Basic graph is sine or cosine - — Circular (trigonometric) functions

Chapter 6 — Circular (trigonometric) functions

6 Basic graph is sine or cosine

  or π − sin⁻¹ b a

  

 . If θ= 2 then 3θ= 6

so 3θ= π − 6 is also a possibility. Hence θ= 2 3

π _{− could} be a solution.

The corresponding general solutions would be 3θ= sin⁻¹ b

  

  = sin⁻¹ b a

  

 + 2nπ θ = 1 2

3 6 3

nπ

× +

= 2 + 2 3 nπ

or θ= 1 2

( 6)

3 3

nπ π

× − +

= −2 + (2 1) 3 n+ π

If n = 1 then 2 + 2 3

π is a solution.

The answer is D.

6 Basic graph is sine or cosine.

Period is T, so A and E are eliminated.

Amplitude is 2a so C is eliminated.

Check B ⇒ amplitude of −2a produces reflected cosine graph with no horizontal translation.

This fits

Check D ⇒ Need a sine graph translated 2

π to the right.

In this case the translation has to be a function of T so D cannot be correct.

The answer is B.

7 If x = a is one solution to an equation of the form cos(2x) = k, then x = −a will also be a solution. The third solution will be a period beyond x = −a at x = −a + π.

So the two adjacent solutions are x = a and x = −a + π. The difference between them will be (−a + π) – a = π – 2a The answer is E.

Extended response 1 a 500 100

− = 200 birds

b takes 6 months from maximum to minimum. ∴ period is 12 months.

c mean population 500 100 2

+ = 300 birds d Population is a minimum after 7 months e P(t) = a sin(b(t + c)) + d

a is the amplitude so is 200 b is related to the period by T = 2

π _{⇒ 12 =}² b

⇒ b = 6 π

c is the horizontal translation which is two months left so c = 2

d is the vertical translation so d = 300 So P(t) = 200 sin ( 2)

6 t π

 

 + 

  + 300

f Initial population: P(0) = 200 sin 3 π

  

  + 300

= 200 × 3

2 + 300

= 473 birds

g Q(t) = Ae^−kt

(0, 500) ⇒ 500 = Ae^−k^×⁰

A = 500

(60, 370) ⇒ 370 = 500e^−k^×⁶⁰

370 = 500e^−60κ

e^−60κ = 370 500

= 0.74

−60k = loge(0.74)

k = log (0.74)

−

k = 0.005018

So Q(t) = 500e^−0.005018t h Population = ( ) ( )

500 P x Q x

200sin ( 2) 300 500 6 t

  + + ×

   

 

0.005018

500

e⁻ t

= ^0.005018 200sin ( 2) 300

e₋ t× π t+ + 

t = 0 ⇒ Population = 1 × [200 sin 3 π

  

  + 300]

= 200 × 3

2 + 300

= 473

This value is consistent with the sine model and close to the exponential model.

t = 7 ⇒ population = 96.5

This is consistent with the sine model t = 60 ⇒ population = 350

This is consistent with the exponential model.

t = 120 ⇒ population = 259

This is consistent with the exponential model.

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Exercise 7A — Review — gradient and rates of change 1 a m = rise

run = 1

1 = 1

∴ gradient function y = 1

b m = rise run = 3

−1 = −3

∴ gradient function y = −3

c m = rise run = 1

−2 = −1

∴ gradient function y = −1

d m = rise run = 5

2 or 2.5 ∴ gradient function y = 5

e m = 0

2 a m = rise run = 2

1 = 2

The answer is B.

b The graph of the gradient function is given by f ′(x) = 2.

The answer is C.

