Need solutions for Cambridge 3U Year 11 Textbook (1 Viewer)

[x.Exhaust.x]

Could anyone post their answers with their solutions for the following questions? I got my answers to a few of them, but I'm still not sure if I used the right method and the right way of answering the question. They should be quite simple, as neither of them are extension questions . Thx.

Exercise 8D - Q2d
Exercise 8E - Q6
Exercise 8F - Q12-Q15
Exercise 8H - Q7, Q13

 

[lyounamu]

Could anyone post their answers with their solutions for the following questions? I got my answers to a few of them, but I'm still not sure if I used the right method and the right way of answering the question. They should be quite simple, as neither of them are extension questions . Thx.

Exercise 8D - Q2d
Exercise 8E - Q6
Exercise 8F - Q12-Q15
Exercise 8H - Q7, Q13

1) cot^2(x) = cosec x +1
1 + cot x^2(x) = cosec x +2
cosec^2(x) = cosec x +2
cosec^2(x) - cosec x -2 = 0
(cosecx - 2)(cosecx + 1) = 0
cosec x = 2 or cosec x = -1
1/sinx = 2 or 1/sinx = -1
sin x = 1/2 or sinx = -1
x = 30 degrees, 150 degrees or 270 degrees

2) P = x + x + (8-x) + (8-x)
A = x . (8-x)
= 8x - x^2
dA/dx = 8 - 2x
Maximum value occurs when dA/dx = 0
i.e. 8-2x = 0
x=4
When x<4, dA/dx >0
When x>4, dA/dx <0
Therefore, at x=4, there is a maximum turning point. So the maximum value occurs when x=4
A = 4 (8-4)
= 16 m^2

I would love to do more questions but I am currently involved in my English Assignment. 10 minutes was all I could spare at the moment. I am sorry for other questions. As soon as I finish this task, I will come back and finish the rest. Ciao

 

[lsam]

too many questions. I hang myself in vain.

 

[Aplus]

Well we're on this subject, can someone help me with Exercise 8E, the last question:
Give a complete gemoetrical proof that the largest triangle which can fit into a circle is an equilateral triangle.

 

[PC]

EXERCISE 8F

12 (a)
y = 2x² - 7x + 4
dy/dx = 4x - 7
We need the slope of the tangent to be 1
.: 4x - 7 = 1
4x = 8
x= 2
When x = 2, y = -2
At (2,-2), b = -4

12 (b)
y = 2x² + 3x + 1
dy/dx = 4x + 3
Gradient of tangent = -a/b = -2/1 = -2
.: 4x + 3 = -2
x = -5/4
When x = -5/4, y = 3/8
At (-5/4,3/8), b = 17/8

12 (c)
Solving for points of intersection:
y = 3x² + 5x + 7 and y = mx + 4
3x² + (5 - m)x + 3 = 0
Since the curve is a parabola, there will only be one point of intersection, so ∆ = 0.
(5 - m)² - 4.3.3 = 0
m² - 10m - 11 = 0
m = 11 or -1

13
Gradient of tangent = -a/b = -3/-1 = 3
y = x² - 5x -3
dy/dx = 2x - 5
.: 2x - 5 = 3
x = 4
When x = 4, y = -9
Equation of tangent is y - y1 = m(x - x1)
y + 9 = 3(x - 4)
3x - y - 21 = 0

14
y = (2 - x)(1 + 3x)
= 2 + 5x - 3x²
dy/dx = 5 - 6x
Since tangent passes through (1,7), equation of tangent is:
y - y1 = m(x - x1)
y - 7 = m(x - 1)
y = mx + (7 - m)
Now solving for points of intersection:
y = 2 + 5x - 3x² and y = mx + (7 - m)
2 + 5x - 3x² = mx + (7 - m)
3x² + (m - 5)x + (5 - m) = 0
Since ∆ = 0, (m - 5)² - 4.3.(5 - m) = 0
m² + 2m – 35 = 0
(m + 7)(m – 5) = 0
m = 5 or -7

