# IBDP Maths AI: Topic: SL 5.7: Optimisation problems in context: IB style Questions SL Paper 1

## Question

A cuboid has a rectangular base of width $$x$$ cm and length 2$$x$$ cm . The height of the cuboid is $$h$$ cm . The total length of the edges of the cuboid is $$72$$ cm.

The volume, $$V$$, of the cuboid can be expressed as $$V = a{x^2} – 6{x^3}$$.

Find the value of $$a$$.[3]

a.

Find the value of $$x$$ that makes the volume a maximum.[3]

b.

## Markscheme

$$72 = 12x + 4h\;\;\;$$(or equivalent)     (M1)

Note: Award (M1) for a correct equation obtained from the total length of the edges.

$$V = 2{x^2}(18 – 3x)$$     (A1)

$$(a = ){\text{ }}36$$     (A1)     (C3)

a.

$$\frac{{{\text{d}}V}}{{{\text{d}}x}} = 72x – 18{x^2}$$     (A1)

$$72x – 18{x^2} = 0\;\;\;$$OR$$\;\;\;\frac{{{\text{d}}V}}{{{\text{d}}x}} = 0$$     (M1)

Notes: Award (A1) for  $$– 18{x^2}$$  seen. Award (M1) for equating derivative to zero.

$$(x = ){\text{ 4}}$$     (A1)(ft)     (C3)

Note: Follow through from part (a).

OR

Sketch of $$V$$ with visible maximum     (M1)

Sketch with $$x \geqslant 0,{\text{ }}V \geqslant 0$$ and indication of maximum (e.g. coordinates)     (A1)(ft)

$$(x = ){\text{ 4}}$$     (A1)(ft)     (C3)

Notes: Follow through from part (a).

Award (M1)(A1)(A0) for $$(4,{\text{ }}192)$$.

Award (C3) for $$x = 4,{\text{ }}y = 192$$.

b.

## Question

A company sells fruit juices in cylindrical cans, each of which has a volume of $$340\,{\text{c}}{{\text{m}}^3}$$. The surface area of a can is $$A\,{\text{c}}{{\text{m}}^2}$$ and is given by the formula

$$A = 2\pi {r^2} + \frac{{680}}{r}$$ ,

where $$r$$ is the radius of the can, in $${\text{cm}}$$.

To reduce the cost of a can, its surface area must be minimized.

Find $$\frac{{{\text{d}}A}}{{{\text{d}}r}}$$[3]

a.

Calculate the value of $$r$$ that minimizes the surface area of a can.[3]

b.

## Markscheme

$$\left( {\frac{{{\text{d}}A}}{{{\text{d}}r}}} \right) = 4\pi r – \frac{{680}}{{{r^2}}}$$        (A1)(A1)(A1)      (C3)

Note: Award (A1) for $$4\pi r$$ (accept $$12.6r$$), (A1) for $$– 680$$, (A1) for $$\frac{1}{{{r^2}}}$$ or $${r^{ – 2}}$$

Award at most (A1)(A1)(A0) if additional terms are seen.

a.

$$4\pi r – \frac{{680}}{{{r^2}}} = 0$$          (M1)

Note: Award (M1) for equating their $$\frac{{{\text{d}}A}}{{{\text{d}}r}}$$ to zero.

$$4\pi {r^3} – 680 = 0$$          (M1)

Note: Award (M1) for initial correct rearrangement of the equation. This may be assumed if $${r^3} = \frac{{680}}{{4\pi }}$$ or $$r = \sqrt[3]{{\frac{{680}}{{4\pi }}}}$$ seen.

OR

sketch of $$A$$ with some indication of minimum point         (M1)(M1)

Note: Award (M1) for sketch of $$A$$, (M1) for indication of minimum point.

OR

sketch of $$\frac{{{\text{d}}A}}{{{\text{d}}r}}$$ with some indication of zero      (M1)(M1)

Note: Award (M1) for sketch of $$\frac{{{\text{d}}A}}{{{\text{d}}r}}$$, (M1) for indication of zero.

$$(r = )\,\,3.78\,({\text{cm}})\,\,\,\,\,(3.78239…)$$                (A1)(ft) (C3)

Note: Follow through from part (a).

b.

## Question

Maria owns a cheese factory. The amount of cheese, in kilograms, Maria sells in one week, $$Q$$, is given by

$$Q = 882 – 45p$$,

where $$p$$ is the price of a kilogram of cheese in euros (EUR).

Maria earns $$(p – 6.80){\text{ EUR}}$$ for each kilogram of cheese sold.

To calculate her weekly profit $$W$$, in EUR, Maria multiplies the amount of cheese she sells by the amount she earns per kilogram.

Write down how many kilograms of cheese Maria sells in one week if the price of a kilogram of cheese is 8 EUR.[1]

a.

Find how much Maria earns in one week, from selling cheese, if the price of a kilogram of cheese is 8 EUR.[2]

b.

Write down an expression for $$W$$ in terms of $$p$$.[1]

c.

Find the price, $$p$$, that will give Maria the highest weekly profit.[2]

d.

## Markscheme

522 (kg)     (A1)     (C1)[1 mark]

a.

$$522(8 – 6.80)$$ or equivalent     (M1)

Note:     Award (M1) for multiplying their answer to part (a) by $$(8 – 6.80)$$.

626 (EUR) (626.40)     (A1)(ft)     (C2)

Note:     Follow through from part (a).[2 marks]

b.

$$(W = ){\text{ }}(882 – 45p)(p – 6.80)$$     (A1)

OR

$$(W = ) – 45{p^2} + 1188p – 5997.6$$     (A1)     (C1)[1 mark]

c.

sketch of $$W$$ with some indication of the maximum     (M1)

OR

$$– 90p + 1188 = 0$$     (M1)

Note:     Award (M1) for equating the correct derivative of their part (c) to zero.

