Solution of Past Paper of CAIE Math 9709/12 Feb/Mar 2020 Question # 10

CIE/GCE/AS/Math/P1/20/Mar/12/Q#10

Question
1
𝑑𝑦
The gradient of a curve at the point (x, y) is given by = 2(𝑥 + 3)2 − 𝑥. The curve has
𝑑𝑥

a stationary point at (a, 14), where a is a positive constant.

a) Find the value of 𝑎.

b) Determine the nature of the stationary point.

c) Find the equation of the curve.



Solution

a)


We are given that gradient of the curve at the point (x, y) is;

𝑑𝑦 1
= 2 (𝑥 + 3 )2 − 𝑥
𝑑𝑥

We are also given that point at (a, 14) is a stationary point.

A stationary point 𝑆 (𝑥, 𝑦) on the curve 𝑦 is the point where gradient of the curve is
equal to zero;
𝑑𝑦
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝐶𝑢𝑟𝑣𝑒𝐴𝑡 𝑆𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑃𝑜𝑖𝑛𝑡 = | = 0
𝑑𝑥 𝑠

Therefore, at point (a, 14);

𝑑𝑦
=0
𝑑𝑥
Gradient (slope) of the curve at the particular point is the derivative of equation of the
curve at that particular point.

, CIE/GCE/AS/Math/P1/20/Mar/12/Q#10
𝑑𝑦
Gradient (slope) of the curve 𝑦 at a particular point 𝑃(𝑥, 𝑦) can be found by
𝑑𝑥
substituting x-coordinates of that point in the expression for gradient of the curve;
𝑑𝑦
|
𝑑𝑥 𝑥=𝑥𝑃

Hence, gradient at point (a, 14);

𝑑𝑦 1
| = 2(𝑥𝑃 + 3)2 − 𝑥𝑃
𝑑𝑥 𝑥=𝑥𝑃


𝑑𝑦 1
| = 2 (𝑎 + 3 )2 − 𝑎
𝑑𝑥 𝑥=𝑎


Coordinates of stationary point on the curve 𝑦 can be found from the derivative of
equation of the curve by equating it with ZERO. This results in value of x-coordinate of
the stationary point 𝑆(𝑥, 𝑦) on the curve.

Hence;
𝑑𝑦
| =0
𝑑𝑥 𝑥=𝑎

1
2 (𝑎 + 3 )2 − 𝑎 = 0
1
2 (𝑎 + 3 )2 = 𝑎

1 2
(2(𝑎 + 3)2 ) = (𝑎 )2

4 (𝑎 + 3 ) = 𝑎 2

4𝑎 + 12 = 𝑎2

𝑎2 − 4𝑎 − 12 = 0

𝑎2 − 6𝑎 + 2𝑎 − 12 = 0