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These are well-established conventions. But I don't know if you'd be penalised for their use in your HSC exams. If I were a marker, I'd accept it. But it is possible some markers are themselves not familiar with this shorthand and be confused; but most "smart enough" markers should be able to...

(1 + 1)^2n has 2n+1 (odd number!) of terms. The expression 2nC0 + 2nC1 + . . . + 2nCn has n+1 terms. By adding an additional 2nCn to the remaining expression, I make it equal to the 1st half. I did not notice this at the beginning and that's why I thought the identity was incorrect at first.

Proof: 2^ {2n} = (1+1)^{2n} = \left [ \binom {2n} 0 + \binom {2n} 1 + \binom {2n} 2 + \cdots \binom {2n} n \right ] \\ \\ + \left [\binom {2n} n + \binom {2n} {n+1} + \binom {2n} {n+2} + \cdots + \binom {2n}{2n} \right] - \binom {2n} n \\ \\ = 2 \times \left [ \binom {2n} 0 + \binom {2n} 1 +...

Q22 P(x) = ax^{n+1} + bx^n + 1\\ \\ P'(x) = (n+1)ax^n + nbx^{n-1}\\ \\ \therefore P(1) = a + b + 1 = 0 \implies a+ b = -1 \\ \\ P'(1) = (n+1)a + nb = 0 \implies n(a+b) + a = 0 \\ \\ \implies -1 \times n + a = 0 \implies a = n \\ \\ \therefore a+b+1 = 0 \implies n+b+1 = 0 \implies b = -(n+1)

Don't forget, sometimes, the base case maybe for, say, n = 3 instead of 1.

Will they allow you to take 1141 with just your MX1? If so, I can help you learn 1141 contents during your December/January break after your HSC, so that you won't be badly handicapped. But you must be pretty good in your MX1.

Nicely set out. Your 'x' looks like an 'n'. Computationally, easier this way. \therefore \binom {15} {10} 5^5 (2x)^{10} = \binom {15}{11}5^4 (2x)^{11}\\ \\ \therefore \frac {(2x)^{11}}{(2x)^{10}} = \frac {\binom {15}{10} 5^5}{\binom {15}{11}5^4} \\ \\ \therefore 2x = \frac {15! \times 11!4...

z^5 + a = 0 is a polynomial equation (of degree 5) whose roots are the points on the circle. Since these points are symmetric about the x-axis, it means every (of the 4) complex roots appear in complex-conjugate pairs. Therefore the coeffs of this polynomial eqn must be all reals; therefore 'a'...

i.e |z - 0| = |z -(1+i)| It means the distances of (all) z from the points O(0,0) and from A(1,1) are equal. So the equation defines the locus of a point which is equidistant from these 2 given points O & A, viz. the perpendicular bisector of the line segment OA. Draw this perpendicular...

(1 + xy+y^2)^n = \sum _{i=0}^n \binom n i (1+xy)^{n-i} (y^2)^i =\sum _{i=0} ^n \binom n i y^{2i}\left[1 + \binom {n-i} 1 (xy)^1 + \binom {n-i} 2 (xy)^2 + \binom {n-i} 3 (xy)^3 + \cdots\right ] Only term in x^3y^5 comes from, where i=1: \binom n 1 y^2 \times \binom {n-1} 3 (xy)^3 $ whose...

Wow. Not straightforward. For only 1 mark? Will post solution later, if I have the time. What puts me off is the amount of LaTeX input required.

No solutions because: The smallest each of the terms: e^{sin^2} $ and $ e^{cos^2} can be is 1. Therefore the sum of these 2 terms cannot be less than 2. But this sum equals 2 only when each term equals 1 simultaneously. Each term equals 1 only when sin^2 = 0 $ or $ cos^2 = 0; but this...

My place is only 10 minutes from yours.

Write (not type) it down. Do it again and again; each time writing down as much as possible by recall, until you can recall the whole essay.

This is 2U Maths. \frac {cos^2\theta - sin^2 \theta}{cos^2 \theta + cos\theta sin\theta} = \frac {(cos \theta+sin\theta)(cos\theta - sin\theta)}{cos\theta (cos \theta + sin \theta)} \\ \\ = \frac {cos\theta}{cos\theta} - \frac {sin\theta}{cos\theta} = 1 - tan\theta

Or (not shorter): \int \frac{e^x}{e^x + e^{-x}} dx = \int \frac {e^x \cdot e^x}{e^{2x} + 1} dx = \frac {1}{2} \int \frac {1}{e^{2x} + 1} d(e^{2x} + 1) = \frac {1}{2} ln(e^{2x} + 1) + C \\ \\ \therefore \int ^a _{-a} \frac {e^x}{e^x + e^{-x}}dx = \frac {1}{2}\left [ ln(e^{2x} + 1)\right ]^a...

You make study notes for your own needs. Different students should have different study notes, tailored to the needs of each. I suggest you prepare your notes by topics. For each, write down materials/items that are likely to be important to the course; include solutions of special relevance...

Q12 Following same approach as for Q19: \overrightarrow {OD} = \overrightarrow {OA} + \overrightarrow {OC} - \overrightarrow {OB}\\ \\ = [3,-8,-2] + [-2,-2,1] - [2,4,5] \\ \\ = [-1,-14,-6](the position vector of point D)

Q19 Don't know how to write a column matrix. So will use [a,b,c] instead. For case A-B-C: \overrightarrow {BD} = \overrightarrow {BA} + \overrightarrow {BC}\\ \\ \therefore \overrightarrow {OD} - \overrightarrow {OB} = (\overrightarrow{OA}-\overrightarrow{OB}) + (\overrightarrow {OC} -...

Q19 If A, B, C are supposed to be 3 consecutive vertices, in cyclic order, (adjacent sides BA and BC), then there is only one D. In cyclic order, B-A-C (adjacent sides AB & AC) will generate another D and finally A-C-B (adjacent sides CA & CB) a 3rd D. A-B-C and C-B-A are treated as the...