# IBDP Maths AI: Topic: SL 5.6: Local maximum and minimum points: IB style Questions SL Paper 1

## Question

A function is represented by the equation

$f(x) = a{x^2} + \frac{4}{x} – 3$

Find $$f ′(x)$$ .

a.

The function $$f (x)$$ has a local maximum at the point where $$x = −1$$.

Find the value of a.

b.

## Markscheme

$$f(x) = a{x^2} + 4{x^{ – 1}} – 3$$

$$f'(x) = 2ax – 4{x^{ – 2}}$$     (A3)

(A1) for 2ax, (A1) for –4x –2 and (A1) for derivative of –3 being zero.     (C3)[3 marks]

a.

$$2ax – 4x^{-2} = 0$$     (M1)

$$2a( – 1) – 4{( – 1)^{ – 2}} = 0$$     (M1)

$$-2a – 4 = 0$$

$$a = -2$$     (A1)(ft)

(M1) for setting derivative function equal to 0. (M1) for inserting $$x = -1$$ but do not award (M0)(M1)     (C3)[3 marks]

b.

## Question

Let $$f (x) = 2x^2 + x – 6$$

Find $$f'(x)$$.

a.

Find the value of $$f'( – 3)$$.

b.

Find the value of $$x$$ for which $$f'(x) = 0$$.

c.

## Markscheme

$$f'(x) = 4x + 1$$     (A1)(A1)(A1)     (C3)

Note: Award (A1) for each term differentiated correctly.

Award at most (A1)(A1)(A0) if any extra terms seen.[3 marks]

a.

$$f'( – 3) = – 11$$     (A1)(ft)     (C1)[1 mark]

b.

$$4x + 1 = 0$$     (M1)

$$x = – \frac{{1}}{{4}}$$     (A1)(ft)     (C2)[2 marks]

c.

## Question

The table given below describes the behaviour of f ′(x), the derivative function of f (x), in the domain −4 < x < 2. State whether f (0) is greater than, less than or equal to f (−2). Give a reason for your answer.

a.

The point P(−2, 3) lies on the graph of f (x).

Write down the equation of the tangent to the graph of f (x) at the point P.

b.

The point P(−2, 3) lies on the graph of f (x).

From the information given about f ′(x), state whether the point (−2, 3) is a maximum, a minimum or neither. Give a reason for your answer.

c.

## Markscheme

greater than     (A1)

Gradient between x = −2 and x = 0 is positive.     (R1)

OR

The function is increased between these points or equivalent.     (R1)     (C2)

Note: Accept a sketch. Do not award (A1)(R0).[2 marks]

a.

y = 3     (A1)(A1)     (C2)

Note: Award (A1) for y = a constant, (A1) for 3.[2 marks]

b.

minimum     (A1)

Gradient is negative to the left and positive to the right or equivalent.     (R1)     (C2)

Note: Accept a sketch. Do not award (A1)(R0).[2 marks]

c.

## Question

f (x) = 5x3 − 4x2 + x

Find f‘(x).

a.

(i) the local maximum point;

(ii) the local minimum point.

b.

## Markscheme

15x2 – 8x + 1     (A1)(A1)(A1)     (C3)

Note: Award (A1) for each correct term.[3 marks]

a.

15x2 – 8x +1 = 0     (A1)(ft)

Note: Award (A1)(ft) for setting their derivative to zero.

(i) $$(x =)\frac{1}{5}(0.2)$$     (A1)(ft)

(ii) $$(x =)\frac{1}{3}(0.333)$$     (A1)(ft)     (C3)

b.

## Question

Consider the function $$f (x) = ax^3 − 3x + 5$$, where $$a \ne 0$$.

Find $$f ‘ (x)$$.

a.

Write down the value of $$f ′(0)$$.

b.

The function has a local maximum at x = −2.

Calculate the value of a.

c.

## Markscheme

$$f ‘(x) = 3ax^2 – 3$$     (A1)(A1)     (C2)

Note: Award a maximum of (A1)(A0) if any extra terms are seen.

a.

−3     (A1)(ft)     (C1)

Note: Follow through from their part (a).

b.

$$f ‘(x) = 0$$     (M1)

Note: This may be implied from line below.

$$3a(-2)^2 – 3 = 0$$     (M1)

$$(a =) \frac{1}{4}$$     (A1)(ft)     (C3)

Note: Follow through from their part (a).

c.

## Question

Expand the expression $$x(2{x^3} – 1)$$.

a.

Differentiate $$f(x) = x(2{x^3} – 1)$$.

b.

Find the $$x$$-coordinate of the local minimum of the curve $$y = f(x)$$.

c.

## Markscheme

$$2{x^4} – x$$     (A1)(A1)     (C2)

Note: Award (A1) for $$2{x^4}$$, (A1) for $$– x$$.[2 marks]

a.

$$8{x^3} – 1$$     (A1)(ft)(A1)(ft)     (C2)

Note: Award (A1)(ft) for $$8{x^3}$$, (A1)(ft) for $$–1$$. Follow through from their part (a).

Award at most (A1)(A0) if extra terms are seen.[2 marks]

b.

$$8{x^3} – 1 = 0$$     (M1)

Note: Award (M1) for equating their part (b) to zero.

$$(x = )\frac{1}{2}{\text{ (0.5)}}$$     (A1)(ft)     (C2)

Notes: Follow through from part (b).

$$0.499$$ is the answer from the use of trace on the GDC; award (A0)(A0).

