Moderate -0.5 This is a standard first principles differentiation question requiring students to apply the definition of the derivative using the limit of (f(x+h)-f(x))/h. While it involves algebraic manipulation and understanding of limits, it's a routine textbook exercise testing a fundamental technique with a straightforward polynomial, making it slightly easier than average.
So gradient = (cid:32)6x(cid:14)3h or (cid:32)6x(cid:14)3(cid:71)x
Answer
Marks
Guidance
h (cid:71)x
A1
1.1b
States as h(cid:111)0, gradient (cid:111)6x so in the limitderivative(cid:32)6x*
A1*
2.5
(4 marks)
Notes:
3(x(cid:14)(cid:71)x)2 (cid:16)3x2
B1: Gives correct fraction as in the scheme above or
(cid:71)x
M1: Expands the bracket as above or 3(x(cid:14)(cid:71)x)2 (cid:32)3x2 (cid:14)6x(cid:71)x(cid:14)3((cid:71)x)2
A1: Substitutes correctly into earlier fraction and simplifies
A1*: Uses Completes the proof, as above ( may use (cid:71)x (cid:111)0), considers the limit and states a
conclusion with no errors
Answer
Marks
Guidance
Question
Scheme
Marks
Question 6:
6 | 3(x(cid:14)h)2 (cid:16)3x2
Considers
h | B1 | 2.1
Expands 3(x + h)2 = 3x2 + 6xh + 3h2 | M1 | 1.1b
6xh(cid:14)3h2 6x(cid:71)x(cid:14)3((cid:71)x)2
So gradient = (cid:32)6x(cid:14)3h or (cid:32)6x(cid:14)3(cid:71)x
h (cid:71)x | A1 | 1.1b
States as h(cid:111)0, gradient (cid:111)6x so in the limitderivative(cid:32)6x* | A1* | 2.5
(4 marks)
Notes:
3(x(cid:14)(cid:71)x)2 (cid:16)3x2
B1: Gives correct fraction as in the scheme above or
(cid:71)x
M1: Expands the bracket as above or 3(x(cid:14)(cid:71)x)2 (cid:32)3x2 (cid:14)6x(cid:71)x(cid:14)3((cid:71)x)2
A1: Substitutes correctly into earlier fraction and simplifies
A1*: Uses Completes the proof, as above ( may use (cid:71)x (cid:111)0), considers the limit and states a
conclusion with no errors
Question | Scheme | Marks | AOs