A2 June 2025 Q6
6.
| Scheme | Marks | AO |
|---|---|---|
| \(\left\{\lim\limits_{x \to 0}\right\}\dfrac{1 - \cos 7x}{x\sin 9x} = \left\{\lim\limits_{x \to 0}\right\}\dfrac{A\sin 7x}{\sin 9x + Bx\cos 9x}\) | M1 | 1.1b |
| \(\left\{\lim\limits_{x \to 0}\right\}\dfrac{1 - \cos 7x}{x\sin 9x} = \left\{\lim\limits_{x \to 0}\right\}\dfrac{7\sin 7x}{\sin 9x + 9x\cos 9x}\) | A1 | 1.1b |
| \(\left\{\lim\limits_{x \to 0}\dfrac{7\sin 7x}{\sin 9x + 9x\cos 9x} =\right\}\left\{\lim\limits_{x \to 0}\right\}\dfrac{49\cos 7x}{9\cos 9x + 9\cos 9x - 81x\sin 9x}\) | M1 | 3.1a |
| \(\lim\limits_{x \to 0}\dfrac{49\cos 7x}{9\cos 9x + 9\cos 9x - 81x\sin 9x} = \dfrac{49}{9 + 9} = \dfrac{49}{18}\ *\) | A1* | 1.1b |
| (4) |
Notes
M1: Differentiates numerator and denominator to the correct form, may be done separately
A1: Correct derivatives
M1: Recognises the requirement to differentiate numerator and denominator again and achieves the correct form of the derivatives may be seen separately or as a fraction \(\dfrac{A\cos 7x}{B\cos 9x + C\cos 9x + Dx\sin 9x}\)
A1*: Completes the proof, including correct limit notation seen when \(x = 0\) is substituted with as a minimum \(\dfrac{49}{9 + 9} = \dfrac{49}{18}\) or \(\lim\limits_{x \to 0}\dfrac{49\cos 7x}{18\cos 9x - 81x\sin 9x} = \dfrac{49}{18}\)
| Scheme | Marks | AO |
|---|---|---|
| \(u = x^2 \Rightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x} = 2x, \dfrac{\mathrm{d}^2 u}{\mathrm{d}x^2} = 2, \left\{\dfrac{\mathrm{d}^3 u}{\mathrm{d}x^3} = 0\right\}\) | M1 | 1.1b |
| \(v = \mathrm{e}^{3x} \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}x} = 3\mathrm{e}^{3x}, \dfrac{\mathrm{d}^2 v}{\mathrm{d}x^2} = 9\mathrm{e}^{3x}, \dfrac{\mathrm{d}^3 v}{\mathrm{d}x^3} = 27\mathrm{e}^{3x}, \ldots, \left\{\dfrac{\mathrm{d}^k v}{\mathrm{d}x^k} = 3^k\mathrm{e}^{3x}\right\}\) | M1 | 2.1 |
| \(\dfrac{\mathrm{d}^{k}y}{\mathrm{d}x^{k}} = 3^k\mathrm{e}^{3x}x^2 + k \times 2x \times 3^{k-1}\mathrm{e}^{3x} + \dfrac{k(k - 1)}{2} \times 2 \times 3^{k-2}\mathrm{e}^{3x}\) | M1 A1 | 3.1a 1.1b |
| \(= 3^{k-2}\mathrm{e}^{3x}\left(9x^2 + 6kx + k(k - 1)\right)\) | A1 | 2.2a |
| (5) |
Notes
M1: Differentiates \(u = x^2\) twice, may be seen as part of the expression for the \(k\)th derivative
M1: Uses \(v = \mathrm{e}^{3x}\) to establish the form of the derivatives \(\dfrac{\mathrm{d}^k v}{\mathrm{d}x^k} = 3^k\mathrm{e}^{3x}\). Look for multiples of \(\mathrm{e}^{3x}\) increasing by a factor of 3 each time, need at least three. This may be seen as part of the expression for the \(k\)th derivative
M1: A correct strategy for the \(k\)th derivative. This requires the correct derivative of \(x^2\) combined with the correct derivative of \(\mathrm{e}^{3x}\) in terms of \(k\) together with the correct binomial coefficient. Condone using \(k(k - 1)\) for the third term
A1: A correct unsimplified expression
A1: Correct expression in the required form with correct values of \(A\), \(B\) and \(C\). Condone missing trailing bracket
(NB \(A = 9\), \(B = 6\), \(C = 1\))
Repeated differentiation to find at least the third derivative
\(y = x^2\mathrm{e}^{3x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\mathrm{e}^{3x} + 3x^2\mathrm{e}^{3x} \Rightarrow \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 2\mathrm{e}^{3x} + 6x\mathrm{e}^{3x} + 6x\mathrm{e}^{3x} + 9x^2\mathrm{e}^{3x}\)
\(\Rightarrow \dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = 6\mathrm{e}^{3x} + 12\mathrm{e}^{3x} + 36x\mathrm{e}^{3x} + 18x\mathrm{e}^{3x} + 27x^2\mathrm{e}^{3x}\)
they can score the first M1M1 for differentiating \(u\) twice and \(v\) three times. Then will score M0A0A0.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}^{9}y}{\mathrm{d}x^{9}} = 3^{9-2}\mathrm{e}^{3x}\left(9x^2 + 6 \times 9x + 9 \times 8\right) = 0\) \(\Rightarrow 9x^2 + 54x + 72 = 0 \Rightarrow x = \ldots\) | M1 | 1.1b |
| \(x = -4, -2\) cso | A1 | 2.2a |
| (2) | ||
| (11 marks) |
Notes
M1: Substitutes \(k = 9\) into their result from part (a) and solves the resulting 3TQ which leads to real roots
A1: These values only following full marks in (a)