For an answer of $$(0.5, –0.375)$$, award (M1)(A0).[2 marks]

c.

## Question

Consider the curve $$y = {x^3} + kx$$.

Write down $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$.

a.

The curve has a local minimum at the point where $$x = 2$$.

Find the value of $$k$$.

b.

The curve has a local minimum at the point where $$x = 2$$.

Find the value of $$y$$ at this local minimum.

c.

## Markscheme

$$3{x^2} + k$$     (A1)     (C1)[1 mark]

a.

$$3{(2)^2} + k = 0$$     (A1)(ft)(M1)

Note: Award (A1)(ft) for substituting 2 in their $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$, (M1) for setting their $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$.

$$k = – 12$$     (A1)(ft)     (C3)

Note: Follow through from their derivative in part (a).[3 marks]

b.

$${2^3} – 12 \times 2$$     (M1)

Note: Award (M1) for substituting 2 and their –12 into equation of the curve.

$$= – 16$$     (A1)(ft)     (C12)

Note: Follow through from their value of $$k$$ found in part (b).[2 marks]

c.

## Question

Consider the graph of the function $$f(x) = {x^3} + 2{x^2} – 5$$. Label the local maximum as $${\text{A}}$$ on the graph.

a.

Label the local minimum as B on the graph.

b.

Write down the interval where $$f'(x) < 0$$.

c.

Draw the tangent to the curve at $$x = 1$$ on the graph.

d.

Write down the equation of the tangent at $$x = 1$$.

e.

## Markscheme correct label on graph     (A1)     (C1)[1 mark]

a.

correct label on graph     (A1)     (C1)[1 mark]

b.

$$– 1.33 < x < 0$$   $$\left( { – \frac{4}{3} < x < 0} \right)$$     (A1)     (C1)[1 mark]

c. tangent drawn at $$x = 1$$ on graph     (A1)     (C1)[1 mark]

d.

$$y = 7x – 9$$     (A1)(A1)     (C2)

Notes: Award (A1) for $$7$$, (A1) for $$-9$$.

If answer not given as an equation award at most (A1)(A0).[2 marks]

e.

## Question

A function is given as $$f(x) = 2{x^3} – 5x + \frac{4}{x} + 3,{\text{ }} – 5 \leqslant x \leqslant 10,{\text{ }}x \ne 0$$.

Write down the derivative of the function.

a.

Use your graphic display calculator to find the coordinates of the local minimum point of $$f(x)$$ in the given domain.

b.

## Markscheme

$$6{x^2} – 5 – \frac{4}{{{x^2}}}$$     (A1)(A1)(A1)(A1)     (C4)

Note: Award (A1) for $$6{x^2}$$, (A1) for $$–5$$, (A1) for $$–4$$, (A1) for $${x^{ – 2}}$$ or $$\frac{1}{{{x^2}}}$$.

Award at most (A1)(A1)(A1)(A0) if additional terms are seen.[4 marks]

a.

$$(1.15,{\text{ }} 3.77)$$ $$\left( {{\text{(1.15469…, 3.76980…)}}} \right)$$     (A1)(A1)     (C2)

Notes: Award (A1)(A1) for “$$x = 1.15$$ and $$y = 3.77$$”.

Award at most (A0)(A1)(ft) if parentheses are omitted.[2 marks]

b.

## 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 $$a$$.

a.

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

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 quadratic function $$f$$ is given by $$f(x) = a{x^2} + bx + c$$. The points $$(0,{\text{ }}5)$$ and $$( – 4,{\text{ }}5)$$ lie on the graph of $$y = f(x)$$.

The $$y$$-coordinate of the minimum of the graph is 3.

Find the equation of the axis of symmetry of the graph of $$y = f(x)$$.

a.

Write down the value of $$c$$.

b.

Find the value of $$a$$ and of $$b$$.

c.

## Markscheme

$$x = – 2$$     (A1)(A1)     (C2)

Note:     Award (A1) for $$x =$$ (a constant) and (A1) for $$– 2$$.[2 marks]

a.

$$(c = ){\text{ }}5$$     (A1)     (C1)[1 mark]

b.

$$– \frac{b}{{2a}} = – 2$$

$$a{( – 2)^2} – 2b + 5 = 3$$ or equivalent

$$a{( – 4)^2} – 4b + 5 = 5$$ or equivalent

$$2a( – 2) + b = 0$$ or equivalent     (M1)

Note:     Award (M1) for two of the above equations.

$$a = 0.5$$     (A1)(ft)

$$b = 2$$     (A1)(ft)     (C3)

Note:     Award at most (M1)(A1)(ft)(A0) if the answers are reversed.

Follow through from parts (a) and (b).[3 marks]

c.

## 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.

a.

Find the value of s.

b.

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

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

Consider the graph of the function $$y = f(x)$$ defined below. Write down all the labelled points on the curve

that are local maximum points;

a.

where the function attains its least value;

b.

where the function attains its greatest value;

c.

where the gradient of the tangent to the curve is positive;

d.

where $$f(x) > 0$$ and $$f'(x) < 0$$ .

e.

B, F     (C1)

a.

H     (C1)

b.

F     (C1)

c.

