June 2023 Paper 1 Q9

OCR MEICurrent spec10 marksIntegration Logs & Exponentials

9 The gradient of a curve is given by \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{x} - 4\mathrm{e}^{-x}\).

(a) Show that the \(x\)-coordinate of any point on the curve at which the gradient is 3 satisfies the equation \(\left(\mathrm{e}^{x}\right)^2 - 3\mathrm{e}^{x} - 4 = 0\). [2]

(b) Hence show that there is only one point on the curve at which the gradient is 3, stating the exact value of its \(x\)-coordinate. [3]

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{x} - 4\mathrm{e}^{-x} = 3\)	M1	2.1
\(\left(\mathrm{e}^{x}\right)^2 - 3\mathrm{e}^{x} - 4 = 0\)	E1	2.1
	[2]

Notes

M1: Equate to 3

E1: AG Rearrange to quadratic in \(\mathrm{e}^{x}\) Also allow \(\mathrm{e}^{2x} - 3\mathrm{e}^{x} - 4 = 0\)
Expression \(= 0\) must be seen

Mark scheme (b)

Scheme	Marks	AO
Solve to give \(\mathrm{e}^{x} = [-1],\ 4\)	M1	2.1
When \(\mathrm{e}^{x} = 4,\ x = \ln 4\)	A1	2.1
When \(\mathrm{e}^{x} = -1\) there are no real values of \(x\), so no other points on the curve.	E1	2.1
	[3]

Notes

M1: May be BC giving at least one root of the quadratic equation (a)

A1: \(x = \ln 4\) must be seen explicitly

E1: must explain why they reject the value \(-1\) for \(\mathrm{e}^{x}\), or state \(\mathrm{e}^{x} + 1\) is never zero

Mark scheme (c)

Scheme	Marks	AO
Equation \(y = \displaystyle\int \left(\mathrm{e}^{x} - 4\mathrm{e}^{-x}\right)\mathrm{d}x\) \([y =]\ \mathrm{e}^{x} + 4\mathrm{e}^{-x}\ [+c]\)	B1	3.1a
When \(x = 0,\ y = 0 = 1 + 4 + c\)	M1	3.1a
So \(c = -5\) \([y = \mathrm{e}^{x} + 4\mathrm{e}^{-x} - 5]\)	A1	1.1b
When \(x = 1,\ y = \mathrm{e}^{1} + 4\mathrm{e}^{-1} - 5\)	M1	2.1
\(y = -0.810 < 0\) so below the \(x\)-axis	E1	2.1
	[5]

Notes

B1: Condone missing \(+c\) in their integral

M1: Attempt to evaluate \(c\)

M1: Substituting \(x = 1\) into their expression

E1: AG must argue below the axis from correct \(y\) value. Must be clear that \(-0.81\) is a \(y\)-coordinate