Edexcel A-Level Mathematics AS P2.7 Differentiation Questions

Practise applying differentiation to curves and practical models, finding stationary points, ranges of increase or decrease, and justified extrema.

Exam points

set dy/dx=0 to find stationary coordinates or unknown values in a model
use d²y/dx² or sign changes to justify a maximum or minimum
solve derivative inequalities to state where a function is increasing or decreasing

Covered syllabus points

P2.7.1 - Applications of differentiation

[Maximum number: 3]

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

f(x)=4 x^{3}+13 x^{2}-10 x+8

Find the range of values of x for which f(x) is decreasing.

$f'(x)=12x^2+26x-10$ $12x^2+26x-10=0$ $2(6x^2+13x-5)=0$ (2x+5)(3x-1)=0 $x=-\frac52,\ \frac13$ $f(x)\text{ is decreasing for }-\frac52<x<\frac13$

[Maximum number: 6]

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Figure 2

Figure 2 shows the plan view of the design for a stage at a trade fair.
The shape of the stage ABCDEFA, consists of a rectangle ACDF joined to two congruent sectors of circles. ABC is a sector of a circle centre A and FDE is a sector of a circle centre F.

Given that A C=2 r metres, C D=r metres, angle $D F E=\theta$ radians and the area of the stage is $30 \mathrm{~m}^{2}$ ,

(a)

Use calculus to find the minimum value for P, giving your answer in the form $a \sqrt{b}$ , where a and b are integers to be found.

[ 4 ]

$\frac{dP}{dr}=4-\frac{30}{r^2}$ $\frac{dP}{dr}=0\Rightarrow r=\sqrt{\frac{15}{2}}$ $P=4\sqrt{\frac{15}{2}}+\frac{30}{\sqrt{\frac{15}{2}}}=4\sqrt{30}$

(b)

Justify that the value of P found in part (b) is the minimum.

[ 2 ]

$\frac{d^2P}{dr^2}=\frac{60}{r^3}$
At $r=\sqrt{\frac{15}{2}}>0$ ,
$\frac{d^2P}{dr^2}>0$
Hence the value of P is a minimum.

[Maximum number: 7]

Figure 2

Figure 2 shows a sketch of an open container.
The sides CDEF and ABFE are rectangles.
The ends ADE and BCF are congruent (identical) right-angled triangles.
The container is made from metal of negligible thickness.
Given that
- A E=B F=3 x metres
- D E=C F=2 x metres
- A B=D C=E F=L metres
and the capacity of the container is $12 \mathrm{~m}^{3}$

(a)

use algebraic calculus to find the minimum value of S, giving your answer to one decimal place.

[ 5 ]

Condone other letters used instead of S and x for $\frac{\mathrm{d} S}{\mathrm{~d} x}$ including y for S.

Accept use of e.g. $S^{\prime}$
M1: Correct form for the derivative $\left(\frac{\mathrm{d} S}{\mathrm{~d} x}=\right) a x+\frac{b}{x^{2}}$
A1ft: Follow through on the derivative of their $S=P x^{2}+\frac{Q}{x} \Rightarrow 2 P x-\frac{Q}{x^{2}}$
dM 1: Solves $\frac{\mathrm{d} S}{\mathrm{~d} x}=0 \Rightarrow x=\ldots$ . Do not be concerned by the mechanics of the rearrangement as long as a value for x is achieved. =0 may be implied by further work or if just a correct value for x then you may need to check this. It is dependent on the previous method mark.
ddM1: Substitutes their value of x in S to find S. Condone slips. It is dependent on the previous two method marks.
If only a value is stated you may need to check this.
A1: awrt $25.3\left(\mathrm{~m}^{2}\right)$ following correct differentiation. Units not required.

(b)

Justify that the value of S found in part (b) is a minimum.

[ 2 ]

M1: Differentiate to find $\frac{d^2S}{dx^2}$ of the form $M+\frac{N}{x^3}$ , or use sign-change values for $\frac{dS}{dx}$ .
A1*: Justify $\frac{d^2S}{dx^2}>0$ from a correct expression and give a minimal conclusion, or justify a change of sign for $\frac{dS}{dx}$ .
For $S=6x^2+\frac{20}{x}$ , $\frac{d^2S}{dx^2}=12+\frac{40}{x^3}>0$ for x>0, so the stationary value is a minimum.

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)