C1 Maths AS aqa 2016 (unofficial mark scheme new)
Unoffical Mark scheme for C1 AQA 2016
It would help if you could link answers to questions as i cant remember them
Questions:
1) a)Asked to work out gradient of a tangent, m= - 5/3 [2]
b)Asked to find co-ordinates of B B( - 3,4) [3]
c)Asked to find K K= - 30 [2]
2) a) simplify (3√5)^2 = 45 [1]
b) i cant remember the question but the answer was 75 - 32√5 [4]
3) a) y=(x-7/2)^2 - 41/4
b) min value = -41/4 [1]
c) Translation of (1/2 , 41/4)
4) a) show that (x+3) was a factor of x^3 - 5x^2 -8x + 48, p(-3) = 0, (x+3) is a factor of p(x).
b) three linear factors of p(x) = (x+3)(x-4)(x-4)
c) find remainder when p(x) was divided by (x+2) R=20
d) Factorise p(x) using (x+2) R = (x-2)(x^2-3x-14) + 20 [3]
5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2) [3]
b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7) [2]
c) asked to find equation of tangent at A 7x-4y+18=0 [5]
d) asked to find length of CT = 9 [2]
6) a) y=-32x-40
b)Q(-5/4) [1]
c) upside down positive graph passing through y axis at 8 [2]
d) x = -1±√5
7) a)
b)
c) definite integral =
= 81 /4 [5]
d) area of shaded region =
= 45/4 [3]
8) a) d^2y/dx^2= - 2x - 9x^2 [2]
b) verify that P was a minimum point sub x co-ordinate of P into dy/dx (given in the question) to prove it was a stationary point, then sub in x-coord of p into d^2y/dx^2 to get +45, d^2y/dx^2>0 therefore minimum point [4]
c) show that y is decreasing, so make your dy/dx<0 and rearrange to get into the form they wanted (note when you multiply by -1 the inequality sign flips)
d) cant remember anything except the answers were k>6 and k<-3/2
Answers that need a question to be assigned to
- k=4 and k=20
It would help if you could link answers to questions as i cant remember them
Questions:
1) a)Asked to work out gradient of a tangent, m= - 5/3 [2]
b)Asked to find co-ordinates of B B( - 3,4) [3]
c)Asked to find K K= - 30 [2]
2) a) simplify (3√5)^2 = 45 [1]
b) i cant remember the question but the answer was 75 - 32√5 [4]
3) a) y=(x-7/2)^2 - 41/4
b) min value = -41/4 [1]
c) Translation of (1/2 , 41/4)
4) a) show that (x+3) was a factor of x^3 - 5x^2 -8x + 48, p(-3) = 0, (x+3) is a factor of p(x).
b) three linear factors of p(x) = (x+3)(x-4)(x-4)
c) find remainder when p(x) was divided by (x+2) R=20
d) Factorise p(x) using (x+2) R = (x-2)(x^2-3x-14) + 20 [3]
5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2) [3]
b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7) [2]
c) asked to find equation of tangent at A 7x-4y+18=0 [5]
d) asked to find length of CT = 9 [2]
6) a) y=-32x-40
b)Q(-5/4) [1]
c) upside down positive graph passing through y axis at 8 [2]
d) x = -1±√5
7) a)
b)
c) definite integral =
= 81 /4 [5]d) area of shaded region =
= 45/4 [3]8) a) d^2y/dx^2= - 2x - 9x^2 [2]
b) verify that P was a minimum point sub x co-ordinate of P into dy/dx (given in the question) to prove it was a stationary point, then sub in x-coord of p into d^2y/dx^2 to get +45, d^2y/dx^2>0 therefore minimum point [4]
c) show that y is decreasing, so make your dy/dx<0 and rearrange to get into the form they wanted (note when you multiply by -1 the inequality sign flips)
d) cant remember anything except the answers were k>6 and k<-3/2
Answers that need a question to be assigned to
- k=4 and k=20
(edited 7 years ago)
Scroll to see replies
Original post by beanigger
Unoffical Mark scheme for C1 AQA 2016
It would help if you could link answers to questions as i cant remember them
Questions:
1) a)Asked to work out gradient of a tangent, m= - 3/2
b)Asked to find co-ordinates of B B( - 3,4)
c)Asked to find K K= - 30
2) a) simplify (3√5)^2 = 45
b) i cant remember the question but the answer was 75 - 32√5
3) a)
4) a)
5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2)
b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7)
c) asked to find equation of tangent at A 7x-4y+18=0
d) asked to find length of CT = 9
