O-LevelAdditional MathematicsDifferentiation and integration4037/11 · May/June 2019 · Q119 Marks
It is given that y = (x²+1)(2x-3)^(1/2).
- (i)Show that dy/dx = (Px²+Qx+1)/(2x-3)^(1/2), where P and Q are integers.[5]
- (ii)Hence find the equation of the normal to the curve y = (x²+1)(2x-3)^(1/2) at the point where x = 2, giving your answer in the form ax + by + c = 0, where a, b and c are integers.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.
- [M1] differentiation of a product
- [B1] for (2x-3)^(-1/2) = 1/2 × 2(2x-3)^(-1/2) oe
- [A1] all else correct i.e. dy/dx = (x²+1)f'(x)+2x(2x-3)^(1/2)
- [M1] correctly taking out a factor of (2x-3)^(-1/2) or correctly using (2x-3)^(1/2) as denominator
- [A1] 5x²-6x+1 / (2x-3)^(1/2)
- [B1] When x = 2, y=5
- [M1] substitution to obtain gradient and correct method for gradient of normal
- [M1] DepM1 for equation of normal
- [A1] Must be in this form x+9y-47=0 or -x-9y+47=0
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