OR

$$(p = ){\text{ }}\frac{{ – 1188}}{{2 \times ( – 45)}}$$     (M1)

Note:     Award (M1) for correct substitution into the formula for axis of symmetry.

$$(p = ){\text{ }}13.2{\text{ (EUR)}}$$     (A1)(ft)     (C2)

Note:     Follow through from their part (c), if the value of $$p$$ is such that $$6.80 < p < 19.6$$.[2 marks]

d.

## Question

A factory produces shirts. The cost, C, in Fijian dollars (FJD), of producing x shirts can be modelled by

C(x) = (x − 75)2 + 100.

The cost of production should not exceed 500 FJD. To do this the factory needs to produce at least 55 shirts and at most s shirts.

Find the cost of producing 70 shirts.[2]

a.

Find the value of s.[2]

b.

Find the number of shirts produced when the cost of production is lowest.[2]

c.

## Markscheme

(70 − 75)2 + 100     (M1)

Note: Award (M1) for substituting in x = 70.

125     (A1) (C2)[2 marks]

a.

(s − 75)2 + 100 = 500     (M1)

Note: Award (M1) for equating C(x) to 500. Accept an inequality instead of =.

OR

(M1)

Note: Award (M1) for sketching correct graph(s).

(s =) 95    (A1) (C2)[2 marks]

b.

(M1)

Note: Award (M1) for an attempt at finding the minimum point using graph.

OR

$$\frac{{95 + 55}}{2}$$     (M1)

Note: Award (M1) for attempting to find the mid-point between their part (b) and 55.

OR

(C’(x) =) 2x − 150 = 0     (M1)

Note: Award (M1) for an attempt at differentiation that is correctly equated to zero.

75     (A1) (C2)[2 marks]

c.

## Question

A small manufacturing company makes and sells $$x$$ machines each month. The monthly cost $$C$$ , in dollars, of making $$x$$ machines is given by
$C(x) = 2600 + 0.4{x^2}{\text{.}}$The monthly income $$I$$ , in dollars, obtained by selling $$x$$ machines is given by
$I(x) = 150x – 0.6{x^2}{\text{.}}$$$P(x)$$ is the monthly profit obtained by selling $$x$$ machines.

Find $$P(x)$$ .[2]

a.

Find the number of machines that should be made and sold each month to maximize $$P(x)$$ .[2]

b.

Use your answer to part (b) to find the selling price of each machine in order to maximize $$P(x)$$ .[2]

c.

## Markscheme

$$P(x) = I(x) – C(x)$$     (M1)
$$= – {x^2} + 150x – 2600$$     (A1)    (C2)

a.

$$– 2x + 150 = 0$$     (M1)

Note: Award (M1) for setting $$P'(x) = 0$$ .

OR

Award (M1) for sketch of $$P(x)$$ and maximum point identified.     (M1)
$$x = 75$$     (A1)(ft)     (C2)

b.

$$\frac{{7875}}{{75}}$$     (M1)

Note: Award (M1) for $$7875$$ seen.

$$= 105$$ (A1)(ft)     (C2)

c.

[MAI 5.8] MONOTONY – CONCAVITY – OPTIMIZATION-lala-ready

### Question

[Maximum mark: 12]
Differentiate the following functions and hence determine whether each function is
increasing or decreasing (or neither).

(i) $$f(x)=x^{5}+e^{x}+1$$,                                          (ii) $$f(x)=x^{3}+ln\;x, \;x>0$$
(iii) $$f(x)=5-3e^{2x}$$                                                 (iv) $$f(x)=\frac{e^{x}-1}{e^{x}+1}$$

Ans.

(i) $${f}'(x)=5x^{4}+e^{x}>0$$, $$f$$ increasing                             (ii) $${f}'(x)=3x^{2}+\frac{1}{x}>0$$, $$f$$ increasing
(iii) $${f}'(x)=-6e^{2x}<0$$, $$f$$ decreasing                                    (iv) $${f}'(x)=\frac{2e^{x}}{(e^{x}+1)^{2}}>0$$, $$f$$ increasing

### Question

[Maximum mark: 8]
Consider $$f(x)=\frac{1}{3}x^{3}+2x^{2}-5x$$. Part of the graph of $$f$$ is shown below. There is a
maximum point at M, and a point of inflexion at N.

(a) Find $${f}'(x)$$
(b) Find the x – coordinate of M.
(c) Find the x – coordinate of N.
(d) Write down the equation of the normal line at M .

Ans.

(a)$${f}'(x)=x^{2}+4x-5$$
(b)$${f}'(x)=0\Leftrightarrow x=-5$$, $$x=1$$
so $$x=-5$$
(c) $${f}”(x)=2x+4$$
2x+4=0
$$x=-2$$
(d) $$x=-5$$

### Question

[Maximum mark: 4]
It is given that $${f}”(x)=(x-1)(x-3)(x-4)^{2}$$.
Find the points of inflection of $$f$$; justify your answer.

Ans.

x = 1 and x = 3 are points of inflection ( x = 4 is not)

### Question

[Maximum mark: 5]
Let $${f}'(x)=-24x^{3}+9x^{2}+3x+1$$.
(a) There are two points of inflexion on the graph of $$f$$ . Write down the x-coordinates
of these points.
(b) Let $$g(x)={f}”(x)$$. Explain why the graph of $$g$$ has no points of inflexion.

Ans.

