Alevel_C1C2
Alevel_C1C2
Alevel_C1C2
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7<br />
Exam-style mark scheme<br />
Differentiation<br />
Question Scheme Marks<br />
Number<br />
B1<br />
1. Substitute (1, 2) in y = ax 2 + bx + 3; 2 = a + b + 3 B1<br />
a + b = -1 …. (*)<br />
B1<br />
dy<br />
= 2 ax + b<br />
dx<br />
M1 M1<br />
dy<br />
at A is 2a + b = 1 ….. (**)<br />
dx Subtract (*) from (**): a = 2, b = -3<br />
B2<br />
dy<br />
= 4x<br />
−3, then gradient at B is -7<br />
dx<br />
B1<br />
8<br />
2. (a) y = 4x + x -1 B1<br />
dy<br />
= − x<br />
− 2<br />
4<br />
M1 M1 (3)<br />
dx<br />
(b) 4 - x -2 = 0 B1<br />
4x 2 - 1 = 0<br />
B1<br />
x =± 1 2<br />
A1 (3)<br />
(c)<br />
2<br />
d y<br />
2<br />
dx = 2x-3 M1<br />
y<br />
x = ± 1 d<br />
⇒ = ± 16<br />
2 2<br />
dx<br />
2<br />
A2 (3)<br />
9<br />
3. (a) dy<br />
2<br />
= 3x<br />
− 2x<br />
dx<br />
M2<br />
Gradient at x = -1 is 5<br />
B1<br />
If x = -1 then y = -2, A(-1, -2)<br />
B1<br />
Equation of the tangent at A is y + 2 = 5(x + 1)<br />
B1oe<br />
y = 5x + 3 A1 (6)<br />
(b) gradient of the normal at A is − 1 B1<br />
5<br />
equation of the normal at A is y + 2 = − 1 (x + 1)<br />
5 B1oe<br />
5y + x + 11 = 0 A1 (3)<br />
9<br />
© Oxford University Press 2008<br />
Core C1
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