 + 5\right)^{-2}\)
M1: Simplifies to the form \(\mathrm{f}'(x) = \dfrac{A + Bx^2}{\left(x^2 + 5\right)^2}\) oe. This is not dependent so could be scored from \(\dfrac{v'u - u'v}{v^2}\)
When the product rule has been used the \(A\) of \(A\left(x^2 + 5\right)^{-1}\) must be adapted.
A1: CAO. Accept exact equivalents such as \(\left(\mathrm{f}'(x) =\right)\dfrac{4\left(5 - x^2\right)}{\left(x^2 + 5\right)^2}\), \(-\dfrac{4x^2 - 20}{\left(x^2 + 5\right)^2}\) or \(\dfrac{-4\left(x^2 - 5\right)}{x^4 + 10x^2 + 25}\)
Remember to isw after a correct answer
| Scheme | Marks |
|---|---|
| \(\dfrac{20 - 4x^2}{\left(x^2 + 5\right)^2} < 0 \Rightarrow x^2 > \dfrac{20}{4}\) Critical values of \(\pm\sqrt{5}\) | M1 |
| \(x < -\sqrt{5},\ x > \sqrt{5}\) or equivalent | dM1A1 |
| (3) | |
| (7 marks) |
Notes
M1: Sets their numerator either \(= 0\), \(< 0\), \(\leqslant 0\), \(> 0\), \(\geqslant 0\) and proceeds to at least one value for \(x\)
For example \(20 - 4x^2 \geqslant 0 \Rightarrow x \geqslant \sqrt{5}\) will be M1 dM0 A0.
It cannot be scored from a numerator such as 4 or indeed \(20 + 4x^2\)
dM1: Achieves two critical values for their numerator \(= 0\) and chooses the outside region
Look for \(x <\) smaller root, \(x >\) bigger root. Allow decimals for the roots.
Condone \(x \leqslant -\sqrt{5}\), \(x \geqslant \sqrt{5}\) and expressions like \(-\sqrt{5} > x > \sqrt{5}\)
If they have \(4x^2 - 20 < 0\) following an incorrect derivative they should be choosing the inside region
A1: Allow \(x < -\sqrt{5}, x > \sqrt{5}\) \(x < -\sqrt{5}\) or \(x > \sqrt{5}\) \(\left\{x : -\infty < x < -\sqrt{5} \cup \sqrt{5} < x < \infty\right\}\) \(|x| > \sqrt{5}\)
Do not allow for the A1 \(x < -\sqrt{5}\) and \(x > \sqrt{5}\). \(\sqrt{5} < x < -\sqrt{5}\) or \(\left\{x : -\infty < x < -\sqrt{5} \cap \sqrt{5} < x < \infty\right\}\)
but you may isw following a correct answer.