6) a)
7) a) i think there was 4 parts to this question i cant remember part a and b
b)
c) definite integral = 81/4
d) area of shaded region = 45/4
8) a) cant remember anything except the answers were k>6 and k<-3/2
Answers that need a question to be assigned to
- Translation of (1/2 , 41/4) and y=(x-7/2)^2 - 41/4
- Q(-5/4), y=-32x-40, upside down positive graph passing through y axis at 8
- k=4 and k=20
- x = -1±√5
- remainder = 20
- three linear factors (x+3)(x-4)(x-4)
- d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
It would help if you could link answers to questions as i cant remember them
Questions:
1) a)Asked to work out gradient of a tangent, m= - 3/2
b)Asked to find co-ordinates of B B( - 3,4)
c)Asked to find K K= - 30
2) a) simplify (3√5)^2 = 45
b) i cant remember the question but the answer was 75 - 32√5
3) a)
4) a)
5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2)
b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7)
c) asked to find equation of tangent at A 7x-4y+18=0
d) asked to find length of CT = 9
6) a)
7) a) i think there was 4 parts to this question i cant remember part a and b
b)
c) definite integral = 81/4
d) area of shaded region = 45/4
8) a) cant remember anything except the answers were k>6 and k<-3/2
Answers that need a question to be assigned to
- Translation of (1/2 , 41/4) and y=(x-7/2)^2 - 41/4
- Q(-5/4), y=-32x-40, upside down positive graph passing through y axis at 8
- k=4 and k=20
- x = -1±√5
- remainder = 20
- three linear factors (x+3)(x-4)(x-4)
- d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
Love you bruv, i love people who make mark schemes, i imagine them as snails working away slowly on their computer, for hours on end.
You are mark scheme champion
You are mark scheme champion
Original post by UbaidS
how did you get 65 for 5a
Many people in our 6th form got Root65 for the Radius of the Circle.
Do you remember the equation given for the very first question?
Original post by Rit101
Many people in our 6th form got Root65 for the Radius of the Circle.
Do you remember the equation given for the very first question?
Do you remember the equation given for the very first question?
The q 'stated' (x-a)^2 + (y-b)^2 = k
where 'k' = r^2 = 65
anyone who put sqroot65 is wrong and would of lost 1 mark
Original post by Rit101
Many people in our 6th form got Root65 for the Radius of the Circle.
Do you remember the equation given for the very first question?
Do you remember the equation given for the very first question?
AB 5x+3y+3=0
Original post by Sniperdon227
The q 'stated' (x-a)^2 + (y-b)^2 = k
where 'k' = r^2 = 65
anyone who put sqroot65 is wrong and would of lost 1 mark
where 'k' = r^2 = 65
anyone who put sqroot65 is wrong and would of lost 1 mark
Yeah I get that but look at what I said. The radius of the circle was infact root65.
I agree with all that you have put.
Original post by Sniperdon227
The q 'stated' (x-a)^2 + (y-b)^2 = k
where 'k' = r^2 = 65
anyone who put sqroot65 is wrong and would of lost 1 mark
where 'k' = r^2 = 65
anyone who put sqroot65 is wrong and would of lost 1 mark
shouldnt it be 34?!
Original post by UbaidS
how did you get 65 for 5a
It should be 34right?
Original post by saaidq
It should be 34right?
Exactly i keep seeing 65 making me worried and that
Original post by saaidq
It should be 34right?
Nope pretty sure the top students in my Sixth Form got root65 as the radius
Original post by UbaidS
shouldnt it be 34?!
If you can justify why it was 34 ill give you a cookie!
You can only complete the sq when they give you x^2 +y^2 +6x+8y+3=0 to find the radius, in this case they gave two coordinates!
by the way that x^2 + y^2 .... was made up
(edited 7 years ago)
Original post by xs4
AB 5x+3y+3=0
This means that the gradient of AB was -3/5
Hence the parallel line should have the same gradient correct?
Original post by Sniperdon227
If you can justify why it was 34 ill give you a cookie!
