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Date	May 2021	Marks available	2	Reference code	21M.2.AHL.TZ2.11
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	State	Question number	11	Adapted from	N/A

Question

A function $f$ is defined by $f (x) = \frac{k e^{\frac{x}{2}}}{1 + e^{x}}$ , where $x \in ℝ, x \geq 0$ and $k \in ℝ^{+}$ .

The region enclosed by the graph of $y = f (x)$ , the $x$ -axis, the $y$ -axis and the line $x = \ln 16$ is rotated $360 °$ about the $x$ -axis to form a solid of revolution.

Pedro wants to make a small bowl with a volume of $300 {cm}^{3}$ based on the result from part (a). Pedro’s design is shown in the following diagrams.

The vertical height of the bowl, $BO$ , is measured along the $x$ -axis. The radius of the bowl’s top is $OA$ and the radius of the bowl’s base is $BC$ . All lengths are measured in $cm$ .

For design purposes, Pedro investigates how the cross-sectional radius of the bowl changes.

Show that the volume of the solid formed is $\frac{15 k^{2} π}{34}$ cubic units.

[6]

Find the value of $k$ that satisfies the requirements of Pedro’s design.

[2]

Find $OA$ .

[2]

c.i.

Find $BC$ .

[2]

c.ii.

By sketching the graph of a suitable derivative of $f$ , find where the cross-sectional radius of the bowl is decreasing most rapidly.

[4]

d.i.

State the cross-sectional radius of the bowl at this point.

[2]

d.ii.

Markscheme

attempt to use $V = π \int_{a}^{b} {(f (x))}^{2} d x$ (M1)

$V = π \int_{0}^{\ln 16} {(\frac{k e^{\frac{x}{2}}}{1 + e^{x}})}^{2} d x (V = k^{2} π \int_{0}^{\ln 16} \frac{e^{x}}{{(1 + e^{x})}^{2}} d x)$

EITHER

applying integration by recognition (M1)

$= k^{2} π {[- \frac{1}{1 + e^{x}}]}_{0}^{\ln 16}$ A3

$u = 1 + e^{x} \Rightarrow d u = e^{x} d x$ (A1)

attempt to express the integral in terms of $u$ (M1)

when $x = 0, u = 2$ and when $x = \ln 16, u = 17$

$V = k^{2} π \int_{2}^{17} \frac{1}{u^{2}} d u$ (A1)

$= k^{2} π {[- \frac{1}{u}]}_{2}^{17}$ A1

$u = e^{x} \Rightarrow d u = e^{x} d x$ (A1)

attempt to express the integral in terms of $u$ (M1)

when $x = 0, u = 1$ and when $x = \ln 16, u = 16$

$V = k^{2} π \int_{1}^{16} \frac{1}{{(1 + u)}^{2}} d u$ (A1)

$= k^{2} π {[- \frac{1}{1 + u}]}_{1}^{16}$ A1

Note: Accept equivalent working with indefinite integrals and original limits for $x$ .

THEN

$= k^{2} π (\frac{1}{2} - \frac{1}{17})$ A1

so the volume of the solid formed is $\frac{15 k^{2} π}{34}$ cubic units AG

Note: Award (M1)(A0)(M0)(A0)(A0)(A1) when $\frac{15}{34}$ is obtained from GDC

[6 marks]

a valid algebraic or graphical attempt to find $k$ (M1)

$k^{2} = \frac{300 \times 34}{15 π}$

$k = 14.7 (= 2 \sqrt{\frac{170}{π}} = \sqrt{\frac{680}{π}})$ (as $k \in ℝ^{+}$ ) A1

Note: Candidates may use their GDC numerical solve feature.

[2 marks]

attempting to find $OA = f (0) = \frac{k}{2}$

with $k = 14.712 \dots (= 2 \sqrt{\frac{170}{π}} = \sqrt{\frac{680}{π}})$ (M1)

$OA = 7.36 (= \sqrt{\frac{170}{π}})$ A1

[2 marks]

c.i.

attempting to find $BC = f (\ln 16) = \frac{4 k}{17}$

with $k = 14.712 \dots (= 2 \sqrt{\frac{170}{π}} = \sqrt{\frac{680}{π}})$ (M1)

$BC = 3.46 (= \frac{8}{17} \sqrt{\frac{170}{π}} = \frac{8 \sqrt{10}}{\sqrt{17 π}})$ A1

[2 marks]

c.ii.

EITHER

recognising to graph $y = f' (x)$ (M1)

Note: Award M1 for attempting to use quotient rule or product rule differentiation. $f' (x) = \frac{k e^{\frac{x}{2}} (1 - e^{x})}{2 {(1 + e^{x})}^{2}}$

for $x > 0$ graph decreasing to the local minimum A1

before increasing towards the $x$ -axis A1

recognising to graph $y = f'' (x)$ (M1)

Note: Award M1 for attempting to use quotient rule or product rule differentiation. $f'' (x) = \frac{k e^{\frac{x}{2}} (e^{2 x} - 6 e^{x} +1)}{4 {(1 + e^{x})}^{3}}$