O-LevelAdditional MathematicsDifferentiation and integration4037/22 · Oct/Nov 2017 · Q79 Marks

The gradient of the normal to a curve at the point with coordinates (x, y) is given by √x / (1 - 3x).

(i)Find the equation of the curve, given that the curve passes through the point (1, -10).[5]
(ii)Find, in the form y = mx + c, the equation of the tangent to the curve at the point where x = 4.[4]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

[B1] (Gradient or dy/dx) = (3x-1)/√x; Gradient = Negative reciprocal. Can be implied.
[B1] = 3x^(1/2) - x^(-1/2); + One correct term
[M1] y = 2x^(3/2) - 2x^(1/2) (+C); at least 1 fractional power increased by1.
[A1] -10=2-2+ C → C = -10; one term correct with simplified coefficients
[A1] y = 2x^(3/2) - 2x^(1/2) - 10; For C from correct working.
[B1] x = 4 → y = 16-4-10=2
[B1] dy/dx = 6-1/2 = 5.5
[M1] Eqn with their grad and point (4, ...)
[A1] Eqn of tangent: (y-2)/(x-4) = 5.5 → y = 5.5x-20 oe; Must be in the form y = mx + c but accept 2y = 11x-40

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.

Topic: Differentiation and integration, O-Level Additional Mathematics (syllabus 4037). Structured question, 9 marks.
Source: O-Level Additional Mathematics, Oct/Nov 2017 Paper 2 Variant 2, question 7.

Questions and mark schemes are from past examination papers and remain the copyright of the exam board. How Oracle Prep builds its content