| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2018 |
| Session | September |
| Marks | 10 |
| Topic | Differential equations |
| Type | Separable variables - standard (requires integration by parts) |
| Difficulty | Challenging +1.2 This is a separable differential equations question requiring integration of trigonometric functions and use of a boundary condition. While it involves multiple steps (separation, integration with substitution for both sides, applying initial condition, solving for specific value), the techniques are standard A-level methods. The trigonometric simplification (2cos²4y - 1 = cos8y) and integrations are straightforward once separated, making this moderately above average but not requiring novel insight. |
| Spec | 1.07s Parametric and implicit differentiation1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(\int(2\cos^2 4y - 1)dy = \int x^2 \sin 2x dx\) | M1 |
| \(2\cos^2 4y - 1 = \cos 8y\) | M1 | Attempt use of double angle formula (Obtain \(\pm \cos 8y\)) |
| \(\int \cos 8y dy = \frac{1}{8}\sin 8y + c_1\) | A1 | Obtain correct integral (Condone no \(+c_1\)) |
| \(\int x^2 \sin 2x dx = -\frac{1}{2}x^2 \cos 2x + \int x \cos 2x dx\) | M1 | Attempt integration by parts once |
| \(= -\frac{1}{2}x^2 \cos 2x + \frac{1}{2}x \sin 2x - \int \frac{1}{2}\sin 2x dx\) | M1 | Attempt second integration by parts |
| \(= -\frac{1}{2}x^2 \cos 2x + \frac{1}{2}x \sin 2x + \frac{1}{4}\cos 2x + c_2\) | A1 | Obtain correct integral (Condone no \(+c_2\)) |
| \(\frac{1}{8}\sin 8y = -\frac{1}{2}x^2 \cos 2x + \frac{1}{2}x \sin 2x + \frac{1}{4}\cos 2x + c\) |
Total: [6]
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | \(\sin\frac{8}{12}\pi = -4(\frac{1}{6})\cos\frac{1}{6} - 4(\frac{1}{12})\sin\frac{1}{6} - 2\cos\frac{1}{6} - c\) | M1 |
| \(c = \frac{1}{2}\sqrt{3} - \pi\) | A1 | Obtain correct value for \(c\) for their correct equation (Condone decimal equiv (– 2.276) if fractions not yet cleared, or \(c = \pm \frac{1}{16}\sqrt{3}\pi - \frac{8}{8}\) if fractions not yet cleared) |
| \(\sin 8y = 2 + \frac{1}{2}\sqrt{3} - 8y = (-0.279), 3.421\) | M1 | Attempt positive value for \(y\) when \(x = 0\) |
| \(y = (-0.035), 0.428\) | A1 | Obtain correct value for \(y\) (A0 if extra values) |
| \(y = 0.428\) |
Total: [4]
(i) | $\int(2\cos^2 4y - 1)dy = \int x^2 \sin 2x dx$ | M1 | Separate variables
| $2\cos^2 4y - 1 = \cos 8y$ | M1 | Attempt use of double angle formula (Obtain $\pm \cos 8y$)
| $\int \cos 8y dy = \frac{1}{8}\sin 8y + c_1$ | A1 | Obtain correct integral (Condone no $+c_1$)
| $\int x^2 \sin 2x dx = -\frac{1}{2}x^2 \cos 2x + \int x \cos 2x dx$ | M1 | Attempt integration by parts once
| $= -\frac{1}{2}x^2 \cos 2x + \frac{1}{2}x \sin 2x - \int \frac{1}{2}\sin 2x dx$ | M1 | Attempt second integration by parts
| $= -\frac{1}{2}x^2 \cos 2x + \frac{1}{2}x \sin 2x + \frac{1}{4}\cos 2x + c_2$ | A1 | Obtain correct integral (Condone no $+c_2$)
| $\frac{1}{8}\sin 8y = -\frac{1}{2}x^2 \cos 2x + \frac{1}{2}x \sin 2x + \frac{1}{4}\cos 2x + c$ | | |
Total: [6]
(ii) | $\sin\frac{8}{12}\pi = -4(\frac{1}{6})\cos\frac{1}{6} - 4(\frac{1}{12})\sin\frac{1}{6} - 2\cos\frac{1}{6} - c$ | M1 | Attempt $c$, using $x = \frac{1}{6}\pi$, $y = \frac{1}{12}\pi$
| $c = \frac{1}{2}\sqrt{3} - \pi$ | A1 | Obtain correct value for $c$ for their correct equation (Condone decimal equiv (– 2.276) if fractions not yet cleared, or $c = \pm \frac{1}{16}\sqrt{3}\pi - \frac{8}{8}$ if fractions not yet cleared)
| $\sin 8y = 2 + \frac{1}{2}\sqrt{3} - 8y = (-0.279), 3.421$ | M1 | Attempt positive value for $y$ when $x = 0$
| $y = (-0.035), 0.428$ | A1 | Obtain correct value for $y$ (A0 if extra values)
| $y = 0.428$ | | |
Total: [4]12 The gradient function of a curve is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 2 } \sin 2 x } { 2 \cos ^ { 2 } 4 y - 1 }$.\\
(i) Find an equation for the curve in the form $\mathrm { f } ( y ) = g ( x )$.
The curve passes through the point $\left( \frac { 1 } { 4 } \pi , \frac { 1 } { 12 } \pi \right)$.\\
(ii) Find the smallest positive value of $y$ for which $x = 0$.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR H240/01 2018 Q12 [10]}}