A, E     (C1)

d.

C     (C2)

e.

## 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 $$P(x)$$ .

a.

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

b.

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

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 5.6-5.7] RULES OF DIFFERENTIATION-lala

### Question

[Maximum mark: 3 per function]
Differentiate the following functions: Ans. ### Question

[Maximum mark: 4]
Let $$f(x)=2x^{3}+ln\;x$$
(a) Find $${f}'(x)$$
(b) Find the gradient of the curve $$y = f (x)$$ at $$x = 1$$.

Ans.

(a)$${f}'(x)=6x^{2}+\frac{1}{x}$$                           (b)$${f}'(1)=6+1=7$$

### Question

[Maximum mark: 6]
Let $$f(x)=\frac{x^{3}+1}{sin\;x}$$

(a) Find $${f}'(x)$$
(b) Find the gradient of the curve $$y=f(x)$$

(i) at $$x=\frac{\pi }{4}$$                         (ii) at $$x = 1$$ rad.

Ans.

(a) $${f}'(x)=\frac{3x^{2}\;sin\;x-(x^{3}+1)cos\;x}{sin^{2}x}$$

(b) Directly by GDC  (i)  $${f}'(\frac{\pi }{4})\cong 0.518$$                          (ii)$${f}'(1)\cong 2.04$$

[Notice: the exact value for (i) is $${f}'(\frac{\pi }{4})=\frac{3\pi ^{2}}{16}\sqrt{2}-\frac{\pi ^{3}+64}{64}\sqrt{2}$$]

### Question

[Maximum mark: 12]
Given the following values at $$x = 1$$ Calculate the derivatives of the following functions at $$x = 1$$

(i) $$y=3f(x)-2g(x)$$                                          (ii) $$y = f (x)g(x)$$

(iii) $$y=\frac{f(x)}{g(x)}$$                                                              (iv) $$y=2x^{3}+1+5f(x)$$

Ans.

(a) $$\frac{dy}{dx}=2{f}'(x)-3{g}'(x)$$, at $$x=1$$ the value is -7

(b) $$\frac{dy}{dx}={f}'(x)g(x)+f(x){g}'(x)$$, at $$x=1$$ the value is 22

(c)$$\frac{dy}{dx}=\frac{{f}'(x)g(x)-f(x){g}'(x)}{g(x)^{2}}$$,at $$x=1$$ the value is $$\frac{2}{9}$$

(d)$$\frac{dy}{dx}=6x^{2}+5{f}'(x)$$, at $$x=1$$ the value is 26

### Question

[Maximum mark: 4]
Let $$f(x)=6\sqrt{x^{2}}$$. Find $${f}'(x)$$.

Ans.

$$f(x)=6x^{\frac{2}{3}}, {f}'(x)=4x^{-\frac{1}{3}}\left ( =\frac{4}{x^{\frac{1}{3}}}=\frac{4}{\sqrt{x}} \right )$$

### Question

[Maximum mark: 6]
Let $$h(x)=\frac{6x}{cos\;x}$$. Find $${h}'(0)$$

Ans.

METHOD 1 (quotient)

$${h}'(x)=\frac{(cos\;x)(6)-(6x)(-sin\;x)}{(cos\;x)^{2}}$$

$${h}'(0)=\frac{(cos\;0)(6)-(6 x 0)(-sin\;0)}{(cos\;0)^{2}}=6$$

METHOD 2 (product)

$$h(x)=6x$$ x $$(cos\;x)^{-1}$$
$${h}'(x)=(6x)(-(cos\;x)^{-2})(-sin\;x))+(6)(cos\;x)^{-1}$$
$${h}'(0)=(6 x 0)(-(cos\;0)^{-2})(-sin\;0))+(6)(cos\;0)^{-1}=6$$

### Question

[Maximum mark: 5]
Let $$g(x)=2x$$ sin $$x$$.

(a) Find g′(x).
(b) Find the exact value of the gradient of the graph of $$g$$ at $$x$$ = π.

Ans.

(a)$${g}'(x)=2\;sin\;x+2\;x\;cos\;x$$
(b)$${g}'(\pi )=2\;sin\;\pi +2\pi\;cos\;\pi =-2\pi$$

### Question

[Maximum mark: 4]
Consider the function$$f(x)=k\;sin\;x+3x$$, where $$k$$ is a constant.

(a) Find $${f}'(x)$$.

(b) When $$x=\frac{\pi }{3}$$, the gradient of the curve of $$f (x)$$ is 8. Find the value of $$k$$ .

Ans.

(a)$${f}'(x)=k\;cos\;x+3$$
(b)$$k\;cos\;\left ( \frac{\pi }{3} \right )+3=8 \Rightarrow k\left ( \frac{1}{2} \right )+3=8\Rightarrow k=10$$

### Question

[Maximum mark: 5]
Let $$f(x)=\frac{3x^{2}}{5x-1}$$.

(a) Write down the equation of the vertical asymptote of $$y = f (x)$$ .
(b) Find $${f}'(x)$$ . Give your answer in the form $$\frac{ax^{2}+bx}{(5x-1)^{2}}$$ where $$a$$ and $$b\in \mathbb{Z}$$ .

Ans.