(a) f″(x) = 0 OR the max and min of f′ gives the points of inflexion on f
–0.114, 0.364
(b) graph of g is a quadratic function, so it does not have any points of inflexion
OR graph of g is concave down over entire domain therefore no change in concavity
OR g″(x) = –144, therefore no points of inflexion as g″(x) ≠ 0

### Questioin

[Maximum mark: 8]
The graph of a function $$g$$ is given in the diagram below.

The gradient of the curve has its maximum value at point B and its minimum value at
point D. The tangent is horizontal at points C and E.

(a) Complete the table below, by stating whether the first derivative g′ is positive or
negative, and whether the second derivative g″ is positive or negative.

(b) Complete the table below by noting the points on the graph described by the
following conditions.

(c) Write down the values of g′ (e) and g′′ (e).
(d) Write down the number of points of inflexion for this curve.

Ans.

### Question

[Maximum mark: 7]
Consider the function $$f(x)=\frac{3x-2}{2x+5}$$.
The graph of this function has a vertical and a horizontal asymptote.
(a) Write down the equations of the asymptotes
(b) Find $${f}'(x)$$ , simplifying the answer as much as possible.
(c) Write down the number of stationary points of the graph. Justify your answer.
(d) Write down the number of points of inflection of the graph.

Ans.

(a)       (i) $$x=-\frac{5}{2}$$               (ii) $$y=\frac{3}{2}$$

(b) By quotient rule:  $$\frac{dy}{dx}=\frac{(2x+5)(3)-(3x-2)(2)}{(2x+5)^{2}}=\frac{19}{(2x+5)^{2}}$$

(c) There are no stationary points, since $$\frac{dy}{dx}\neq 0$$ (or by the graph) (A1)

(d) There are no points of inflexion.

### Question

[Maximum mark: 5]
A function $$f$$ has its first derivative given by $${f}'(x)=(x-3)^{3}$$.
(a) Find the second derivative.
(b) Find $${f}'(3)$$ and $${f}”(3)$$ .
(c) Explain why the point P on the graph with x-coordinate 3 is not a point of inflexion.

Ans.

(a) f″(x) = 3(x – 3)2
(b) f′(3) = 0, f″(3) = 0
(c) f″ does not change sign at P

### Question

[Maximum mark: 8]
The diagram below shows the graph of $$f(x)=x^{2}e^{-x}$$ for 0 ≤ x ≤ 6 . There are points of
inflexion at A and C and there is a maximum at B.

(a) Using the product rule for differentiation, find $${f}'(x)$$.
(b) Find the exact value of the y-coordinate of B.
(c) (i) Show that $${f}”(x)=(x^{2}-4x+2)e^{-x}$$.
(ii) Hence, find the exact value of the x – coordinate of C.

Ans.

(a) $${f}'(x)=2xe^{-x}-x^{2}e^{-x}=(2-x)xe^{-x}$$

(b) Maximum occurs at x = 2
Exact maximum value = 4e-2

(c) $${f}”(x)=2e^{-x}+2xe^{-x}-2xe^{-x}+x^{2}e^{-x}=(x^{2}-4x+2)e^{-x}$$
For inflexion,$${f}”(x)=0$$
$$x=\frac{4+\sqrt{8}}{2}(=2+\sqrt{2})$$

### Question

[Maximum mark: 6]
Let $$g(x)=\frac{ln\;x}{x^{2}}$$, for x > 0 . The graph of $$g$$ has a maximum point at A
(a) Use the quotient rule to show that $${g}'(x)=\frac{1-2\;ln\;x}{x^{3}}$$.
(b) Find the x-coordinate of A in the form ea

Ans.

(a) $${g}'(x)=\frac{x^{2}\left ( \frac{1}{x} \right )-2x\;ln\;x}{x^{4}}=\frac{x-2x\;ln\;x}{x^{4}}=\frac{x(1-2\;ln\;x)}{x^{4}}=:\frac{1-2\;ln\;x}{x^{3}}$$

(b) $${g}'(x)=0\Leftrightarrow 1-2ln\;x=0\Leftrightarrow ln\;x=\frac{1}{2}\Leftrightarrow x=e^{\frac{1}{2}}$$

### Question

[Maximum mark: 6]
A function $$f$$ is defined for -4 ≤ x ≤ 3. The graph of $$f$$ is given below.

The graph has a local maximum when $$x$$ = 0 , and local minima when $$x$$ =-3, $$x$$ = 2 .
(a) Write down the x -intercepts of the graph of the derivative function, $${f}’$$.
(b) Write down all values of $$x$$ for which $${f}'(x)$$ is positive.
(c) At point D on the graph of $$f$$, the x-coordinate is – 0.5. Explain why $${f}”(x) < 0$$ at D.

Ans.

(a) x-intercepts at –3, 0, 2
(b) –3 < x < 0, 2 < x < 3
(c) the graph of f is concave-down therefore the second derivative is negative

### Question

[Maximum mark: 8]
Consider the function , $$h:x \mapsto \frac{x-2}{(x-1)^{2}}$$,  $$x\neq 1$$.
A sketch of part of the graph of $$h$$ is given below.

The line (AB) is a vertical asymptote. The point P is a point of inflexion.
(a) Write down the equation of the vertical asymptote.
(b) Find $${h}'(x)$$ writing your answer in the form $$\frac{a-x}{(x-1)^{n}}$$
(c) Given that $${h}”(x)=\frac{2x-8}{(x-1)^{4}}$$ , calculate the coordinates of P.

Ans.

