C3 January 2013 Q1
1. The curve \(C\) has equation
\[y = (2x - 3)^5\]
The point \(P\) lies on \(C\) and has coordinates \((w, -32)\).
Find
| Scheme | Marks |
|---|---|
| \(-32 = (2w - 3)^5 \Rightarrow w = \dfrac{1}{2}\) oe | M1A1 |
| (2) |
Notes
M1 Substitute y=-32 into \(y = (2w - 3)^5\) and proceed to w=….. [Accept positive sign used of \(y\), ie y=+32]
A1 Obtains \(w\) or \(x = \dfrac{1}{2}\) oe with no incorrect working seen. Accept alternatives such as 0.5.
Sight of just the answer would score both marks as long as no incorrect working is seen.
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 5 \times (2x - 3)^4 \times 2\) or \(10(2x - 3)^4\) | M1A1 |
| When \(x = \dfrac{1}{2}\), Gradient = 160 | M1 |
| Equation of tangent is \('160' = \dfrac{y - (-32)}{x - '\frac{1}{2}'}\) oe | dM1 |
| \(y = 160x - 112\) cso | A1 |
| (5) | |
| (7 marks) |
Notes
M1 Attempts to differentiate \(y = (2x - 3)^5\) using the chain rule.
Sight of \(\pm A(2x - 3)^4\) where \(A\) is a non- zero constant is sufficient for the method mark.
A1 A correct (un simplified) form of the differential.
Accept \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 5 \times (2x - 3)^4 \times 2\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 10(2x - 3)^4\)
M1 This is awarded for an attempt to find the gradient of the tangent to the curve at P
Award for substituting their numerical value to part (a) into their differential to find the numerical gradient of the tangent
dM1 Award for a correct method to find an equation of the tangent to the curve at P. It is dependent upon the previous M mark being awarded.
Award for \('\text{their } 160' = \dfrac{y - (-32)}{x - \text{their } '\frac{1}{2}'}\)
If they use \(y = mx + c\) it must be a full method, using m= ‘their 160’, their ‘\(\frac{1}{2}\)’ and -32.
An attempt must be seen to find c=…
A1 cso \(y = 160x - 112\). The question is specific and requires the answer in this form.
You may isw in this question after a correct answer.