You can only complete the sq when they give you x^2 +y^2 +6x+8y+3=0 to find the radius, in this case they gave two coordinates!
by the way that x^2 + y^2 .... was made up
You can only complete the sq when they give you x^2 +y^2 +6x+8y+3=0 to find the radius, in this case they gave two coordinates!
by the way that x^2 + y^2 .... was made up
You got root65 like me? For the radius?
Original post by rit101
nope pretty sure the top students in my sixth form got root65 as the radius
the top students were sitting on the floor below me, i was top, i was sitting on the 2nd floor, they were on ground floor, so whos top now bruv ?
Yalll mandem
Original post by Rit101
You got root65 like me? For the radius?
got k= r^2= 65
where r= sqroot65 correct
Okay, I've added a little bit.
1) a)Asked to work out gradient of a parallel line, m= - 3/2
b) Asked to find co-ordinates of B, B( - 3,4)
c) Asked to find K, K= - 30
2) a) simplify (3√5)^2 = 45
b) Put ((3√5)^2 + √5) / (7 + 3 √5) into m + n√5 = 75 - 32√5
3) a) y=(x-7/2)^2 - 41/4
b) Translation of (1/2 , 41/4)
4) a) show that (x+3) was a factor of x^3 - 5x^2 -8x + 48. p(-3) = -3^3 - 5 * 3-^2 - 8 * -3 + 48 = 0. As p(-3) = 0, (x+3) is a factor of p(x).
b) Three linear factors goes to (x+3)(x-4)(x-4)
c) find the remainder when x^3 - 5x^2 -8x + 48 was divided by (x+2) R=20
d) factorise p(x) using your answer to part c) as R = (x-2)(x^2-3x+14) + 20
5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2)
b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7)
c) asked to find the equation of the tangent at A 7x-4y+18=0
d) asked to find the length of CT = 9
6) a)
7) a) I think there were 4 parts to this question I can't remember part a and b
b)
c) Definite integral (from 1 to -2) = 81/4 (20.25)
d) Area of shaded region (integral - triangle area) = 45/4 (11.25)
8) a) can't remember anything except the answers were k>6 and k<-3/2 (NOT 6 > k > -3/2 )
Answers that need a question to be assigned to
- Q(-5/4), y=-32x-40, upside down positive graph passing through y axis at 8
- k=4 and k=20
- x = -1±√5
- d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
1) a)Asked to work out gradient of a parallel line, m= - 3/2
b) Asked to find co-ordinates of B, B( - 3,4)
c) Asked to find K, K= - 30
2) a) simplify (3√5)^2 = 45
b) Put ((3√5)^2 + √5) / (7 + 3 √5) into m + n√5 = 75 - 32√5
3) a) y=(x-7/2)^2 - 41/4
b) Translation of (1/2 , 41/4)
4) a) show that (x+3) was a factor of x^3 - 5x^2 -8x + 48. p(-3) = -3^3 - 5 * 3-^2 - 8 * -3 + 48 = 0. As p(-3) = 0, (x+3) is a factor of p(x).
b) Three linear factors goes to (x+3)(x-4)(x-4)
c) find the remainder when x^3 - 5x^2 -8x + 48 was divided by (x+2) R=20
d) factorise p(x) using your answer to part c) as R = (x-2)(x^2-3x+14) + 20
5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2)
b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7)
c) asked to find the equation of the tangent at A 7x-4y+18=0
d) asked to find the length of CT = 9
6) a)
7) a) I think there were 4 parts to this question I can't remember part a and b
b)
c) Definite integral (from 1 to -2) = 81/4 (20.25)
d) Area of shaded region (integral - triangle area) = 45/4 (11.25)
8) a) can't remember anything except the answers were k>6 and k<-3/2 (NOT 6 > k > -3/2 )
Answers that need a question to be assigned to
- Q(-5/4), y=-32x-40, upside down positive graph passing through y axis at 8
- k=4 and k=20
- x = -1±√5
- d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
(edited 7 years ago)
Original post by Rit101
Nope pretty sure the top students in my Sixth Form got root65 as the radius
root65 is the radius however writing it in the form asked for, it is r squared
Original post by Rit101
This means that the gradient of AB was -3/5
Hence the parallel line should have the same gradient correct?
Hence the parallel line should have the same gradient correct?
-5/3 but I can't even remember the question I just remember rearranging to get 3y=-5x-3.
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