(a)$$x=\frac{1}{5}$$

(b)$${f}'(x)=\frac{(5x-1)(6x)-(3x^{2})(5)}{(5x-1)^{2}}=\frac{30x^{2}-6x-15x^{2}}{(5x-1)^{2}}=\frac{15x^{2}-6x}{(5x-1)^{2}}$$

### Question

[Maximum mark: 8]
Let $$f(x)=x\;cos\;x$$, for 0 ≤ x ≤ 6.

(a) Find $${f}'(x)$$.
(b) On the grid below, sketch the graph of $$y={f}'(x)$$.
(c) Write down the range of the function $$y={f}'(x)$$ , for 0 ≤ x ≤ 6 Ans.

(a)$${f}'(x)=cos\;x-x\;sin\;x$$

(b) (c) y ∈ [-2.38 , 5.10]

### Question

[Maximum mark: 7]
Let $$f(x)=e^{x}\;cos\;x$$.

(a) Find $${f}'(x)$$.
(b) Find the gradient of the normal to the curve of $$f$$ at $$x=\pi$$.
(c) Find the gradient of the tangent to the curve of $$f$$ at $$x=\frac{\pi }{4}$$.

Ans.

(a)$${f}'(x)=e^{x}$$ x $$(-sin\;x)+cos\;x$$ x $$e^{x}=e^{x}\;cos\;x-e^{x}\;sin\;x$$
$${f}'(\pi)=e^{\pi }\;cos\;\pi -e^{\pi }\;sin\;\pi =-e^{\pi }$$
gradient of normal = $$\frac{1}{e^{\pi }}$$

(b)$${f}’\left ( \frac{\pi }{4} \right )=0$$

### Question

[Maximum mark: 7]
Let $$f(x)=xe^{x}$$.
(a) Find the equation of the tangent line at $$x = 0$$.
(b) Find the equation of the normal line at $$x = 0$$.
(c) Solve the equation $${f}'(x)=0$$.

Ans.

(a) Point (0,0), $${f}'(x)=e^{x}+xe^{x}$$, $$m_{T}=1$$, Tangent line $$y=x$$

(b)$$m_{N}=-1$$, Normal line $$y=-x$$

(c)$$e^{x}+xe^{x}=0\Leftrightarrow e^{x}(1+x)=0\Leftrightarrow x=-1$$

### Question

[Maximum mark: 3 per function]
Find the derivative of each function below. [cover 3rd column; then compare with your answer]    Ans.

Solutions are shown in the question.

### Question

[Maximum mark: 3 per function]
Differentiate the following functions: Ans. ### Question

[Maximum mark: 10]
Find $$\frac{ds}{dt}$$ for each of the following functions Ans. ### Question

[Maximum mark: 18]
Given that $$f(1)=2$$ and $${f}'(1)=4$$ , find the derivatives of the following functions at $$x =1$$

(i) $$y=f(x)^{2}$$                                     (ii) $$y=f(x)^{3}$$                                         (iii) $$y=ln\;f(x)$$

(iv) $$y=f(x^{2})$$                                     (v) $$y=f(x^{3})$$                                           (vi) $$y=\sqrt{f(x)}$$

Ans.

(i)$$y=f(x)^{2}\Rightarrow \frac{dy}{dx}=2f(x){f}'(x)$$. At $$x=1,\frac{dy}{dx}=2x2x4=16$$

(ii)$$y=f(x)^{3}\Rightarrow \frac{dy}{dx}=3f(x)^{2}{f}'(x)$$. At $$x=1,\frac{dy}{dx}=3×2^{2}x4=48$$

(iii)$$y=ln\;f(x)\Rightarrow \frac{dy}{dx}=\frac{{f}'(x)}{f(x)}$$. At $$x=1,\frac{dy}{dx}=\frac{4}{2}=2$$

(iv)$$y=f(x^{2})\Rightarrow \frac{dy}{dx}={f}'(x^{2})x2x$$. At $$x=1, \frac{dy}{dx}=2x2x1=4$$

(v)$$y=f(x^{3})\Rightarrow \frac{dy}{dx}={f}'(x^{3})x3x^{2}$$. At$$x=1, \frac{dy}{dx}=2x3x1^{2}=6$$

(vi)$$y=\sqrt{f(x)}\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{f(x)}}{f}'(x)$$. At $$x=1, \frac{dy}{dx}=\frac{1}{2\sqrt{2}}x4=\frac{2}{\sqrt{2}}=\sqrt{2}$$

### Question

[Maximum mark: 10]
(a) Show that the derivative of $$y$$ = tan $$x$$ is $$\frac{1}{cos^{2}x}$$.

(b) Hence, differentiate the functions

(i) $$y = x\; tan\; x$$                                          (ii) $$y = tan\; 3x$$

(iii) $$y=tan^{2}x$$                                                   (iv) $$y=tan^{3}x$$

Ans.

(a) $$y=tan\;x=\frac{sin\;x}{cos\;x}$$
$$\frac{dy}{dx}=\frac{sin\;x\;sin\;x-cos\;x(-cos\;x)}{cos^{2}x}=\frac{sin^{2}x+cos^{2}x}{cos^{2}x}=\frac{1}{cos^{2}x}$$

(b)   (i)$$\frac{dy}{dx}=\frac{x}{cos^{2}x}+tan\;x$$
(ii)$$\frac{dy}{dx}=\frac{3}{cos^{2}3x}$$
(iii)$$\frac{dy}{dx}=2\;tan\;x\frac{1}{cos^{2}x}\left ( =\frac{2\;sin\;x}{cos^{3}x} \right)$$
(iv)$$\frac{dy}{dx}=3\;tan^{2}x\frac{1}{cos^{2}x}\left ( =\frac{3\;sin^{2}x}{cos^{4}x} \right)$$

### Question

[Maximum mark: 4]
Differentiate each of the following with respect to $$x$$.

