Moderate -0.3 This is a straightforward application of the quotient rule followed by solving a quadratic inequality. While it requires multiple steps (differentiation, simplification, and inequality solving), each step is routine for C3/C4 students with no novel insight needed. The algebra is manageable and the question type is standard practice material, making it slightly easier than average.
1.
$$\mathrm { f } ( x ) = \frac { 2 x } { x ^ { 2 } + 3 } , \quad x \in \mathbb { R }$$
Find the set of values of \(x\) for which \(\mathrm { f } ^ { \prime } ( x ) > 0\)
You must show your working.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
M1: Applies Quotient Rule to \(\frac{2x}{x^2+3}\). If rule quoted it must be correct; must attempt to differentiate both terms. Accept unsimplified correct form. Alternatively Product Rule with \(u=2x\), \(v=(x^2+3)^{-1}\) giving \((x^2+3)^{-1}\times A \pm 2x\times(x^2+3)^{-2}\times Bx\)
Critical values: \(6-2x^2 = 0 \Rightarrow x = \pm\sqrt{3}\)
M1A1
M1: Setting numerator of \(f'(x) = 0\) or \(> 0\) and proceeding to find two critical values. A1: Both critical values \(\pm\sqrt{3}\) found. Accept \(x = \pm\sqrt{3}\), \(\pm 1.73\)
Inside region chosen: \(-\sqrt{3} < x < \sqrt{3}\)
dM1A1
dM1: Choosing inside region of their critical values. Inequality must have been of correct form. A1 cao: Accept \((-\sqrt{3}, \sqrt{3})\), \(x < \sqrt{3}\) and \(x > -\sqrt{3}\). Do not accept \(x < \sqrt{3}\) or \(x > -\sqrt{3}\), or \(-1.73 < x < 1.73\). Do not accept correct answer from incorrect inequality (dM0A0).
(6 marks)
## Question 1:
$$f(x) = \frac{2x}{x^2+3}$$
| Working/Answer | Marks | Guidance |
|---|---|---|
| $f'(x) = \frac{(x^2+3)(2) - 2x \times 2x}{(x^2+3)^2} = \frac{6-2x^2}{(x^2+3)^2}$ | M1A1 | M1: Applies Quotient Rule to $\frac{2x}{x^2+3}$. If rule quoted it must be correct; must attempt to differentiate both terms. Accept unsimplified correct form. Alternatively Product Rule with $u=2x$, $v=(x^2+3)^{-1}$ giving $(x^2+3)^{-1}\times A \pm 2x\times(x^2+3)^{-2}\times Bx$ |
| $f'(x) > 0 \Rightarrow \frac{6-2x^2}{(x^2+3)^2} > 0$ | | |
| Critical values: $6-2x^2 = 0 \Rightarrow x = \pm\sqrt{3}$ | M1A1 | M1: Setting numerator of $f'(x) = 0$ or $> 0$ and proceeding to find two critical values. A1: Both critical values $\pm\sqrt{3}$ found. Accept $x = \pm\sqrt{3}$, $\pm 1.73$ |
| Inside region chosen: $-\sqrt{3} < x < \sqrt{3}$ | dM1A1 | dM1: Choosing inside region of their critical values. Inequality must have been of correct form. A1 cao: Accept $(-\sqrt{3}, \sqrt{3})$, $x < \sqrt{3}$ and $x > -\sqrt{3}$. Do **not** accept $x < \sqrt{3}$ or $x > -\sqrt{3}$, or $-1.73 < x < 1.73$. Do not accept correct answer from incorrect inequality (dM0A0). |
| | **(6 marks)** | |
1.
$$\mathrm { f } ( x ) = \frac { 2 x } { x ^ { 2 } + 3 } , \quad x \in \mathbb { R }$$
Find the set of values of $x$ for which $\mathrm { f } ^ { \prime } ( x ) > 0$
You must show your working.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
\hfill \mbox{\textit{Edexcel C34 2014 Q1 [6]}}