15
Equation of parabola is y = ax² + bx + c
Since y-intercept is -3, c = -3
Equation of parabola is y = ax² + bx - 3
Since axis is x = 1/2, -b/2a = 1/2
-2b = 2a
b = -a
Equation of parabola is y = ax² - ax - 3
Solving for points of intersection:
y = ax² - ax - 3 and y = 4x - 7
ax² - ax - 4x - 3 + 7 = 0
ax² + (-a-4)x + 4 = 0
Since ∆ = 0;
(-a-4)² - 4.a.4 = 0
a² - 8a + 16 = 0
a = 4
.: Equation of parabola is y = 4x² - 4x + 3

 

[PC]

EXERCISE 8H

7 (a)
(a - b)²
= a² - 2ab + b²
= a² + 2ab + b² - 4ab
= (a + b)² - 4ab

7 (b)
(i) a + b = 3
ab = 1
(a - b)² = 3² - 4.1 = 5
(ii) a + b = -5
ab = -7
(a - b)² = (-5)² - 4.-7 = 53
(iii) a + b = 7/3
ab = 2/3
(a - b)² = (7/3)² - 4(2/3) = 25/9

7 (c)
(i) √5
(ii) √53
(iii) 5/3

13 (a)
a + b = -(m + 1)/(2m - 1)
= (m + 1)/(1 - 2m)
ab = 1/(2m - 1)

If a = -b, then a + b = 0
(m + 1)/(1 - 2m) = 0
m + 1 = 0
m = -1

13 (b)
If a = 1/b then ab = 1
1/(2m - 1) = 1
1 = 2m - 1
2m = 2
m = 1

13 (c)
If a = 2 then b + 2 = (m + 1)/(1 - 2m) and 2b = 1/(2m - 1)
2[ (m + 1)/(1 - 2m) - 2 ] = 1/(2m - 1)
(2m + 2)/(1 - 2m) - 4 = -1/(2m - 1)
(2m + 2 - 4(1 - 2m))/(1 - 2m) = -1/(2m - 1)
2m + 2 - 4 + 8m = -1
10m - 2 = -1
10m = 1
m = 1/10

13 (d)
If a + b = 2ab then
(m + 1)/(1 - 2m) = -2/(1 - 2m)
m + 1 = -2
m = -3

 

[lyounamu]

Good work there PC. You seem to have filled all the gaps that I left out.

 

[x.Exhaust.x]

Thanks guys. We haven't learned dy/dx yet though . Any simple explanation? Thanks.

 

[Aerath]

Differentiation. It's basically limits.

 

[Continuum]

Thanks guys. We haven't learned dy/dx yet though . Any simple explanation? Thanks.

It's just another way of writing y' and f'(x) - the equation of the tangent.

 

[x.Exhaust.x]

It's just another way of writing y' and f'(x) - the equation of the tangent.

Differentiation. It's basically limits.

Ty . Sorry for being noob lol.

 

[Continuum]

Ty . Sorry for being noob lol.

Nah, just different learning order.

 

[Aplus]

Help with my question

 

[Aerath]

Help with my question

Wait for Lyounamu. I can't do it. =\

 

[lyounamu]

Wait for Lyounamu. I can't do it. =\

LOL. I can't do it either. I tried to find how the equation of the area of the triangle looks like in a circle and differentiated it. But all I can get is the minimum turning point for it.

I guess I need to use the different method.

 

[TBK11]

hey off topic a lil but does Cambridge 3U Yr11 have answers at the back for you to check? thanks

 

[lolman12567]

yeah, it does. but they aren't worked

 

[Dragonmaster262]

i just got this book. It sure is big. I'm just wondering if this book stays on topic with the syllabus? Does it ever you know go beyond the syllabus and start cramming liuke the Jacaranda books?

 

[kaz1]

i just got this book. It sure is big. I'm just wondering if this book stays on topic with the syllabus? Does it ever you know go beyond the syllabus and start cramming liuke the Jacaranda books?

Yeah it does go beyond the syllabus in the extension questions. I remember seeing a z axis question in the locus of the parabola section.

 

[Aquawhite]

Yeah it does go beyond the syllabus in the extension questions. I remember seeing a z axis question in the locus of the parabola section.

Lol, I've never even heard of a z-axis :O

Then again, I'm only starting the course...