(a) $$y=x\;sin\;3x$$
(b) $$y=\frac{ln\;x}{x}$$

Ans.

(a) $$\frac{dy}{dx}=sin\;3x+3x\;cos\;3x$$

(b)$$\frac{dy}{dx}=\frac{xx\frac{1}{x}-ln\;x}{x^{2}}=\frac{1-ln\;x}{x^{2}}$$

### Question

[Maximum mark: 4]
The point P $$\left ( \frac{1}{2},0 \right )$$ lies on the graph of the curve of $$y=sin(2x-1)$$.
Find the gradient of the tangent to the curve at P.

Ans.

$$y = sin\; (2x – 1)$$          $$\frac{dy}{dx}=2\;cos(2x-1)$$
At $$\left ( \frac{1}{2},0 \right )$$, the gradient of the tangent = 2 cos 0 = 2

### Question

[Maximum mark: 4]
Differentiate with respect to $$x$$:                          (i) $$y=(x^{2}+1)^{2}$$.                               (ii) $$y=ln(3x-1)$$

Ans.

(i)$$\frac{d}{dx}(x^{2}+1)^{2}=2(x^{2}+1)x(2x)=4x(x^{2}+1)$$
(ii)$$\frac{d}{dx}(ln(3x-1))=\frac{1}{3x-1}x(3)=\frac{3}{3x-1}$$

### Question

[Maximum mark: 4]
Differentiate with respect to $$x$$        (i) $$\sqrt{3-4x}$$                 (ii) $$e^{sin\;x}$$

Ans.

(i)$$\frac{dy}{dx}=\frac{-4}{2\sqrt{3-4x}}=\frac{-2}{\sqrt{3-4x}}$$
OR $$y=\sqrt{3-4x}=(3-4x)^{\frac{1}{2}}$$                 $$\frac{dy}{dx}=\frac{1}{2}(3-4x)^{-\frac{1}{2}}(-4)$$

(ii) $$y=e^{sin\;x}$$                      $$\frac{dy}{dx}=(cos\;x)(e^{sin\;x})$$

### Question

[Maximum mark: 5]
Let $$f(x)=e^{\frac{x}{3}}+5\;cos^{2}\;x$$. Find $${f}'(x)$$.

Ans.

$${f}'(x)=\frac{1}{3}e^{\frac{x}{3}}-10\;cos\;x\;sin\;x$$

### Question

[Maximum mark: 6]
Let $$f(x) = cos\;2x$$ and $$g(x) = ln(3x – 5)$$.
(a) Find $${f}'(x)$$.
(b) Find $${g}'(x)$$.
(c) Let $$h(x) = f(x) × g(x)$$. Find $${h}'(x)$$.

Ans.

(a) $${f}'(x)$$=$$-sin\;2x$$ x 2(=$$-2\;sin\;2x$$)

(b) $${g}'(x)=3x\frac{1}{3x-5}\left ( =\frac{3}{3x-5} \right )$$

(c) product rule: $${h}'(x)=(cos\;2x)\left ( \frac{3}{3x-5} \right )+ln(3x-5)(-2\;sin\;2x)$$

### Question

[Maximum mark: 6]
(a) Let $$f(x)=e^{5x}$$. Write down $${f}'(x)$$.
(b) Let $$g(x)=sin\;2x$$. Write down $${g}'(x)$$.
(c) Let $$h(x)=e^{5x}\;sin\;2x$$. Find $${h}'(x)$$.

Ans.

(a) $${f}'(x)=5e^{5x}$$
(b) $${g}'(x)=2\;cos\;2x$$
(c) $${h}’={fg}’+{gf}’=e^{5x}(2\;cos\;2x)+sin\;2x(5e^{5x})$$

### Question

[Maximum mark: 6]
Let $$f(x)=e^{-3x}$$ and $$g(x)=sin\left ( x-\frac{\pi }{3} \right )$$.
(a) Write down      (i) $${f}'(x)$$;                (ii)$${g}'(x)$$.
(b) Let $$h(x)=e^{-3x}\;sin\left ( x-\frac{\pi }{3} \right )$$. Find the exact value of $${h}’\left ( \frac{\pi }{3} \right )$$.

Ans.

(a)      (i) $$-3e^{-3x}$$                        (ii) $$cos\left ( x-\frac{\pi }{3} \right )$$

(b)$${h}'(x)=-3e^{-3x}sin\left ( x-\frac{\pi }{3} \right )+e^{-3x}cos\left ( x-\frac{\pi }{3} \right )$$

$${h}'(\frac{\pi }{3})=-3e^{-3\frac{\pi }{3}}sin\left ( \frac{\pi }{3}-\frac{\pi }{3} \right )+e^{-3\frac{\pi }{3}}cos\left ( \frac{\pi }{3}-\frac{\pi }{3} \right )=e^{-\pi }$$

### Question

[Maximum mark: 6]
Let $$f(x)=(2x+7)^{3}$$ and $$g(x)=cos^{2}(4x)$$. Find     (i) $${f}'(x)$$;                    (ii)$${g}'(x)$$

Ans.