3 a g′(x) < 0 if x < 0 g′(x) = 0 if x = 0 g′(x) > 0 if x > 0

b g′(x) > 0 if x < 0 g′(x) = 0 if x = 0 g′(x) < 0 if x > 0

c g′(x) < 0 if x < 0 g′(x) = 0 if x = 0 g′(x) > 0 if x > 0

d g′(x) < 0 if x < 3 g′(x) = 0 if x = 3 g′(x) > 0 if x > 3

e g′(x) > 0 if x < −2 g′(x) = 0 if x = −2 g′(x) < 0 if x > −2

4 a f ′(x) > 0 if x < 0 and the graph is becoming less sleep

f ′(x) = 0 if x = 0 f ′(x) < 0 if x > 0

Hence f ′(x) is always decreasing.

The answer is B.

b f ′(x) > 0 if x < 0 f ′(x) = 0 if x = 0 f ′(x) < 0 if x > 0 The answer is A.

5 a f ′(x) > 0 for −3 < x < 2 f ′(x) = 0 for x = −3 and x = 2 f ′(x) < 0 for x < −3 and x > 2 Since f (x) is cubic, f ′(x) is

quadratic (parabola).

Chapter 7 — Differentiation

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b f ′(x) > 0 for x < 1 and x > 4 f ′(x) = 0 for x = 1 and x = 4 f ′(x) < 0 for 1 < x < 4 Since f(x) is cubic, f ′(x) is

quadratic (parabola).

c f ′(x) > 0 for 0 < x < 5 f ′(x) = 0 for x = 0 and x = 5 f ′(x) < 0 for x < 0 and x > 5 Since f(x) is cubic, f ′(x) is

quadratic (parabola).

d f ′(x) > 0 for x < 0 and x > 0 f ′(x) = 0 for x = 0

Since f(x) is cubic, f ′(x) is quadratic (parabola).

e f ′(x) < 0 for x < 0 and x > 0 f ′(x) = 0 for x = 0

Since f(x) is cubic, f ′(x) is quadratic (parabola).

f f ′(x) > 0 for −3 < x < 0 f ′(x) = 0 for x = −3 and x = 0 f ′(x) < 0 for x < −3 and x > 0 Since f(x) is cubic, f ′(x) is

quadratic (parabola).

g f ′(x) > 0 for x < 0 and x > 0 f ′(x) = 0 for x = 0

Since f(x) is cubic, f ′(x) is quadratic (parabola)

6 a Positive gradient occurs when f(x) slopes upward from left to right, that is x < −1 and x > 2.

The answer is D.

b Negative gradient occurs when f(x) slopes downward from left to right that is −1 < x < 2.

The answer is C.

c f ′(x) > 0 for x < −1 and x > 2 f ′(x) < 0 for −1 < x < 2 Gradient function has a unique

value throughout its domain (no sharp points)

since f(x) is smooth and continuous.

The answer is E.

7 a i n/a since f(x) has no stationary points.

ii R since gradient is always positive.

iii n/a since gradient is always positive.

iv n/a since f(x) is smooth and continuous.

b i n/a since f(x) has no stationary points.

ii n/a since gradient is always negative.

iii R since gradient is always negative.

iv n/a since f(x) is smooth and continuous.

c i x = −1 since local maximum ii (−∞, −1)

iii (−1, ∞)

iv n/a since g(x) is smooth and continuous

d i x = 4 since local minimum ii (4, ∞)

iii (−∞, 4)

iv n/a since g(x) is smooth and continuous.

e i x = 0 since stationary point of inflection

ii n/a since gradient is negative either side of x = 0

iii R\{0}

iv n/a since g(x) is smooth and continuous.

f i x = −2 and x = 3 since local minimum and local maximum respectively.

ii (−2, 3) since f(x) slopes upward from left to right.

iii (−∞, −2) ∪ (3, ∞) since f(x) slopes downward for x < −2 and x > 3 moving left to right.

iv n/a since f(x) is smooth and continuous.

g i x = 2 since stationary point of inflection.

ii R\{2} since gradient is positive either side of x = 2.

iii n/a since f(x) slopes upward from left to right.

iv n/a since f(x) is smooth and continuous.