(a) x = 1
(b) Using quotient rule

$${h}'(x)=\frac{(x-1)^{2}(1)-(x-2)[2(x-1)]}{(x-1)^{4}}=\frac{(x-1)-(2x-4)}{(x-1)^{3}}=\frac{3-x}{(x-1)^{3}}$$

(c) at point of inflexion $${g}”(x)=0$$
$$x = 4$$

$$y=\frac{2}{9}=0.222$$ ie $$P\left ( 4,\frac{2}{9} \right )$$

### Question

[Maximum mark: 6]
The diagram below shows the graph of $$y = f (x)$$ . The tangent lines on this curve at
points A, B, C are parallel to the x-axis.

(a) Sketch the graph of the function $$y={f}'(x)$$ , by indicating the x-intercepts.

(b) Sketch the graph of the function $$y={f}”(x)$$ , by indicating the x-intercepts.

Ans.

### Question

[Maximum mark: 4]
The following diagram shows the graph of a function $$f$$.

Ans.

### Question

[Maximum mark: 6]
The first diagram below shows part of the graph of the gradient function $$y={f}'(x)$$.
(a) On the grid on the right, sketch a graph of $${f}”(x)$$; indicate the x-intercept.

(b) Complete the table, for the graph of $$y = f (x)$$.

Ans.

### Question

[Maximum mark: 8]
Let $$y = g(x)$$ be a function of $$x$$ for 1 ≤ x ≤ 7 . The graph of $$g$$ has an inflexion point at
P, and a minimum point at M.
Partial sketches of the curves of $${g}’$$ and $${g}”$$ are shown below.

Use the above information to answer the following.
(c) Given that g(4) = 0, sketch the graph of $$g$$ .Mark the points P and M.

Ans.

(a) x = 4  g″ changes sign at x = 4 or concavity changes
(b) x = 2
EITHER g′ goes from negative to positive
OR g′ (2) = 0 and g″ (2) is positive

### Question

[Maximum mark: 7]
Consider all rectangles of constant perimeter 4a .
Find the dimensions of the rectangle of maximum area and hence the maximum area.

Ans.

Let $$x$$ be one side of the rectangle.
The other side will be $$y=\frac{4a-2x}{2}=2a-x$$
Then the area is given by $$A=x(2a-x)=2ax-x^{2}$$

$$\frac{dA}{dx}=2a-2x=0\Leftrightarrow x=a$$
$$\frac{d^{2}A}{dx^{2}}=-2<0$$ so $$x=a$$ gives a maximum.
It is a square of side $$x=a$$ and the maximum area is $$A=a^{2}$$

### Question

[Maximum mark: 7]
Consider all rectangles of constant area A = a2 .

Ans.

Let x be one side of the rectangle.
The other side will be $$y=\frac{a^{2}}{x}$$.
Then the perimeter is given by $$P=2x+\frac{2a^{2}}{x}$$

$$\frac{dP}{dx}=2-\frac{2a^{2}}{x^{2}}=0\Leftrightarrow x^{2}=a^{2}\Leftrightarrow x=a$$

$$\frac{d^{2}P}{dx^{2}}=\frac{400}{x^{3}}>0$$ for $$x=a$$, so it gives a minimum.
It is a square of side $$x=a$$ and the minimum perimeter is $$P = 4a$$

### Question

[Maximum mark: 9]
Two cubes of edges x and y respectively respectively are shown below.
The total surface area of two cubes is 300 cm2.

The total surface area of the two cubes is 300 cm2.
(a) Find an expression of y in terms of x .
(b) Given that there is a minimum value of the total volume of the cubes, find this
minimum value and the corresponding dimensions of the two cubes.

Ans.

(a) $$6x^{2}+6y^{2}=300\Leftrightarrow x^{2}+y^{2}=50\Leftrightarrow y=\sqrt{50-x^{2}}$$

(b) $$V=x^{3}+(50-x^{2})^{\frac{3}{2}}$$.
$$\frac{dV}{dx}=3x^{2}-\frac{3}{2}(50-x^{2})^{\frac{1}{2}}2x=3x^{2}-3x\sqrt{50-x^{2}}$$
$$\frac{dV}{dx}=0\Leftrightarrow 3x^{2}-3x\sqrt{50-x^{2}}=0\Leftrightarrow x=\sqrt{50-x^{2}}\Leftrightarrow x^{2}=50-x^{2}\Leftrightarrow x=5$$

Then y = 5. We have two similar cubes of total volume $$V=5^{3}+5^{3}=250$$.

Notice: By using GDC, graph mode, the minimum of the function $$V=x^{3}+(50-x^{2})^{\frac{3}{2}}$$ is (5,250)
So the minimum value is 250

### Question

[Maximum mark: 9]
A point on the curve $$y=x^{2}$$ has coordinates P(a,a2). The point A$$\left ( 2,\frac{1}{2} \right )$$ does not lie
in the curve.

(a) Express the distance D between P and A in terms of $$a$$ .
(b) Find $$\frac{dD}{da}$$
(c) Hence find
(i) the coordinates of the point P on the line which is closest to A;
(ii) the minimum distance D between the point A and the curve $$y=x^{2}$$.

Ans.