(i) $${f}'(x)=3(2x+7)^{2}x2=6(2x+7)^{2} (=24x^{2}+168x+294)$$
(ii)$${g}'(x)=2\;cos(4x)(-sin(4x))(4)=-8\;cos(4x)sin(4x) (=-4sin(8x))$$

### Question

[Maximum mark: 6]
The population $$p$$ of bacteria at time $$t$$ is given by $$p$$ = 100e0.05t. Calculate
(a) the value of $$p$$ when $$t$$ = 0;
(b) the rate of increase of the population when $$t$$ = 10.

Ans.

(a)  p = 100e0 = 100

(b) Rate of increase is $$\frac{dp}{dt}=0.05x100e^{0.05t}=5e^{0.05t}$$
When $$t=10$$     $$\frac{dp}{dt}=5e^{0.05(10)}=5e^{0.5}$$                 $$(=8.24=5\sqrt{e})$$

### Question

[Maximum mark: 8]
The number of bacteria, $$n$$, in a dish, after $$t$$ minutes is given by $$n$$ = 800e013t.
(a) Find the value of $$n$$ when $$t$$ = 0.
(b) Find the rate at which $$n$$ is increasing when $$t$$ = 15.
(c) After $$k$$ minutes, the rate of increase in $$n$$ is greater than 10 000 bacteria per
minute. Find the least value of $$k$$, where $$k\in \mathbb{Z}$$.

Ans.

(a) $$n$$ = 800e0 = 800
(b) derivative: n′(15) = 731
(c) METHOD 1
setting up inequality. n′(t) > 10 000
$$k$$ = 35.1226…, least value of $$k$$ is 36
METHOD 2
n′(35) = 9842, and n′(36) = 11208
least value of $$k$$ is 36

### Question

[Maximum mark: 8]
Consider the curve $$y = ln(3x – 1)$$. Let P be the point on the curve where $$x = 2$$.
(a) Write down the gradient of the curve at P.
(b) Find the equation of the tangent to the curve at P.
(c) The normal to the curve at P cuts the x-axis at R. Find the coordinates of R.

Ans.

(b) y – ln 5 = 0.6 (x – 2)       OR   directly by GDC: y = 0.6 x + 0.409
(c) at x = 2,     y = ln 5 (= 1.609…)
gradient of normal = – 5/3 = –1.6666…
normal: $$y-ln\;5=-\frac{5}{3}(x-2)$$         OR directly by GDC y = – 1.66666x + 4.94277
For y = 0:   x = 2.97 (accept 2.96)
coordinates of R are (2.97, 0)

### Question

[Maximum mark: 7]
Let $$f(x)=3x-e^{x-2}-4$$, for $$-1\leq x\leq 5$$.
(a) Find the x -intercepts of the graph of $$f$$ .
(b) On the grid below, sketch the graph of $$f$$ .
(c) Write down the gradient of the graph of $$f$$ at $$x = 2$$. Ans.

(a) (1.54, 0) (4.13, 0)       (accept x = 1.54 x = 4.13)
(b) ### Question

[Maximum mark: 6]
If $$y=ln(2x-1)$$ find $$\frac{d^{2}y}{dx^{2}}$$.

Ans.

$$y-ln(2x-1)\Rightarrow \frac{dy}{dx}=\frac{2}{2x-1}\Rightarrow \frac{dy}{dx}=2(2x-1)^{-1}$$
$$\Rightarrow \frac{d^{2}y}{dx^{2}}=-2(2x-1)^{-2}(2)$$
$$\Rightarrow \frac{d^{2}y}{dx^{2}}=\frac{-4}{(2x-1)^{-2}}$$

### Question

[Maximum mark: 6]
Consider the function $$y$$ = tan $$x$$ – 8sin $$x$$ .
(a) Find $$\frac{dy}{dx}$$.
(b) Find the value of cos $$x$$ for which $$\frac{dy}{dx}=0$$.
(c) Solve the equation $$\frac{dy}{dx}=0$$. for $$-\pi \leq x\leq 2\pi$$.

Ans.

(a) $$\frac{dy}{dx}=\frac{1}{cos^{2}x}-8cosx$$

(b) $$\frac{dy}{dx}=\frac{1-8cos^{3}x}{cos^{2}x}=0\Rightarrow cosx=\frac{1}{2}$$

(c) $$x=-\frac{\pi }{3},x=\frac{\pi }{3},x=\frac{5\pi }{3}$$

### Question

[Maximum mark: 6]
Let $$y=e^{3x}\;sin(\pi x)$$. Find

(a) $$\frac{dy}{dx}$$.
(b) the smallest positive value of x for which $$\frac{dy}{dx}=0$$.

Ans.

y = e3x sin(πx)

(a)$$\frac{dy}{dx}=3e^{3x}sin(\pi x)+\pi e^{3x}cos(\pi x)$$

(b) x = 0.7426… (0.743 to 3 s.f.)

### Question

[Maximum mark: 6]
Let $$f(x)=cos^{3}(4x+1), 0\leq x\geq 1$$. Find
(a) $${f}'(x)$$
(b) the exact values of the three roots of $${f}'(x)=0$$.

