Edexcel IAL P4 Maths Revision: Pure Mathematics 4

\(a \cdot b = (x_1 \times x_2) + (y_1 \times y_2)\). M1
\(a \cdot b = (3 \times 4) + (-2 \times 5)\). M1A1
\(= 12 - 10 = 2\). A1

M1 for setting up the partial fraction decomposition: $\frac{A}{x-1} + \frac{B}{x+2}$.
M1 for setting up the algebraic identity: $A(x+2) + B(x-1) = x+5$.
M1 for substituting a strategic value, e.g., let $x = 1 \implies 3A = 6$.
A1 for finding $A = 2$.
A1 for substituting $x = -2 \implies -3B = 3$, giving $B = -1$, and writing the final expression $\frac{2}{x-1} - \frac{1}{x+2}$.

Velocity is the derivative of the position vector with respect to time. [M1]
\(\vec{v}(t) = \frac{d\vec{r}}{dt} = \begin{pmatrix} 3 \\ -1 \\ 4t \end{pmatrix}\) [A1][A1]
At \(t=2\), \(\vec{v}(2) = \begin{pmatrix} 3 \\ -1 \\ 8 \end{pmatrix} \text{ m/s}\). [A1]

Separate the variables by multiplying both sides by \(y\) and \(dx\): M1
\(y \, dy = x \, dx\).
Integrate both sides: \(\int y \, dy = \int x \, dx\). M1
\(\frac{y^2}{2} = \frac{x^2}{2} + C\). A1
Multiply entirely by 2 to isolate \(y^2\):
\(y^2 = x^2 + 2C\). (Note: \(2C\) can just be written as a new constant \(K\)). A1

1. M1 for assuming the opposite: $\sqrt{5}$ is rational, so it can be written as $\frac{a}{b}$ where $a$ and $b$ are integers with no common factors.
2. M1 for squaring both sides to get $5 = \frac{a^2}{b^2}$, leading to $a^2 = 5b^2$, and deducing that $a^2$ is a multiple of $5$, so $a$ must also be a multiple of $5$.
3. M1 for letting $a = 5k$, substituting to get $(5k)^2 = 5b^2 \implies 25k^2 = 5b^2 \implies b^2 = 5k^2$, and deducing $b$ must also be a multiple of $5$.
4. A1 for stating that both $a$ and $b$ being multiples of $5$ contradicts the assumption that they have no common factors, hence $\sqrt{5}$ must be irrational.

1. M1 for stating the formula with $n=-\frac{1}{2}$ and using $(-x)$.
2. M1 for calculating the $x^2$ coefficient: $\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-x)^2$.
3. M1 for calculating the $x^3$ coefficient: $\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3!}(-x)^3$.
4. A1 for the completely simplified answer: $1 + \frac{1}{2}x + \frac{3}{8}x^2 + \frac{5}{16}x^3$.

1. M1 for rearranging to $\sec t = \frac{x}{2}$ and $\tan t = \frac{y}{2}$.
2. M1 for stating the trigonometric identity $\sec^2 t - \tan^2 t = 1$.
3. M1 for substituting the expressions: $(\frac{x}{2})^2 - (\frac{y}{2})^2 = 1$.
4. A1 for multiplying by 4 to obtain $x^2 - y^2 = 4$.

1. M1 for differentiating to find $\frac{dx}{dt} = -2\sin t$ and $\frac{dy}{dt} = 3\cos t$, yielding $\frac{dy}{dx} = -\frac{3\cos t}{2\sin t} = -\frac{3}{2}\cot t$.
2. M1 for evaluating the gradient at $t = \frac{\pi}{4}$ to get $m = -\frac{3}{2}$.
3. M1 for finding the coordinates at $t = \frac{\pi}{4}$: $x = \sqrt{2}$, $y = \frac{3\sqrt{2}}{2}$ and substituting into $y - y_1 = m(x - x_1)$.
4. A1 for the correct equation, e.g., $y - \frac{3\sqrt{2}}{2} = -\frac{3}{2}(x - \sqrt{2})$ or $3x + 2y - 6\sqrt{2} = 0$.

1. M1 for differentiating to find $du = \cos x \, dx$.
2. M1 for correctly changing the limits: when $x=0, u=0$ and when $x=\frac{\pi}{2}, u=1$.
3. M1 for setting up and evaluating the new definite integral: $\int_{0}^{1} u^4 \, du = [\frac{1}{5}u^5]_0^1$.
4. A1 for concluding the exact value is $\frac{1}{5}$.