weeping_fire
Badges: 8
Rep:
?
#1
Report Thread starter 3 years ago
#1
1) Draw the graph that goes through (0,1) with a gradient of -2. The line y=-2x+1 [2 marks]

2)
i) (1 7/9)^-1/2 = 3/4 [3 marks]
ii) (6x^5y^2)^3/18y^10 = 12x^15/y^4 [2 marks]

3) 6-x > 5(x-3) = x < 21/6 [3 marks]

4) Intersection of 2x+5y=5 and x-2y = (10/3, -1/3) [4 marks]

5)
i) Centre: (-2, 3) Radius: 5^1/2 [2 marks]
ii) Line parallel to 5x + y = 4 and touches centre: y = -5x -7 [2 marks]

6)
r = (V/(a+b))^1/2
r^2 = V/(a+b)
r^2(a+b) = V
ar^2 + br^2 = V
br^2 = V - ar^2
b = (V-ar^2)/r^2 [4 marks]

7)
i) (5-2root7)/3+root7 = (29 -11root7)/2 [3 marks]
ii) 12/root2 + root98 = 13root2 [2 marks]

8) (a+bx)^5. Constant is 32. x^3 coefficient is -1080. a = 2, b = -3 [5 marks]

9) (n+2)^2 - n^2
= n^2 + 4n + 4 - n^2
= 4n + 4
= 4(n + 1) [4 marks]

10)
i) AB = root(5^2 + 5^2)
= root50
BC = (7^2 + 1^2)^1/2
= root50 [2 marks]
ii) Line perpendicular to AC which passes through B.
Grad of AC is -2, perp gradient is 1/2. y = 1/2x-1 [4 marks]
iii) (10, 4) [2 marks]
iv) Equation of line AD 7y = x + 18
Sub in x coordinate of point E (8, 3.8)
7y = 8 + 18
7y = 26
y = 26/7
26/7 > 3.8
Lies outside [4 marks]

11)
i) Cubic graph, cutting x-axis at -5, 3/2, 2, and y-axis at 30 [3 marks]
ii) f(x) = (x-2)(2x-3)(x+5)
g(x) = f(x+3)
g(x) = ((x+3)-2)(2(x+3)-3)((x+3)+5)
= 2x^3 +21x^2 +43x +24 [3 marks]
iii) x = -2 is a root of g(x) = 6. Divide g(x) = 6 by (x+2) to get other roots: (-17+ root217)/4 or (-17- root217)/4 [6 marks]

12)
i) (x+1/2)^2 + 11/4. Min point (-1/2, 11/4). Min y value is 11/4 - Above x-axis [4 marks]
ii) Intersection of x^2+x+3 and 2x^2-3x-9: (6, 45) and (-2, 5) [4 marks]
iii) For what values of k is there no intersection of x^2 + x + k and 2x^2 - 3x- 9
Use b^2 - 4ac<0
k< -13 [4 marks]
2
reply
Lxxa14
Badges: 17
Rep:
?
#2
Report 3 years ago
#2
https://www.thestudentroom.co.uk/sho....php?t=4718410
0
reply
sammyboyg101
Badges: 5
Rep:
?
#3
Report 3 years ago
#3
What about the rearranging the forumla one?
0
reply
LokiiR
Badges: 9
Rep:
?
#4
Report 3 years ago
#4
1. Sketched y= -2x + 1 (0,1) and gradient was -2 (2 marks)
2.
i) (1 7/9)^-1/2 = 3/4 (3 marks)
ii) (6x^5y^2)^3/18y^10 = 12x^15/y^4 (2 marks)
3. 6-x > 5(x-3) = 6 - x > 5x - 15 = -6x > -21 = x< 21/6 (3 marks)
4. Intersection of 2x+5y=5 and x-2y = (10/3, -1/3) (4 marjs)
5. (x+2)^2 + (y-3)^2 = 5
i) radius is root 5, centre is (-2,3) (2 marks)
ii) Line parallel to 5x + y = 4 and touches centre is y= -5x-7 (2 marks)
6. r = root (v/a+b) - Make B the subject : Two possible answers are b = (v-ar^2)/r^2 or b = v/r^2 - a (4 marks)
7.
i) (5-2root7)/3+root7 = (29-11root7)/2 (3 marks)
ii) 12/root2 + root 98 = 13root2 (2 marks)
8. (a+bx)^5 x^3 coefficient is -1080, constant is 32. a^5 = 32, so a = 2. b= -3 (5 marks)
9. Three consecutive integers and n is the smallest (therefore n, n+1 and n+2) show difference between squares of smallest and largest is 4 * middle. (n+2)^2 - n^2 = 4n+4. 4n+4 = 4(n+1). (4 marks.