METHOD 1

(a) Let $$D=\sqrt{(a-2)^{2}+(a^{2}-\frac{1}{2})^{2}}$$

(b) $$\frac{dD}{da}=\frac{1}{2\sqrt{—}}\left [ 2(a-2)+2(a^{2}-\frac{1}{2})2a \right ]=\frac{1}{2\sqrt{—}}\left [ 2a-4+4x^{3}-2a \right ]=\frac{1}{2\sqrt{—}}\left [ 4a^{3}-4 \right ]$$
$$=\frac{2}{\sqrt{—}}\left [ a^{3}-1 \right ]$$

(c) $$\frac{dD}{da}=0\Leftrightarrow a^{3}-1=0\Leftrightarrow a=1$$

(i) The point is (1, 12) i.e. (1, 1)
(ii) The minimum distance is $$D=\frac{\sqrt{5}}{2}(\cong 1.12)$$

Notice :
we can also use the GDC graph for the function $$D=\sqrt{(x-2)^{2}+(x^{2}-\frac{1}{2})^{2}}$$
It has a minimum at (1, 1.12)
Hence (i) The point is (1, 12) i.e. (1, 1)
(ii) The minimum distance is D = 1.12

### Question

[Maximum mark: 9]
The following diagram shows a cuboid of square base of side x and height y . The
volume of the cuboid is 125 cm2

(a) Express y in term of x .
(b) Hence, express the surface area S of the cuboid in terms of x .
(c) Use $$\frac{dS}{dx}$$ to find the minimum value of S ; justify your answer.

Ans.

(a) $$x^{2}y=125\Leftrightarrow y=\frac{125}{x^{2}}$$

(b) $$S=2x^{2}+4xy=2x^{2}+4x\frac{125}{x^{2}}=2x^{2}+\frac{500}{x}$$

(c) $$\frac{dS}{dx}=4x-\frac{500}{x^{2}}$$
$$4x-\frac{500}{x^{2}}=0\Leftrightarrow 4x=\frac{500}{x^{2}}\Leftrightarrow x^{3}=125\Leftrightarrow x=5$$
$$\frac{d^{2}S}{dx^{2}}=4+\frac{1000}{x^{3}}$$
For $$x=5$$, $$\frac{d^{2}S}{dx^{2}}<0$$ hence max
$$S_{max}=150$$

### Question

[Maximum mark: 12]
Let $$f(x)=ax^{3}+bx^{2}+cx$$
(a) Find the first and the second derivative of $$f (x)$$, in terms of a, b, c.
(b) The graph
passes through the point P(1,4)
has a local maximum at P
has a point of inflection at $$x$$ = 2 .
Write down three linear equations representing this information.
(c) Hence find the values of a, b, c.
(d) The function has a local minimum at $$x = d$$ . Find the value of $$d$$ and justify that it
is a minimum.

Ans.

(a)$${f}'(x)=3ax^{2}+2bx+c$$
$${f}”(x)=6ax+2b$$

(b) $$f(1)=4\Rightarrow a+b+c=4$$
$${f}'(1)=0\Rightarrow 3a+2b+c=0$$
$${f}”(2)=0\Rightarrow 12a+2b=0$$

(c) a = 1,   b =-6,   c = 9

(d) $${f}'(x)=3x^{2}-12x+9$$, stationary points: x = 1, x = 3
minimum at x = 3 since $${f}”(x)=6x-12$$ and $${f}”(3)=6>0$$

### Question

[Maximum mark: 12]
The function $$f$$ is given by $$f(x)=\frac{ln\;2x}{x},\;x>0$$

(a) (i) Show that $${f}'(x)=\frac{1-ln\;2x}{x^{2}}$$.
Hence
(ii) prove that the graph of $$f$$ can have only one local max or min point;
(iii) find the coordinates of the maximum point on the graph of $$f$$.

(b) By showing that the second derivative $${f}”(x)=\frac{2\;ln\;2x-3}{x^{3}}$$ or otherwise,
find the coordinates of the point of inflexion on the graph of $$f$$ .

Ans.

(a)    (i) $${f}'(x)=\frac{\left ( x\;x\;\frac{1}{2x}\;x\;2 \right )-(ln\;2x\;x1)}{x^{2}}=\frac{1-ln\;2x}{x^{2}}$$

(ii) $${f}'(x)=0\Leftrightarrow \frac{1-ln\;2x}{x^{2}}=0$$ only at 1 point, when $$x=\frac{e}{2}$$

(iii) Maximum point when $${f}'(x)=0$$.

$${f}'(x)=0$$ for $$x=\frac{e}{2}(=1.36)$$
$$y=f\left ( \frac{e}{2} \right )=\frac{2}{e}(=0.736)$$

(b) $${f}”(x)=\frac{-\frac{1}{2x}x2\;x\;x^{2}-(1-ln\;2x)2x}{x^{4}}=\frac{2\;ln\;2x-3}{x^{3}}$$

Inflexion point $$\Rightarrow {f}”(x)=0\Rightarrow 2ln\;2x=3\Rightarrow x=\frac{e^{1.5}}{2}(=2.24)$$

### Question

[Maximum mark: 13]
The following diagram shows the graph of $$f(x)=e^{-x^{2}}$$.

The points A, B, C, D and E lie on the graph of $$f$$ . Two of these are points of inflexion.
(a) Identify the two points of inflexion.
(b) (i) Find $${f}'(x)$$.
(ii) Show that $${f}”(x)=(4x^{2}-2)e^{-x^{2}}$$.
(c) Find the x -coordinate of each point of inflexion.
(d) Use the second derivative to show that one of these points is a point of inflexion.

Ans.

(a) B, D
(b) (i)$${f}'(x)=-2xe^{-x^{2}}$$
(ii) product rule

$${f}”(x)=-2e^{-x^{2}}-2x\;x\;-2xe^{-x^{2}}=-2e^{-x^{2}}+4x^{2}e^{-x^{2}}=(4x^{2}-2)e^{-x^{2}}$$

(c) $${f}”(x)=0\Leftrightarrow (4x^{2}-2)=0$$
$$p=0.707\left ( =\frac{1}{\sqrt{2}} \right )$$, $$q=-0.707\left ( =-\frac{1}{\sqrt{2}} \right )$$

(d) checking sign of f ′′ on either side of POI
sign change of f ′′(x)

### Question

[Maximum mark: 10]
Let the function $$f$$ be defined by $$f(x)=\frac{2}{1+x^{3}}$$,  $$x\neq -1$$

(a) (i) Write down the equation of the vertical asymptote of the graph of $$f$$ .
(ii) Write down the equation of the horizontal asymptote of the graph of $$f$$.
(iii) Sketch the graph of $$f$$ in the domain -3 ≤ x ≤ 3.

