A2 June 2023 Paper 2 Q14

AQACurrent spec10 marksGraphs & Inequalities Integration

14 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = \frac{1}{4x^2 + 16x + 19} \qquad (x \in \mathbb{R})\]

(a) Show, without using calculus, that the graph of \(y = \mathrm{f}(x)\) has a stationary point at \(\left(-2, \dfrac{1}{3}\right)\) [3 marks]

(b) Show that \(\displaystyle\int_{-2}^{-\frac{1}{2}} \mathrm{f}(x)\,\mathrm{d}x = \frac{\pi\sqrt{3}}{18}\) [5 marks]

Fully justify your answer. [2 marks]

Mark scheme (a)

Scheme	Marks	AO
Completes the square for denominator. Or Sets \(\mathrm{f}(x) = k\) and forms a quadratic equation in \(x\)	M1	3.1a
Explains that \(\mathrm{f}\) has a stationary point when \(x = -2\) Or Equates the discriminant of the quadratic equation to 0 and solves for \(k\)	E1	2.4
Completes a reasoned argument, without using calculus, to show that the stationary point is at \(\left(-2, \dfrac{1}{3}\right)\) to obtain the required result.	R1	2.1
	(3)

Typical solution

\[\mathrm{f}(x) = \frac{1}{4\left(x^2 + 4x + \frac{19}{4}\right)} = \frac{1}{4\left((x + 2)^2 + \frac{3}{4}\right)}\]

\(\mathrm{f}\) is maximum when the denominator is minimum, that is when \(x = -2\)

\[\text{and } y = \frac{1}{4\left(\frac{3}{4}\right)} = \frac{1}{3}\]

So the graph of \(y = \mathrm{f}(x)\) has a stationary point at \(\left(-2, \dfrac{1}{3}\right)\)

Mark scheme (b)

Scheme	Marks	AO
Expresses the denominator of the integrand in completed square form.	M1	3.1a
Uses inverse tan to integrate their integrand of the form \(\dfrac{1}{(x + k)^2 + a^2}\) Or Makes a correct substitution	M1	3.1a
Integrates to obtain \(A\tan^{-1}\dfrac{2(x + 2)}{\sqrt{3}}\)	A1	1.1b
Substitutes the upper limit correctly into their integrated expression which includes \(\tan^{-1}\)	M1	1.1a
Completes a reasoned argument to obtain the required result. Must see substitution of \(-2\) in the integrated expression.	R1	2.1
	(5)

Typical solution

\[\int_{-2}^{-\frac{1}{2}} \mathrm{f}(x)\,\mathrm{d}x = \frac{1}{4}\int_{-2}^{-\frac{1}{2}} \frac{1}{(x + 2)^2 + \frac{3}{4}}\,\mathrm{d}x\]\[\begin{aligned} &= \frac{1}{4} \times \frac{2}{\sqrt{3}}\left[\tan^{-1}\left(\frac{2(x + 2)}{\sqrt{3}}\right)\right]_{-2}^{-\frac{1}{2}} \\ &= \frac{1}{2\sqrt{3}}\left(\tan^{-1}\sqrt{3} - \tan^{-1}0\right) \\ &= \frac{1}{2\sqrt{3}}\left(\frac{\pi}{3} - 0\right) = \frac{\pi}{6\sqrt{3}} = \frac{\pi\sqrt{3}}{18} \end{aligned}\]

Mark scheme (c)

Scheme	Marks	AO
Replaces \(\infty\) by a letter (\(N\)) and considers \(\displaystyle\lim_{N \to \infty}\) in the integral or integrated expression.	E1	3.1a
Obtains the correct exact value. ACF	B1	2.2a
	(2)
	(10 marks)

Typical solution

\[\int_{-2}^{\infty} \mathrm{f}(x)\,\mathrm{d}x = \frac{1}{2\sqrt{3}}\lim_{N \to \infty}\left[\tan^{-1}\left(\frac{2(x + 2)}{\sqrt{3}}\right)\right]_{-2}^{N}\]\[= \frac{1}{2\sqrt{3}}\left(\frac{\pi}{2} - 0\right)\]\[= \frac{\pi\sqrt{3}}{12}\]