Section B

10. A(3,3) B (-2,-2) C (5,-1)
i) Show AB is BC. AB = ROOT 50, BC = ROOT 50 (2 marks)
ii)Line perpendicular to AC which passes by B
Grad of AC is -2, perp gradient is 1/2. Line is y= 1/2x-1 (4 marks)
iii) D is (10,4) because difference between B and C is 7, 1. Add 7 to x of A and 1 to y of A to get 10,4 (2 marks)
iv) 3.8 < 26/7. Find equation of lines and prove 8, 3.8 does not belong in them therefore it is outside rhombus (4 marks)
11. f(x) = (x-2)(2x-3)(x+5)
i) Sketch graph (3 marks)
ii) Show when (-3 0) occurs that g(x) is 2x^3 + 21x^2 + 43x + 24. You use f(x+3) and from that x+3 = 2, 3/2, -5
x = -1, -3/2, -8 are new roots
factors are (x+1)(2x+3)(x+8), multiply out to get the answer (3 marks)
iii)x = -2 is root of g(x)=6, prove this and find other roots. Other roots = (-17+-root217)/4 - Not sure of this (6 marks)
12.
i) Show y= x^2 + x + 3 as (x+a)^2 + b and it doesnt touch the x axis. Completed square is (x+1/2)^2 + 11/4 So minimum point is -1/2, 11/4 which is above x axis (4 marks)
ii) Intersection of x^2+x+3 and 2x^2-3x-9 are (6,45) and (-2,5) (4 marks)
iii) For what values of k is there no intersection of x^2 + x + k and 2x^2 - 3x- 9
Use b^2 - 4ac<0
k< -13ImageImageImage
1
reply
economy
Badges: 7
Rep:
?
#5
Report 3 years ago
#5
Our version can be found here.
0
reply
RebeccaP10
Badges: 10
Rep:
?
#6
Report 3 years ago
#6
(Original post by LokiiR)
1. Sketched y= -2x + 1 (0,1) and gradient was -2 (2 marks)
2.
i) (1 7/9)^-1/2 = 3/4 (3 marks)
ii) (6x^5y^2)^3/18y^10 = 12x^15/y^4 (2 marks)
3. 6-x > 5(x-3) = 6 - x > 5x - 15 = -6x > -21 = x< 21/6 (3 marks)
4. Intersection of 2x+5y=5 and x-2y = (10/3, -1/3) (4 marjs)
5. (x+2)^2 + (y-3)^2 = 5
i) radius is root 5, centre is (-2,3) (2 marks)
ii) Line parallel to 5x + y = 4 and touches centre is y= -5x-7 (2 marks)
6. r = root (v/a+b) - Make B the subject : Two possible answers are b = (v-ar^2)/r^2 or b = v/r^2 - a (4 marks)
7.
i) (5-2root7)/3+root7 = (29-11root7)/2 (3 marks)
ii) 12/root2 + root 98 = 13root2 (2 marks)
8. (a+bx)^5 x^3 coefficient is -1080, constant is 32. a^5 = 32, so a = 2. b= -3 (5 marks)
9. Three consecutive integers and n is the smallest (therefore n, n+1 and n+2) show difference between squares of smallest and largest is 4 * middle. (n+2)^2 - n^2 = 4n+4. 4n+4 = 4(n+1). (4 marks.