(b) (i) Using the fact that $${f}'(x)=\frac{-6x^{2}}{(1+x^{3})^{2}}$$, show that the second derivative is

$${f}”(x)=\frac{12x(2x^{3}-1)}{(1+x^{3})^{3}}$$.

(ii) Find the x -coordinates of the points of inflexion of the graph of $$f$$.

Ans.

(a)      (i) Vertical asymptote x = –l                                    (ii) Horizontal asymptote y = 0

(iii)

(b)   (i) $${f}'(x)=\frac{-6x^{2}}{(1+x^{3})^{2}}$$
$${f}”(x)=\frac{(1+x^{3})^{2}(-12x)+6x^{2}(2)(1+x^{3})^{1}(3x^{2})}{(1+x^{3})^{4}}$$
$$=\frac{(1+x^{3})(-12x)+36x^{4}}{(1+x^{3})^{3}}=\frac{-12-12x^{4}+36x^{4}}{(1+x^{3})^{3}}=\frac{12x(2x^{3}-1)}{(1+x^{3})^{3}}$$

(ii) Point of inflexion => $${f}”(x)=0=>x=0$$ or $$x=\sqrt[3]{\frac{1}{2}}$$
$$x=0$$ or  $$x=0.794$$ (3 sf)

### Question

[Maximum mark: 14]
The function $$f$$ is defined as $$f(x)=(2x+1)e^{-x}$$, 0 ≤ x ≤ 3 . The point P(0, 1) lies on the
graph of $$f (x)$$ , and there is a maximum point at Q.
(a) Sketch the graph of $$y = f (x)$$ , labelling the points P and Q.
(b) (i) Show that $${f}'(x)=(1-2x)e^{-x}$$.
(ii) Find the exact coordinates of Q.
(c) The equation $$f (x) = k$$ , where $$k\in \mathbb{R}$$ , has two solutions. Write down the range of
values of $$k$$ .
(d) Given that $${f}”(x)=e^{-x}(-3+2x)$$ , show that the curve of $$f$$ has only one point of
inflexion.

Ans.

(a)

(b)      (i) $${f}'(x)=2e^{-x}+(2x+1)(-e^{-x})=(1-2x)e^{-x}$$

(ii) At Q, $${f}'(x)=0$$

x = 0.5,  y = 2e-0.5     Q is (0.5, 2e-0.5)

(c) 1 ≤ k < 2e-0.5
(d) $${f}”(x)=0\Leftrightarrow e^{-x}(-3+2x)=0$$
This equation has only one root. So $$f$$ has only one point of inflexion.

### Question

[Maximum mark: 10]
Consider the function $$f$$ given by $$f(x)=\frac{2x^{2}-13x+20}{(x-1)^{2}}$$,    $$x\neq 1$$.
A part of the graph of $$f$$ is given below (vertical and horizontal asymptotes are shown)

(a) Write down the equation of the vertical asymptote.
(b) It is given that $$f (100) = 1.91$$, $$f (-100) = 2.09$$, $$f (1000)= 1.99$$
Evaluate $$f (-1000)$$ and write down the equation of the horizontal asymptote.
(c) Show that $${f}'(x)=\frac{9x-27}{(x-1)^{3}}$$,   $$x\neq 1$$.
The second derivative is given by $${f}”(x)=\frac{72-18x}{(x-1)^{4}}$$,  $$x\neq 1$$.
(d) Using values of $${f}'(x)$$ and $${f}”(x)$$ explain why a minimum must occur at $$x = 3$$.
(e) There is a point of inflexion on the graph. Write down the coordinates of this point.

Ans.

(a) x = 1

(b)  (i)  f (–1000) = 2.01                  (ii) y = 2

(c) $${f}'(x)=\frac{(x-1)^{2}(4x-13)-2(x-1)(2x^{2}-13x+20)}{(x-1)^{4}}$$
$$=\frac{(4x^{2}-17x+13)-(4x^{2}-26x+40)}{(x-1)^{3}}=\frac{9x-27}{(x-1)^{3}}$$

(d) f ′(3) = 0                         ⇒ stationary point
$${f}”(3)$$ = $$\frac{18}{16}>0$$                  ⇒ minimum

(e) Point of inflexion  ⇒ $${f}”(x)$$ = 0 ⇒ x = 4
x = 4 ⇒ y = 0               ⇒ Point of inflexion = (4, 0)

### Question

[Maximum mark: 13]
Consider $$f(x)=x\;ln(4-x^{2})$$, for -2 < x < 2 . The graph of $$f$$ is given below.

(a) Let P and Q be the stationary points on the curve of $$f$$ .
(i) Find the x -coordinate of P and of Q.
(ii) Write down all values of $$k$$ for which $$f (x) = k$$ has exactly two solutions.

Let $$g(x)=x^{3}\;ln(4-x^{2})$$, for -2 < x < 2 .

(b) Show that $${g}'(x)=\frac{-2x^{4}}{4-x^{2}}+3x^{2}\;ln(4-x^{2})$$.
(c) By observing the graph of $${g}’$$ , write down all values of $$w$$ for which $${g}'(x) = w$$ has
exactly two solutions.

Ans.

