A2 June 2024 Paper 1 Q7

Notes

M1: Use correct definitions of \(\cosh u\) and \(\sinh u\) in terms of exponentials. Condone omission of \(2\sinh u\cosh u\) for this mark. (Note \(2\sinh u\cosh u\) or \(LHS\) must be included for subsequent A1.) M1 can be implied for stating that \(2\sinh u\cosh u \equiv \frac{1}{2}(\mathrm{e}^{u} - \mathrm{e}^{-u})(\mathrm{e}^u + \mathrm{e}^{-u})\) (but not for the A mark). (Corrected from the printed mark scheme: the first bracket is printed as \((\mathrm{e}^{-u} - \mathrm{e}^{-u})\).)

A1: Complete proof with no errors. Allow \(LHS\) and \(RHS\) in place of \(2\sinh u\cosh u\) and \(\sinh 2u\) respectively. Minimally acceptable proof is of the form \(RHS \equiv 2\left(\dfrac{\mathrm{e}^u - \mathrm{e}^{-u}}{2}\right)\left(\dfrac{\mathrm{e}^u + \mathrm{e}^{-u}}{2}\right) \equiv \dfrac{\mathrm{e}^{2u} - \mathrm{e}^{-2u}}{2} \equiv LHS\) in either direction.
Do not condone sin for sinh or cos for cosh or \(x\) for \(u\) (unless “let \(x = u\)” stated) for A1.
If candidate shows that both \(LHS\) and \(RHS\) are equal to some third expression, then a conclusion must be clearly stated for the A1 mark.
SC B1 If candidate works with given identity and concludes, for example, that 1 = 1.

Alternative method (working left to right)

M1: Use definition of \(\sinh 2u\) in terms of exponentials. Condone omission of \(\sinh 2u\) for this mark. (Note \(\sinh 2u\) or \(RHS\) must be included for subsequent A1).

A1: Complete proof must see \(2 \times \dfrac{1}{2}(\mathrm{e}^u - \mathrm{e}^{-u}) \times \dfrac{1}{2}(\mathrm{e}^u + \mathrm{e}^{-u})\) before stating correct \(RHS\).

Notes

M1*: Differentiating to get \(\pm 16\sinh x \pm k\cosh 2x\) for some \(k \neq 0\). May be in exponential form \(\frac{\mathrm{d}y}{\mathrm{d}x} = 8\left(\mathrm{e}^x - \mathrm{e}^{-x}\right) - (\mathrm{e}^{2x} + \mathrm{e}^{-2x})\) - if using exponential form then the derivative must be of the form \(\frac{\mathrm{d}y}{\mathrm{d}x} = \pm 8\left(\mathrm{e}^x - \mathrm{e}^{-x}\right) \pm k(\mathrm{e}^{2x} + \mathrm{e}^{-2x})\) for some \(k \neq 0\).

A1: Correct second derivative, condone un-simplified. May be in exponential form. \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(\mathrm{e}^x + \mathrm{e}^{-x}) - 2(\mathrm{e}^{2x} - \mathrm{e}^{-2x})\) (oe).

M1dep*: Correct use of the result from part (a) in their second derivative or their \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(\mathrm{e}^x + \mathrm{e}^{-x}) - 2(\mathrm{e}^x - \mathrm{e}^{-x})(\mathrm{e}^x + \mathrm{e}^{-x})\) using difference of two squares (oe e.g. setting equal to zero and converting to a quartic in \(\mathrm{e}^x\) which if correct is \(\mathrm{e}^{4x} - 4\mathrm{e}^{3x} - 4\mathrm{e}^x - 1 = 0\)). If converting to exponentials at any point, then correct exponential expression(s) for sinh and cosh must be used.
Or in exponential form \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2(\mathrm{e}^x + \mathrm{e}^{-x})(4 - \mathrm{e}^x + \mathrm{e}^{-x})\)

A1: Must state \(\cosh x \gt 0\) or \(\cosh x \geqslant 1\) or show that there is no solution of the equation \(\cosh x = 0\) (oe e.g. ‘cosh cannot be 0’, ‘cosh is at least 1’, ‘cosh has a minimum at (0, 1)’, ‘\(\cosh^{-1}\) only valid for \(x \geqslant 1\)’ but not just ‘cosh \(x = 0\) has no solutions’) - allow stating that the only solution satisfies the equation \(\sinh x = 2\). If using exponentials then must show clearly that there is only one solution of the correct equation (which for reference is \(\ln(2 + \sqrt{5})\)).

Alternative method for second and third marks

M1dep*: Use of \(\cosh 2x = \cosh^2 x + \sinh^2 x\) in their first derivative. \(\frac{\mathrm{d}y}{\mathrm{d}x} = 8(\mathrm{e}^x - \mathrm{e}^{-x}) - \frac{1}{2}(\mathrm{e}^x + \mathrm{e}^{-x})^2 - \frac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x})^2\)

A1: Correct second derivative, condone un-simplified. May use exponential form: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(\mathrm{e}^x + \mathrm{e}^{-x}) - (\mathrm{e}^x + \mathrm{e}^{-x})(\mathrm{e}^x - \mathrm{e}^{-x}) - (\mathrm{e}^x - \mathrm{e}^{-x})(\mathrm{e}^x + \mathrm{e}^{-x})\)

Alternative method for first three marks

B1: Using the result from part (a). Or \(y = 8(\mathrm{e}^x + \mathrm{e}^{-x}) - 0.5(\mathrm{e}^x - \mathrm{e}^{-x})(\mathrm{e}^x + \mathrm{e}^{-x})\)

M1: Differentiating to get \(\pm 16\sinh x \pm k\cosh^2 x \pm k\sinh^2 x\) or \(\frac{\mathrm{d}y}{\mathrm{d}x} = \pm 8(\mathrm{e}^x - \mathrm{e}^{-x}) \pm k(\mathrm{e}^x + \mathrm{e}^{-x})^2 \pm k(\mathrm{e}^x - \mathrm{e}^{-x})^2\) for some \(k \neq 0\).

A1: Correct second derivative, condone un-simplified. May use exponential form: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(\mathrm{e}^x + \mathrm{e}^{-x}) - (\mathrm{e}^x + \mathrm{e}^{-x})(\mathrm{e}^x - \mathrm{e}^{-x}) - (\mathrm{e}^x - \mathrm{e}^{-x})(\mathrm{e}^x + \mathrm{e}^{-x})\)

Notes

B1: Condone \(\ln\left|2 + \sqrt{5}\right|\).

B1FT: Finding the value of \(\cosh x\) in a form not involving logs or exponentials following through their \(x\) of the form \(\ln(a + \sqrt{b})\) where \(a\) is non-zero and \(b \gt 0\) and \(a + \sqrt{b} \gt 0\) – allow un-simplified. The correct \(y\)-coordinate implies this and the next B mark. For reference: \(\cosh(\ln(a + \sqrt{b})) = \frac{1}{2}\left(a + \sqrt{b} + \dfrac{1}{a + \sqrt{b}}\right)\).
Or from using \(\cosh^2 x - \sinh^2 x \equiv 1\) with their value of \(\sinh x\) from part (b).

B1: Need not be stated as a coordinate. Accept any exact one term equivalent.