Section B

10. A(3,3) B (-2,-2) C (5,-1)
i) Show AB is BC. AB = ROOT 50, BC = ROOT 50 (2 marks)
ii)Line perpendicular to AC which passes by B
Grad of AC is -2, perp gradient is 1/2. Line is y= 1/2x-1 (4 marks)
iii) D is (10,4) because difference between B and C is 7, 1. Add 7 to x of A and 1 to y of A to get 10,4 (2 marks)
iv) 3.8 < 26/7. Find equation of lines and prove 8, 3.8 does not belong in them therefore it is outside rhombus (4 marks)
11. f(x) = (x-2)(2x-3)(x+5)
i) Sketch graph (3 marks)
ii) Show when (-3 0) occurs that g(x) is 2x^3 + 21x^2 + 43x + 24. You use f(x+3) and from that x+3 = 2, 3/2, -5
x = -1, -3/2, -8 are new roots
factors are (x+1)(2x+3)(x+8), multiply out to get the answer (3 marks)
iii)x = -2 is root of g(x)=6, prove this and find other roots. Other roots = (-17+-root217)/4 - Not sure of this (6 marks)
12.
i) Show y= x^2 + x + 3 as (x+a)^2 + b and it doesnt touch the x axis. Completed square is (x+1/2)^2 + 11/4 So minimum point is -1/2, 11/4 which is above x axis (4 marks)
ii) Intersection of x^2+x+3 and 2x^2-3x-9 are (6,45) and (-2,5) (4 marks)
iii) For what values of k is there no intersection of x^2 + x + k and 2x^2 - 3x- 9
Use b^2 - 4ac<0
k< -13ImageImageImage
Do you think they'll allow both options for rearranging the formula?
0
reply
LokiiR
Badges: 9
Rep:
?
#7
Report 3 years ago
#7
(Original post by RebeccaP10)
Do you think they'll allow both options for rearranging the formula?
they should, both are correct
1
reply
username1909405
Badges: 9
Rep:
?
#8
Report 3 years ago
#8
(Original post by RebeccaP10)
Do you think they'll allow both options for rearranging the formula?
For 11 part 3, the part inside of the root should say (17*17)-4(2)(18) which would be 289-144= 145, so in total it should read (if i'm correct) -17+-SQRT(145)/4
0
reply
fluffybunny333
Badges: 1
Rep:
?
#9
Report 3 years ago
#9
(Original post by Libaan)
For 11 part 3, the part inside of the root should say (17*17)-4(2)(18) which would be 289-144= 145, so in total it should read (if i'm correct) -17+-SQRT(145)/4
Hey I think the equation for g(x) in q11)iii) has been written out wrong! Here is my working for the question... I hope this is right
Name:  unnamed.jpg
Views: 872
Size:  506.6 KB
1
reply
bellaboo8
Badges: 8
Rep:
?
#10
Report 3 years ago
#10
Ah, this is brill thank you!

I think/hope i got most of these.

Although i messed up the binomial expansion and the rhombus questions.

how many marks do you think i'd get for writing out the nCk x a^(n-k) x b^(k) for x^3 ?

Also, for the rhombus questions, i attempted to find the mid point for the 2 mark question. Then the 4 mark part, I didn't find an equation, i just said that point E was bigger than the max points of each coord so it's outside. How many marks for that?

ahh, I'm just hoping for at least 62!
0
reply
weeping_fire
Badges: 8
Rep:
?
#11
Report Thread starter 3 years ago
#11
For the binomial Q, you might get method marks if you're lucky, but won't get the answer marks if you're final answer was wrong.

For the rhombus Q you might get some marks for the second part since you use you're answer from the last question (followed through) but probably not all of them.

I'm sure you did fine!

(Original post by gabriellenlewis)
Ah, this is brill thank you!

I think/hope i got most of these.

Although i messed up the binomial expansion and the rhombus questions.

how many marks do you think i'd get for writing out the nCk x a^(n-k) x b^(k) for x^3 ?

Also, for the rhombus questions, i attempted to find the mid point for the 2 mark question. Then the 4 mark part, I didn't find an equation, i just said that point E was bigger than the max points of each coord so it's outside. How many marks for that?

ahh, I'm just hoping for at least 62!
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

If you haven't confirmed your firm and insurance choices yet, why is that?

I don't want to decide until I've received all my offers (39)
40.63%
I am waiting until the deadline in case anything in my life changes (21)
21.88%
I am waiting until the deadline in case something in the world changes (ie. pandemic-related) (5)
5.21%
I am waiting until I can see the unis in person (9)
9.38%
I still have more questions before I made my decision (5)
5.21%
No reason, just haven't entered it yet (6)
6.25%
Something else (let us know in the thread!) (11)
11.46%

Watched Threads

View All