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MF26


Question 1b
Question 1b

Differentiate x=e^t with respect to t.
\frac{dx}{dt} = e^t
Substitute x = e^t into given differential equation,
\Rightarrow e^t \frac{dy}{dt} = \mathrm{cos}(\mathrm{ln}e^t)\mathrm{sin}(\mathrm{ln}(e^t)^3)
e^t \frac{dy}{dt} = \mathrm{cos}t \mathrm{sin} (3t)
Since \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}
\frac{dy}{dt} = \frac{1}{e^t} \mathrm{cos}t \mathrm{sin} (3t) \times e^t
\frac{dy}{dt} = \mathrm{cos}t \mathrm{sin} (3t)
\int \frac{dy}{dt} dt = \int \mathrm{cos}t \mathrm{sin} (3t) dy
Using the product to sum formula as shown here, we have
\int \frac{dy}{dt} dt = \frac{1}{2} \int \mathrm{sin} 4t + \mathrm{sin} 2t dt
y = \frac{1}{2}(-\frac{\mathrm{cos}4t}{4} - \frac{\mathrm{cos}2t}{2}) + C
y = \frac{1}{2}(-\frac{\mathrm{cos}(4 \mathrm{ln}x)}{4} - \frac{\mathrm{cos}(2\mathrm{ln}x)}{2}) + C


Question 2(iv)
Question 2(iv)

Note: i = \begin{bmatrix}1\\ 0\\ 0\end{bmatrix} and j = \begin{bmatrix}\\ 1\\ 0\end{bmatrix}
\pi: 2x + 3y = -6 — (1)
\pi \prime: x + y + z= 5 — (2)
\pi_1: x = 0 — (3)
Using GC, \vec{OA} = \begin{bmatrix}0\\ -2\\ 7\end{bmatrix}

\pi: 2x + 3y = -6 — (1)
\pi \prime: x + y + z= 5 — (2)
\pi_1: y = 0 — (3)
Using GC, \vec{OA} = \begin{bmatrix}-3\\ 0\\ 8\end{bmatrix}


Qn 11
Qn 11

Qn11
(i)
l: r = \begin{bmatrix}-1\\ -1\\ 0\end{bmatrix} + \alpha \begin{bmatrix}1\\ 1\\ 1\end{bmatrix}, \alpha \in \mathbb{R}
(ii) Since l is the common line of intersection on \pi_1 and \pi_2, we need l to be on \pi_3 too. For that to happen,
1. l must be parallel to \pi_3, that is, direction of l is perpendicular to normal to \pi_3
\Rightarrow \begin{bmatrix}1\\ 1\\ 1\end{bmatrix} \bullet \begin{bmatrix}5\\ {\lambda}\\ 17\end{bmatrix} = 0
\lambda = -22
2. Given that l is parallel to \pi_3 (since \lambda = -22), we need l to be on \pi_3, so we need \begin{bmatrix}-1\\ -1\\ 0\end{bmatrix} to be on \pi_3
\Rightarrow \begin{bmatrix}-1\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}5\\ -22\\ 17\end{bmatrix} = \mu
\mu = 17
(iii)
For 3 planes to have nothing in common, then l must be parallel to \pi_3 (Note: if l is not parallel, l will cut \pi_3 at a point, which means that 3 planes will cut at a point)
\Rightarrow \lambda = -22 from (ii)
But \mu \neq 17, \mu \in \mathbb{R}


Qn12
Qn12

12.
(i)
\pi_1 : r = \begin{bmatrix}4\\ 2\\ 12\end{bmatrix} + \lambda \begin{bmatrix}2\\ 2\\ 1\end{bmatrix} + \mu \begin{bmatrix}5\\ -1\\ -5\end{bmatrix}, \mu , \lambda \in \mathbb{R}
\vec{OP} = \begin{bmatrix}4\\ 2\\ 12\end{bmatrix} is on \pi_1 when \lambda = \mu = 0

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