h i x = −1 and x = 2 since local maximum and local minimum respectively.

ii (−∞, − 1) ∪ (2, ∞) since g(x) slopes upward for x < −1 and x > 2 moving left to right.

iii (−1, 2) since g(x) slopes downward for −1 < x < 2 moving left to right.

iv n/a since g(x) is smooth and continuous.

i i n/a since no stationary points exist

ii (−∞, 0) as f(x) slopes upward for x < 0 moving left to right iii (0, ∞) as f(x) slopes downward

for x > 0 moving left to right.

iv x = 0 since f(x) is discontinuous at x = 0.

j i n/a since no stationary points exist

ii (−∞, 0) as f(x) slopes upward for x < 0 moving left to right.

iii (0, ∞) as f(x) slopes downward for x > 0 moving left to right.

iv x = 0 since f(x) is discontinuous at x = 0.

8 a

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Exercise 7B — Limits and differentiation from first principles

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a Gradient chord PQ

= ( 1 ) ( 1)

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Exercise 7C — The derivative of xⁿ 1 a y = x⁶

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180

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Exercise 7J — Mixed problems on differentiation 1 a f(x) = loge(8x)

1 if 2 0

x x

 + >



− + <



f′(x) = 2 2 if ₂² 2 0

2 2 if 2 0

x x x

 + + >



− − + <



∴ f′(x) = 2 2 if 2 or 0

2 2 if 2 0

x x x

x x

+ < − >



− − − < <

 ii

5 a f(x) = |sin(2x)| for x ∈ [0, π^]

i f′(x) = 2 cos (2x) × 1 if sin(2 ) 0 1 if sin(2 ) 0

x x



− <



f′(x) = 2cos(2 ) if sin(2 ) 0 2cos(2 ) if sin(2 ) 0

x x

− <



f′(x) =

2cos(2 ) if 0 2 2cos(2 ) if

x x

π π

 < <



− < <



b f′(x) = |cos(x)| for x ∈[0, 2π^] i f′(x) = − sin(x) ×

( )

1 if cos 0 1 if cos 0

x x

 >



− <



f′(x) = sin( ) if cos( ) 0 sin( ) if cos( ) 0

x x

− >



 <

f′(x) =

sin( ) if 0 or 3 2

2 2

sin( ) if 3

2 2

x x x

x x

π π

− < < < <



 < <



Chapter review Short answer 1

f ′(x) < 0 if x < −1 f ′(x) > 0 if −1< x < 2 f ′(x) < 0 if x > 2

2 ³ ²

(2 ) 4 lim

h h h

→

+ +

= ²

( 2 4)

lim

h h h

→

+ +

= lim0

h→ (h²+ 2h + 4) h ≠ 0 = 4

3 a f(x) = x³+ 2x f′(x) =

( ) ( )

lim

f x h f x h

→

+ −

= ³ ³

( ) 2( ) ( 2 )

lim

x h x h x x

→

+ + + − +

= ³ ² ² ³ ³

( 3 3 2 2 2

lim

x x h xh h x h x x

→

+ + + + + − −

= ² ² ³

3 3 2

lim

x h xh h h

→

+ + +

= ² ²

(3 3 2)

lim

h x xh h

→

+ + +

= lim0

h→ (3x²+ 3xh + h²+ 2), h ≠ 0 = 3x²+ 2

b When x = 1 Gradient = f′(1) = 3(1)²+ 2

= 5

4 a g(x) = ³ x − 4x 3 g′(x) = (3) 1

3 x^{3 − 1}− 4x^{1 − 1} = x²− 4

b When x = 3 g′(3) = (3)²− 4 = 5 5 h(x) = 3 ⁴

2x + ³ x − 3x 4 a h′(x) = 3

2(4)x⁴^{− 1}+ 1

4(3)x³^{− 1}− 3x¹^{− 1} = 6x³+ 3

4x²− 3

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In document Maths Quest Methods Solutions y11 (Page 157-200)