Standard +0.3 This is a standard P1 tangent problem requiring students to set the line equal to the curve, form a quadratic, and use the discriminant condition (b²-4ac=0) for tangency. While it involves multiple steps, the technique is routine and commonly practiced, making it slightly easier than average.
The straight line with equation \(y = 4x + c\), where \(c\) is a constant, is a tangent to the curve with equation \(y = 2x^2 + 8x + 3\)
Calculate the value of \(c\) [5]
Substitute x = -1 in 2x2 (cid:14)8x(cid:14)3 (cid:32) 4x(cid:14)c
dM1
Attempts to find value of c
dM1
c = 1 or writing y = 4x + 1 cso
A1
(5)
Alternative 2
Answer
Marks
Sets 2x2(cid:14)8x(cid:14)3(cid:32)4x(cid:14)c and collects x terms together
M1
Obtains 2x2(cid:14)4x(cid:14)3(cid:16)c(cid:32)0 or equivalent
A1
States that b2(cid:16)4ac(cid:32)0
dM1
42(cid:16)4(cid:117)2(cid:117)(3(cid:16)c)(cid:32)0 and so c =
dM1
c = 1 cso
A1
(5)
Alternative 3
Answer
Marks
Sets 2x2(cid:14)8x(cid:14)3(cid:32)4x(cid:14)c and collects x terms together
M1
Obtains 2x2(cid:14)4x(cid:14)3(cid:16)c(cid:32)0 or equivalent
A1
Uses 2(x(cid:14)1)2(cid:16)2(cid:14)3(cid:16)c(cid:32)0 or equivalent
dM1
Writes -2 +3 – c = 0
dM1
So c = 1 cso
A1
(5)
(5 marks)
Question 4 continued
Notes:
Method 1A
dy
M1: Attempts to solve their (cid:32)4. They must reach x =... (Just differentiating is M0 A0).
dx
dy
A1: x = –1 (If this follows (cid:32)4x + 8, then give M1 A1 by implication).
dx
dM1: (Depends on previous M mark) Substitutes their x = -1 into f(x) or into “their f(x) from (b)”
to find y.
dM1: (Depends on both previous M marks) Substitutes their x = -1 and their y = -3 values into y =
4x + c to find c or uses equation of line is (y + “3”) = 4(x + “1”) and rearranges to y = mx+c
A1: c = 1 or allow for y = 4x + 1 cso.
Method 1B
M1A1: Exactly as in Method 1A above.
dM1: (Depends on previous M mark) Substitutes their x = -1 into 2x2 (cid:14)8x(cid:14)3 (cid:32) 4x(cid:14)c
dM1: Attempts to find value of c then A1 as before.
Method 2
M1: Sets 2x2(cid:14)8x(cid:14)3(cid:32)4x(cid:14)c and tries to collect x terms together.
A1: Collects terms e.g. 2x2(cid:14)4x(cid:14)3(cid:16)c(cid:32)0or (cid:16)2x2 (cid:16)4x(cid:16)3(cid:14)c(cid:32)0or 2x2 (cid:14)4x(cid:14)3(cid:32)cor even
2x2 (cid:14)4x(cid:32)c(cid:16)3. Allow “=0” to be missing on RHS.
dM1: Then use completion of square 2(x(cid:14)1)2 (cid:16)2(cid:14)3(cid:16)c(cid:32)0 (Allow 2(x+1)2 – k + 3 – c = 0)
where k is non zero. It is enough to give the correct or almost correct (with k) completion of
the square.
dM1: -2 + 3 - c = 0 AND leading to a solution for c (Allow -1 + 3 - c = 0) (x = –1 has been used)
A1: c = 1 cso
Method 3
M1: Sets 2x2 (cid:14)8x(cid:14)3(cid:32)4x(cid:14)c and tries to collect x terms together. May be implied by
2x2 (cid:14)8x(cid:14)3(cid:16)4x(cid:114)con one side.
A1: Collects terms e.g. 2x2(cid:14)4x(cid:14)3(cid:16)c(cid:32)0 or (cid:16)2x2 (cid:16)4x(cid:16)3(cid:14)c(cid:32)0or 2x2 (cid:14)4x(cid:14)3(cid:32)ceven
2x2 (cid:14)4x(cid:32)c(cid:16)3. Allow “=0” to be missing on RHS.
dM1: Then use completion of square 2(x+1)2 – k + 3 – c = 0 (Allow 2(x+1)2 –k + 3 – c = 0)
where k is non zero. It is enough to give the correct or almost correct (with k) completion of
the square.
