O-LevelAdditional MathematicsDifferentiation and integration4037/11 · May/June 2017 · Q15 Marks

The line y = kx – 5, where k is a positive constant, is a tangent to the curve y = x² + 4x at the point A.

(i)Find the exact value of k.[3]
(ii)Find the gradient of the normal to the curve at the point A, giving your answer in the form a + b√5, where a and b are constants.[2]

Mark scheme

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

[M1] equating line and curve equation and collecting terms to form an equation of the form ax² + bx + c = 0
[DM1] correct use of discriminant
[A1] Accept k = 4+√20
[M1] Gradient of line = k; Gradient of curve = dy/dx = 2x + 4; Equating: k = 2x+4
[DM1] substitution of k = 2x + 4 or x = (k-4)/2 in kx-5= x² + 4x and simplify to a quadratic equation in k or x
[A1] Accept k=4+√20
[M1] use of negative reciprocal and attempt to rationalise using a form of a−b√5 or a−√20 or their equivalent from (i)
[A1] 4-2√5 / -4 oe leading to 1 - √5 / 2

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Topic: Differentiation and integration, O-Level Additional Mathematics (syllabus 4037). Structured question, 5 marks.
Source: O-Level Additional Mathematics, May/June 2017 Paper 1 Variant 1, question 1.

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