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Date	November 2018	Marks available	5	Reference code	18N.2.AHL.TZ0.H_5
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Differentiate	Question number	H_5	Adapted from	N/A

Question

Differentiate from first principles the function $f\left( x \right) = 3{x^3} - x$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$= \frac{{\left( {3{{\left( {x + h} \right)}^3} - \left( {x + h} \right)} \right) - \left( {3{x^3} - x} \right)}}{h}$ M1

$= \frac{{3\left( {{x^3} + 3{x^2}h + 3x{h^2} + {h^3}} \right) - x - h - {3x^3} + x}}{h}$ (A1)

$= \frac{{9{x^2}h + 9x{h^2} + 3{h^3} - h}}{h}$ A1

cancelling $h$ M1

$= 9{x^2} + 9xh + 3{h^2} - 1$

then $\mathop {{\text{lim}}}\limits_{h \to 0} \left( {9{x^2} + 9xh + 3{h^2} - 1} \right)$

$= 9{x^2} - 1$ A1

Note: Final A1 dependent on all previous marks.

METHOD 2

$\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$= \frac{{\left( {3{{\left( {x + h} \right)}^3} - \left( {x + h} \right)} \right) - \left( {3{x^3} - x} \right)}}{h}$ M1

$= \frac{{3\left( {{{\left( {x + h} \right)}^3} - {x^3}} \right) + \left( {x - \left( {x + h} \right)} \right)}}{h}$ (A1)

$= \frac{{3h\left( {{{\left( {x + h} \right)}^2} + x\left( {x + h} \right) + {x^2}} \right) - h}}{h}$ A1

cancelling $h$ M1

$= 3\left( {{{\left( {x + h} \right)}^2} + x\left( {x + h} \right) + {x^2}} \right) - 1$

then $\mathop {{\text{lim}}}\limits_{h \to 0} \left( {3\left( {{{\left( {x + h} \right)}^2} + x\left( {x + h} \right) + {x^2}} \right) - 1} \right)$

$= 9{x^2} - 1$ A1

Note: Final A1 dependent on all previous marks.

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.1—Introduction of differential calculus

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