dM1: -2 + 3 - c = 0 AND leading to a solution for c (Allow -1 + 3 - c = 0) (x = -1 has been used)
A1: c = 1 cso
Answer
Marks
Question
Marks
Question 4:
4 | Sets 2x2 + 8x + 3 = 4x + c and collects x terms together | M1
Obtains 2x2 + 4x (cid:14) 3 − c = 0 o.e. | A1
States that b2 – 4ac = 0 | dM1
42 – (cid:23)×2×(cid:11)3 − c) = 0 and so c = | dM1
c = 1 cso | A1
(5)
Alternative 1A
Sets derivative "4x + 8" = 4 (cid:159)x = | M1
x (cid:32) −1 | A1
Substitute x (cid:32) −1 (cid:76)(cid:81) y = 2x2 + 8x + 3 ( (cid:159)y (cid:32) −3(cid:12) | dM1
Substitute x (cid:32) −1 (cid:68)(cid:81)(cid:71) y (cid:32) −3 (cid:76)(cid:81) y = 4x + c or into (y + 3)=4(x + 1) and
expand | dM1
c = 1 or writing y = 4x + 1 cso | A1
(5)
Alternative 1B
Sets derivative"4x(cid:14)8"(cid:32)4 (cid:159) x(cid:32), | M1
x (cid:32) −1 | A1
Substitute x = -1 in 2x2 (cid:14)8x(cid:14)3 (cid:32) 4x(cid:14)c | dM1
Attempts to find value of c | dM1
c = 1 or writing y = 4x + 1 cso | A1
(5)
Alternative 2
Sets 2x2(cid:14)8x(cid:14)3(cid:32)4x(cid:14)c and collects x terms together | M1
Obtains 2x2(cid:14)4x(cid:14)3(cid:16)c(cid:32)0 or equivalent | A1
States that b2(cid:16)4ac(cid:32)0 | dM1
42(cid:16)4(cid:117)2(cid:117)(3(cid:16)c)(cid:32)0 and so c = | dM1
c = 1 cso | A1
(5)
Alternative 3
Sets 2x2(cid:14)8x(cid:14)3(cid:32)4x(cid:14)c and collects x terms together | M1
Obtains 2x2(cid:14)4x(cid:14)3(cid:16)c(cid:32)0 or equivalent | A1
Uses 2(x(cid:14)1)2(cid:16)2(cid:14)3(cid:16)c(cid:32)0 or equivalent | dM1
Writes -2 +3 – c = 0 | dM1
So c = 1 cso | A1
(5)
(5 marks)
Question 4 continued
Notes:
Method 1A
dy
M1: Attempts to solve their (cid:32)4. They must reach x =... (Just differentiating is M0 A0).
dx
dy
A1: x = –1 (If this follows (cid:32)4x + 8, then give M1 A1 by implication).
dx
dM1: (Depends on previous M mark) Substitutes their x = -1 into f(x) or into “their f(x) from (b)”
to find y.
dM1: (Depends on both previous M marks) Substitutes their x = -1 and their y = -3 values into y =
4x + c to find c or uses equation of line is (y + “3”) = 4(x + “1”) and rearranges to y = mx+c
A1: c = 1 or allow for y = 4x + 1 cso.
Method 1B
M1A1: Exactly as in Method 1A above.
dM1: (Depends on previous M mark) Substitutes their x = -1 into 2x2 (cid:14)8x(cid:14)3 (cid:32) 4x(cid:14)c
dM1: Attempts to find value of c then A1 as before.
Method 2
M1: Sets 2x2(cid:14)8x(cid:14)3(cid:32)4x(cid:14)c and tries to collect x terms together.
A1: Collects terms e.g. 2x2(cid:14)4x(cid:14)3(cid:16)c(cid:32)0or (cid:16)2x2 (cid:16)4x(cid:16)3(cid:14)c(cid:32)0or 2x2 (cid:14)4x(cid:14)3(cid:32)cor even
2x2 (cid:14)4x(cid:32)c(cid:16)3. Allow “=0” to be missing on RHS.
dM1: Then use completion of square 2(x(cid:14)1)2 (cid:16)2(cid:14)3(cid:16)c(cid:32)0 (Allow 2(x+1)2 – k + 3 – c = 0)
where k is non zero. It is enough to give the correct or almost correct (with k) completion of
the square.
dM1: -2 + 3 - c = 0 AND leading to a solution for c (Allow -1 + 3 - c = 0) (x = –1 has been used)
A1: c = 1 cso
Method 3
M1: Sets 2x2 (cid:14)8x(cid:14)3(cid:32)4x(cid:14)c and tries to collect x terms together. May be implied by
2x2 (cid:14)8x(cid:14)3(cid:16)4x(cid:114)con one side.
A1: Collects terms e.g. 2x2(cid:14)4x(cid:14)3(cid:16)c(cid:32)0 or (cid:16)2x2 (cid:16)4x(cid:16)3(cid:14)c(cid:32)0or 2x2 (cid:14)4x(cid:14)3(cid:32)ceven
2x2 (cid:14)4x(cid:32)c(cid:16)3. Allow “=0” to be missing on RHS.
dM1: Then use completion of square 2(x+1)2 – k + 3 – c = 0 (Allow 2(x+1)2 –k + 3 – c = 0)
where k is non zero. It is enough to give the correct or almost correct (with k) completion of
the square.
dM1: -2 + 3 - c = 0 AND leading to a solution for c (Allow -1 + 3 - c = 0) (x = -1 has been used)
A1: c = 1 cso
Question | Marks
The straight line with equation $y = 4x + c$, where $c$ is a constant, is a tangent to the curve with equation $y = 2x^2 + 8x + 3$
Calculate the value of $c$ [5]
\hfill \mbox{\textit{Edexcel P1 2018 Q4 [5]}}