(a)  (i) –1.15, 1.15
(ii) it occurs at P and Q (when x = –1.15, x = 1.15)
k = –1.13, k = 1.13

(b) $${g}'(x)=x^{3}$$ x $$\frac{-2x}{4-x^{2}}+ln(4-x^{2})$$ x $$3x^{2}=\frac{-2x^{4}}{4-x^{2}}+3x^{2}ln(4-x^{2})$$

(c)

### Question

[Maximum mark: 13]
Let $$f(x)=3+\frac{20}{x^{2}-4}$$, for $$x$$ = ±2 . The graph of $$f$$ is given below.

The y -intercept is at the point A.
(a)     (i) Find the coordinates of A                   (ii) Show that $${f}'(x)=0$$ at A.

(b) The second derivative of $$f$$ is $${f}”(x)=\frac{40(3x^{2}+4)}{(x^{2}-4)^{3}}$$. Use this to
(i) justify that the graph of $$f$$ has a local maximum at A;
(ii) explain why the graph of $$f$$ does not have a point of inflexion.

(c) Describe the behaviour of the graph of $$f$$ for large $$\left | x \right |$$ .
(d) Write down the range of $$f$$ .

Ans.

(a)  (i) coordinates of A are (0, –2)

(ii) $$f(x)=3+20x(x^{2}-4)^{-1}$$
$${f}'(x)=20$$ x $$(-1)x(x^{2}-4)^{-2}$$ x $$(2x)=-40x(x^{2}-4)^{-2}$$
OR  $$\frac{(x^{2}-4)(0)-(20)(2x)}{(x^{2}-4)^{2}}$$

substituting x = 0 into f′(x) gives f ′(x) = 0

(b) (i) f′(0) = 0 (stationary)
f″(0) = $$\frac{40×4}{(-4)^{3}}\left ( =\frac{-5}{2} \right )$$ negative
then the graph must have a local maximum

(ii) f ″(x) = 0 at point of inflexion,
but the second derivative is never 0 (the numerator is always positive)

(c) getting closer to the line y = 3, horizontal asymptote at y = 3
(d) y ≤ –2,   y > 3

### Question

[Maximum mark: 14]
Let $$f(x)=e^{x}(1-x^{2})$$.
(a) Show that $${f}'(x)=e^{x}(1-2x-x^{2})$$.
Part of the graph of $$y = f (x)$$, for -6 ≤ x ≤ 2 , is shown below. The x -coordinates of
the local minimum and maximum points are r and s respectively.

(b) Write down the equation of the horizontal asymptote.
(c) Write down the value of r and of s .
(d) Show that the normal L to the curve of $$f$$ at P(0, 1) has equation $$x + y = 1$$.
(e) Find the coordinates of the points where the curve and the line L intersect.

Ans.

(a) $${f}'(x)=e^{x}(1-x^{2})+e^{x}(-2x)=e^{x}(1-2x-x^{2})$$
(b) y = 0
(c) at the local maximum or minimum point
$${f}'(x)=0\Leftrightarrow e^{x}(1-2x-x^{2})=0\Rightarrow 1-2x-x^{2}=0$$
r = -2.41 s = 0.414 (OR directly by GDC graph)
(d) f ‘(0) = 1 ⇒ gradient of the normal = -1
$$y-1=-1(x-0)\Leftrightarrow x+y-1$$
(e)   (i)  intersection points at (0,1) and (1,0)

### Question

[Maximum mark: 12]
Let $$f(x)=\frac{1}{3}x^{3}-x^{2}-3x$$. Part of the graph of $$f$$ is shown below.

There is a maximum point at A and a minimum point at B(3, –9).
(a) Find the coordinates of A.
(b) Write down the coordinates of
(i) the image of B after reflection in the y -axis;
(ii) the image of B after translation by the vector $$\binom{-2}{5}$$;
(iii) the image of B after reflection in the x -axis followed by a horizontal stretch
with scale factor $$\frac{1}{2}$$.

Ans.

(a) $${f}'(x)=x^{2}-2x-3$$
$$x^{2}-2x-3=0\Leftrightarrow x=\frac{2\pm \sqrt{16}}{2}\Leftrightarrow x=-1$$ or $$x=3$$
$$x=-1$$ (ignore $$x=3$$)  $$y=-\frac{1}{3}-1+3=\frac{5}{3}$$
coordinates are $$\left ( -1,\frac{5}{3} \right )$$

(b)  (i)  (–3, –9)
(ii)  (1, –4)
(iii) reflection gives (3, 9)                   stretch gives $$\left ( \frac{3}{2},9 \right )$$

### Question

[Maximum mark: 12]
Let $$f(x)=\frac{cos\;x}{sin\;x}$$, for sin x ≠ 0 .

(a) Use the quotient rule to show that $${f}'(x)=\frac{-1}{sin^{2}\;x}$$.

(b) Find $${f}”(x)$$.

In the following table, $${f}’\left ( \frac{\pi }{2} \right )=p$$ and $${f}’\left ( \frac{\pi }{2} \right )=q$$. The table also gives approximate
values of $${f}'(x)$$ and $${f}”(x)$$ near $$x=\frac{\pi }{2}$$.

(c) Find the value of p and of q .
(d) Use information from the table to explain why there is a point of inflexion on the
graph of $$f$$ where $$x=\frac{\pi }{2}$$.

Ans.

(a) quotient rule

$${f}'(x)=\frac{sin\;x(-sin\;x)-cos\;x(cos\;x)}{sin^{2}x}=\frac{-(sin^{2}x+cos^{2}x)}{sin^{2}x}=\frac{-1}{sin^{2}x}$$

(b) $${f}'(x)=-(sin\;x)^{-2}$$
$${f}”(x)=2(sin\;x)^{-3}(cos\;x)\left ( =\frac{2\;cos\;x}{sin^{3}x} \right )$$

(c) substituting $$\frac{\pi }{2}\Rightarrow p=-1, q=0$$

(d) second derivative is zero, second derivative changes sign

### Question

[Maximum mark: 12]
The diagram shows the graph of $$y=e^{x}$$ (cos x + sin x) , -2 ≤ x ≤ 3. The graph has a
maximum turning point at C(a, b) and a point of inflexion at D.