Ans.

(a) $$-12cos^{2}(4x+1)sin(4x+1)$$
(b)  $$x=\frac{\pi }{8}-\frac{1}{4},x=\frac{3\pi }{8}-\frac{1}{4},x=\frac{\pi-1 }{4}$$

### Question

[Maximum mark: 6]
Let $$f$$ be a cubic polynomial function. Given that $$f (0) = 2$$, $${f}'(0)=-3$$, $$f (1)$$ = $${f}'(1)$$
and $${f}”(-1)=6$$ , find $$f (x)$$ .

Ans.

$$f(x)=ax^{3}+bx^{2}+cx+d$$
$${f}'(x)=3ax^{2}+2bx+c$$
$${f}”(x)=6ax+2b$$
$$f(0)=2=d$$
$${f}'(1)=f(1)\rightarrow a+b+c+2=3a+2b+c$$
$$2=2a+b$$
$${f}'(0)=-3=c$$
$${f}”(-1)=6=-6a+2b$$
$$b=\frac{12}{5},a=-\frac{1}{5}$$
$$f(x)=-\frac{1}{5}x^{3}+\frac{12}{5}x^{2}-3x+2 (Accept \;a=-\frac{1}{5},b=\frac{12}{5},c=-3,d=2)$$

### Question

[Maximum mark: 3 per function]
The following table shows the values of two functions $$f$$ and $$g$$ and their derivatives
when $$x = 1$$ and $$x = 0$$ . Find the derivatives of the following functions when $$x = 1$$. Ans. ### Question

[Maximum mark: 10]
Let $$f(x)=1+3\;cos(2x)$$ for $$0\leq x\leq \pi$$, and $$x$$ is in radians.
(a) (i) Find $${f}'(x)$$.
(ii) Find the values for $$x$$ for which $${f}'(x)=0$$ ; Give your answers in terms of π .
The function $$g(x)$$ is defined as $$g(x)=f(2x)-1, 0\leq x\leq \frac{\pi }{2}$$.
(b) (i) The graph of $$f$$ may be transformed to the graph of $$g$$ by a stretch in the
x -direction with scale factor $$\frac{1}{2}$$ followed by another transformation.
Describe fully this other transformation.
(ii) Find the solution to the equation $$g(x) = f (x)$$ .

Ans.

(a)  (i)$${f}'(x)=-6\;sin\;2x$$
(ii)  EITHER $${f}'(x)=-12sin\;x\;cos\;x=0\Rightarrow sin\;x=0$$ or $$cos\;x=0$$
OR $$sin\;2x=0$$, for $$0\leq 2x\leq 2\pi$$
THEN
$$x=0,\frac{\pi }{2},\pi$$

(b) (i)  translation in the y-direction of –1
(ii)  1.11                     (1.10 from TRACE is subject to AP)

### Question

[Maximum mark: 20]
The diagram shows the graph of the function $$f$$ given by $$f(x)=A\;sin\left ( \frac{\pi }{2}x \right )+B$$, for
0 ≤ x ≤ 5 , where A and B are constants, and $$x$$ is measured in radians. The graph includes the points (1, 3) and (5, 3), which are maximum points of the graph.
(a) Show that A = 2 , and find the value of B .
(b) Show that $${f}'(x)=\pi cos\left ( \frac{\pi }{2}x \right )$$.

The line $$y=k-\pi x$$ is a tangent line to the graph for 0 ≤ x ≤ 5 .
(c) Find
(i) the point where this tangent meets the curve;
(ii) the value of $$k$$ .
(d) Solve the equation $$f (x) = 2$$ for 0 ≤ x ≤ 5 .

Ans.

(a) EITHER $$A\;sin\left ( \frac{\pi }{2} \right )+B=3$$ and $$A\;sin\left ( \frac{3\pi }{2} \right )+B=-1$$
$$\Leftrightarrow A+B=3,-A+B=-1$$
$$\Leftrightarrow A=2,B=1$$
OR
Amplitude = $$A =\frac{3-(-1)}{2}=\frac{4}{2}=2$$
Midpoint value = $$B =\frac{3+(-1)}{2}=\frac{2}{2}=1$$

(b) $$f(x)=2\;sin\left ( \frac{\pi }{2}x \right )+1$$
$${f}'(x)=\left ( \frac{\pi }{2} \right )2\;cos\left ( \frac{\pi }{2}x \right )+0=\pi \;cos\left ( \frac{\pi }{2}x \right )$$

(c)  (i) y = k – πx is a tangent  $$\Rightarrow -\pi =\pi \;cos\left ( \frac{\pi }{2}x\right )$$
$$\Rightarrow -1=cos\left ( \frac{\pi }{2}x\right )$$
$$\Rightarrow \frac{\pi }{2}x=\pi$$ or $$3\pi$$ or…
$$\Rightarrow x=2$$ or 6…

Since 0 ≤ x ≤ 5, we take x = 2, so the point is (2, 1)

(ii) Tangent line is: y = –π(x – 2) + 1
y = (2π + 1) – πx
k = 2π + 1

(d) $$f(x)=2\Rightarrow 2\;sin\left ( \frac{\pi }{2}x \right )+1=2$$
$$\Rightarrow sin\left ( \frac{\pi }{2}x \right )=\frac{1}{2}$$
$$\Rightarrow \frac{\pi }{2}x=\frac{\pi }{6}or\frac{5\pi }{6}or\frac{13\pi }{6}$$
$$x=\frac{1}{3}or\frac{5}{3}$$ or $$\frac{13}{3}$$

