Challenging +1.2 This question requires finding a normal line equation (differentiation and negative reciprocal gradient), then solving a transcendental equation involving ln(2x). While it combines multiple techniques across 8 marks, each step is relatively standard: differentiate a quadratic, find normal equation, substitute into the second curve, and solve. The transcendental equation likely has a 'nice' solution by inspection or simple manipulation. It's above average due to length and the ln component, but doesn't require novel insight or particularly sophisticated problem-solving.
\includegraphics{figure_3}
The curve \(C_1\), shown in Figure 3, has equation \(y = 4x^2 - 6x + 4\).
The point \(P\left(\frac{1}{2}, 2\right)\) lies on \(C_1\)
The curve \(C_2\), also shown in Figure 3, has equation \(y = \frac{1}{2}x + \ln(2x)\).
The normal to \(C_1\) at the point \(P\) meets \(C_2\) at the point \(Q\).
Find the exact coordinates of \(Q\).
(Solutions based entirely on graphical or numerical methods are not acceptable.)
[8]
So equation of normal is (y(cid:16)2)(cid:32) x(cid:16) or 4y = 2x+7
(cid:168) (cid:184)
Answer
Marks
Guidance
2(cid:169) 2(cid:185)
A1
1.1b
Eliminates y between y (cid:32) 1 x(cid:14)ln(2x)and their normal equation to
2
Answer
Marks
Guidance
give an equation in x
M1
3.1a
7 1 7
Solves their ln2x (cid:32) so x = e4
Answer
Marks
Guidance
4 2
M1
1.1b
Substitutes to give value for y
M1
1.1b
(cid:167)1 7 1 7 7(cid:183)
Point Q is (cid:168) e4, e4 (cid:14) (cid:184)
2 4 4
Answer
Marks
Guidance
(cid:169) (cid:185)
A1
1.1b
(8 marks)
Notes:
M1: Differentiates correctly
1
M1: Substitutes x = to find gradient (may make a slip)
2
M1: Uses negative reciprocal gradient
A1: Correct equation for normal
M1: Attempts to eliminate y to find an equation in x
M1: Attempts to solve their equation using exp
M1: Uses their x value to find y
A1: Any correct exact form
Answer
Marks
Guidance
Question
Scheme
Marks
Question 15:
15 | dy
Finds (cid:32)8x(cid:16)6
dx | M1 | 3.1a
Gradient of curve at P is –2 | M1 | 1.1b
1 1
Normal gradient is - =
m 2 | M1 | 1.1b
1(cid:167) 1(cid:183)
So equation of normal is (y(cid:16)2)(cid:32) x(cid:16) or 4y = 2x+7
(cid:168) (cid:184)
2(cid:169) 2(cid:185) | A1 | 1.1b
Eliminates y between y (cid:32) 1 x(cid:14)ln(2x)and their normal equation to
2
give an equation in x | M1 | 3.1a
7 1 7
Solves their ln2x (cid:32) so x = e4
4 2 | M1 | 1.1b
Substitutes to give value for y | M1 | 1.1b
(cid:167)1 7 1 7 7(cid:183)
Point Q is (cid:168) e4, e4 (cid:14) (cid:184)
2 4 4
(cid:169) (cid:185) | A1 | 1.1b
(8 marks)
Notes:
M1: Differentiates correctly
1
M1: Substitutes x = to find gradient (may make a slip)
2
M1: Uses negative reciprocal gradient
A1: Correct equation for normal
M1: Attempts to eliminate y to find an equation in x
M1: Attempts to solve their equation using exp
M1: Uses their x value to find y
A1: Any correct exact form
Question | Scheme | Marks | AOs
\includegraphics{figure_3}
The curve $C_1$, shown in Figure 3, has equation $y = 4x^2 - 6x + 4$.
The point $P\left(\frac{1}{2}, 2\right)$ lies on $C_1$
The curve $C_2$, also shown in Figure 3, has equation $y = \frac{1}{2}x + \ln(2x)$.
The normal to $C_1$ at the point $P$ meets $C_2$ at the point $Q$.
Find the exact coordinates of $Q$.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
[8]
\hfill \mbox{\textit{Edexcel AS Paper 1 Q15 [8]}}