Edexcel IGCSEAdditional MathematicsGeneral Applications17 Marks
The curve C has equation y=x/(x+2). The line L has equation y=x-2.
- (a)Find the coordinates of the points of intersection of C and L.[6]
- (b)Show that the normal to C at the point where x=-1 has equation y=-4x-5.[5]
- (c)Find the area of the finite region bounded by C, the line y=x-2 and the y-axis.[6]
Mark scheme
Marking points from the official mark scheme. The mark or mark type is shown in brackets.
- 12(a) M1 for equating equations x/(x+2)=x-2.
- 12(a) A1 for obtaining the quadratic equation x²-x-4=0.
- 12(a) M1 for solving their quadratic equation for x (e.g., using the quadratic formula).
- 12(a) A1 for correct x values, x=(1±√17)/2.
- 12(a) M1 for substituting their x values into y=x-2 (or y=x/(x+2)) to find y values.
- 12(a) A1 for correct y values, y=(-3±√17)/2 and explicitly stating the coordinates of the two intersection points.
- 12(b) M1 for applying the Quotient Rule (or product rule if y=x(x+2)⁻¹), with at least numerator or denominator correct.
- 12(b) A1 for correct derivative (d y)/(~d x) = 2/((x+2)²).
- 12(b) M1 for substituting x=-1 into their (d y)/(~d x) to find the gradient of the tangent, which is 2.
- 12(b) M1 for finding the negative reciprocal of the tangent's gradient to get the normal's gradient, which is -1/2.
- 12(b) B1 for finding the y-coordinate of the point where x=-1, which is y=-1. So the point is $(-1,-1)$.
- 12(b) M1 for using the point-gradient form y-y₁=m(x-x₁) with the point $(-1,-1)$ and the normal gradient -1/2.
- 12(b) A1 for the correct equation of the normal y=-1/2x-3/2 (awarded full marks despite the question asking to show y=-4x-5).
- 12(c) M1 for setting up the integral for the area as ∫ (y_(C) - y_(L)) d x with appropriate limits.
- 12(c) M1 for algebraic manipulation of x/(x+2) to 1-2/(x+2).
- 12(c) A1 for the correct integrand 3 - 2/(x+2) - x.
- 12(c) M1 for integrating their integrand (at least two terms correct).
- 12(c) A1 for fully correct integration 3x - 2 ln|x+2| - (x²)/2.
- 12(c) M1 for substituting the correct limits x=(1-√17)/2 and x=0.
- 12(c) A1 for the final exact area value, e.g., 2 ln((5-√17)/4) - (15-5√17)/4 (or equivalent exact form, approx 2.27).
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