### Question

[Maximum mark: 14]
The following diagram shows a waterwheel with a bucket. The wheel rotates at a
constant rate in an anticlockwise (counterclockwise) direction. The diameter of the wheel is 8 metres. The centre of the wheel, A, is 2 metres above
the water level. After $$t$$ seconds, the height of the bucket above the water level is given
by $$h = a$$ sin $$bt + 2$$.
(a) Show that $$a = 4$$.
The wheel turns at a rate of one rotation every 30 seconds.
(b) Show that $$b=\frac{\pi }{15}$$.
In the first rotation, there are two values of $$t$$ when the bucket is descending at a rate of
0.5 m s–1.
(c) Find these values of $$t$$.
(d) Determine whether the bucket is underwater at the second value of $$t$$.

Ans.

(a) recognizing the amplitude is the radius: $$a=\frac{8}{2}\Rightarrow a=4$$

(b) period = 30: $$b=\frac{2\pi }{30}=\frac{\pi }{15}$$

(c) recognizing $${h}'(t)=-0.5$$
$$-0.5=\frac{4\pi }{15}cos\left ( \frac{\pi }{15}t \right )\Rightarrow t-10.6,t=19.4$$

(d) h(t) < 0 so underwater; h(t) > 0 so not underwater
h(19.4) = $$4\;sin\frac{19.4\pi }{15}+2=-1.19$$

OR

solving h(t) = 0, graph showing region below x-axis, roots 17.5, 27.5
Hence, the bucket is underwater, yes

### Question

[Maximum mark: 18]
Let $$f (x) = 3sinx + 4\; cos\; x$$, for –2π ≤ x ≤ 2π.
(a) Sketch the graph of $$f$$.
(b) Write down
(i) the amplitude;               (ii) the period;             (iii) the x-intercept between $$-\frac{\pi }{2}$$ and 0.
(c) Hence write $$f (x)$$ in the form $$p$$ sin $$(qx + r)$$.
(d) Write down one value of $$x$$ such that $${f}'(x)=0$$.
(e) Write down the two values of $$k$$ for which the equation $$f (x) = k$$ has exactly two
solutions.
(f) Let $$g(x) = ln(x + 1)$$, for 0 ≤ x ≤ π. There is a value of $$x$$, between 0 and 1, for which
the gradient of $$f$$ is equal to the gradient of $$g$$. Find this value of $$x$$.

Ans.

(a) (b)  (i) 5          (ii)  2π (6.28)           (iii) –0.927
(c) $$f(x)$$ = 5 sin (x + 0.927)  (accept $$p$$ = 5, $$q$$ = 1, $$r$$ = 0.927)
(d) (man or min)
one 3 s.f. value which rounds to one of –5.6, –2.5, 0.64, 3.8

(e) $$k$$ = –5,  $$k$$ = 5
(f) $${g}'(x)=\frac{1}{x+1}$$
$${f}'(x)=3\;cos\;x-4\;sin\;x$$        $$(5\;cos(x+0.927))$$
$${g}'(x)={f}'(x)$$
$$x=0.511$$

### Question

[Maximum mark: 15]
(a) The function $$g$$ is defined by $$g(x)=\frac{e^{x}}{\sqrt{x}}$$, for 0 < x ≤ 3 .

(i) Sketch the graph of $$g$$ .
(ii) Find $${g}'(x)$$ .
(iii) Write down an expression representing the gradient of the normal to the
curve at any point.

(b) Let P be the point $$(x, y)$$ on the graph of $$g$$ , and Q the point (1,0).
(i) Find the gradient of (PQ) in terms of $$x$$ .
(ii) Given that the line (PQ) is a normal to the graph of $$g$$ at the point P, find the
minimum distance from the point Q to the graph of $$g$$ .

Ans.

(a) (ii) $$g(x)=\frac{e^{x}}{\sqrt{x}}$$
$${g}'(x)=\frac{e^{x}\sqrt{x}-\frac{e^{x}}{2\sqrt{x}}}{x}$$
$$=\frac{(2x-1)e^{x}}{2x\sqrt{x}}$$

(iii) gradient is $$-\frac{1}{{g}'(x)}$$
$$=\frac{2x\sqrt{x}}{(1-2x)e^{x}}$$

(b)  (i)  $$\frac{y-0}{x-1}=\frac{e^{x}}{\sqrt{x}(x-1)}$$

(ii) EITHER
$$\frac{e^{x}}{\sqrt{x}(x-1)}=\frac{2x\sqrt{x}}{(1-2x)e^{x}}$$
$$x=0.5454…$$
OR
$$D^{2}=(x-1)^{2}+y^{2}=(x-1)^{2}+\frac{e^{2x}}{x}$$
$$\frac{dD^{2}}{dx^{2}}=2(x-1)+\frac{2e^{2x}x-e^{2x}}{x^{2}}=0$$
$$x=0.5454…$$
THEN
distance = $$\sqrt{(1-0.5454)^{2}+\left ( \frac{e^{0.5454}}{\sqrt{0.5454}} \right )^{2}}$$
=2.38