(a) Find $$\frac{dy}{dx}$$.
(b) Find the exact value of a and of b.
(c) Show that at D, $$y=\sqrt{2}e^{\frac{\pi }{4}}$$.

Ans.

(a) $$\frac{dy}{dx}=e^{x}(cos\;x+sin\;x)+e^{x}(-sin\;x+cos\;x)=2e^{x}cos\;x$$

(b) $$\frac{dy}{dx}=0\Rightarrow 2e^{x}cos\;x=0\Rightarrow cos\;x=0\Rightarrow x=\frac{\pi }{2}\Rightarrow a=\frac{\pi }{2}$$

$$y=e^{\frac{\pi }{2}}(cos\frac{\pi }{2}+sin\frac{\pi }{2})=e^{\frac{\pi }{2}}\Rightarrow b=e^{\frac{\pi }{2}}$$

(c) At D, $$\frac{d^{2}y}{dx^{2}}=0\Rightarrow 2e^{x}cos\;x-2e^{x}sin\;x=0\Rightarrow 2e^{x}(cos\;x-sin\;x)=0$$
$$\Rightarrow cos\;x-sin\;x=0\Rightarrow x=\frac{\pi }{4}$$
$$y=e^{\frac{\pi }{4}}(cos\frac{\pi }{4}+sin\frac{\pi }{4})=\sqrt{2}e^{\frac{\pi }{4}}$$

### Question

[Maximum mark: 12]
Part of the graph of  $$f(x)=\frac{1}{1+x^{2}}$$ is shown below.

(a) Write down the equation of the horizontal asymptote of the graph of $$f$$.
(b) Find $${f}'(x)$$.
(c) Show that the second derivative is given by $${f}”(x)=\frac{6x^{2}-2}{(1+x^{2})^{3}}$$.
(d) Let A be the point on the curve of $$f$$ where the gradient of the tangent is a
maximum. Find the x -coordinate of A.

Ans.

(a)  y = 0

(b) $${f}'(x)=\frac{-2x}{(1+x^{2})^{2}}$$

(c) $${f}'(x)=-2x(1+x^{2})^{-2}$$,
$${f}”(x)=-2(1+x^{2})^{-2}+4x(1+x^{2})^{-3}2x=\frac{-2}{(1+x^{2})^{2}}+\frac{8x^{2}}{(1+x^{2})^{3}}$$
$$=\frac{-2(1+x^{2})}{(1+x^{2})^{3}}+\frac{8x^{2}}{(1+x^{2})^{3}}=\frac{6x^{2}-2}{(1+x^{2})^{3}}$$

(d) $${f}”(x)=0\Leftrightarrow 6x^{2}-2=0\Leftrightarrow x=\sqrt[\pm ]{\frac{1}{3}}$$

The maximum gradient is at $$x=\frac{-1}{\sqrt{3}}$$

### Question

[Maximum mark: 11]
Consider the function $$f(x)=e^{(2x-1)}+\frac{5}{2x-1}$$,  $$x\neq \frac{1}{2}$$.

(a) Sketch the curve of $$f$$ for -2 ≤ x ≤ 2 , including any asymptotes.
(b) Write down the equation of the vertical asymptote of $$f$$ .
(c) Find $${f}'(x)$$.
(d) (i) Write down the value of $$x$$ at the minimum point on the curve of $$f$$ .
(ii) The equation $$f (x) = k$$ has no solutions for p ≤ k < q . Write down the value
of p and of q .

Ans.

(a)

(b) $$x=\frac{1}{2}$$    (must be an equation)
(c) $${f}'(x)=2e^{2x-1}-10(2x-1)^{-2}$$
(e)   (i)   x = 1.11        (accept (1.11, 7.49))          (ii)   p = 0, q = 7.49 (0 ≤ k < 7.49)

### Question

[Maximum mark: 12]
The function $$f$$ is defined as $$f(x)=e^{x}\;sin\;x$$, where $$x$$ is in radians. Part of the curve of
$$f$$ is shown below.

There is a point of inflexion at A, and a local maximum point at B. The curve of $$f$$
intersects the x-axis at the point C.
(a) Write down the x -coordinate of the point C.
(b) (i) Find $${f}'(x)$$.
(ii) Write down the value of $${f}'(x)$$ at the point B.
(c) Show that $${f}”(x)=2e^{x}\;cos\;x$$.
(d) (i) Write down the value of $${f}”(x)$$ at A, the point of inflexion.
(ii) Hence, calculate the coordinates of A.

Ans.

(a) π

(b)    (i) $${f}'(x)=e^{x}cos\;x+e^{x}sin\;x=e^{x}(cos\;x+sin\;x)$$

(ii)  At B,  f ‘(x) = 0

(c) $${f}”(x)=e^{x}cos\;x-e^{x}sin\;x+e^{x}sin\;x+e^{x}cos\;x-2e^{x}cos\;x$$

(d)   (i)  At A,  f ”(x) = 0
(ii) $$2e^{x}cos\;x=0\Leftrightarrow cos\;x=0$$
$$x=\frac{\pi }{2}$$, $$y=e^{\frac{\pi }{2}}$$                              Coordinates are $$\left ( \frac{\pi }{2},e^{\frac{\pi }{2}} \right )$$