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(1+kx)^7 =\sum _{i=0} ^7 \binom 7 i (kx)^i\\ \\ \therefore \binom 7 2 k^2 - \binom 7 1 k = \binom 7 3 k^3 - \binom 7 2 k^2 \\ \\ \implies 2 \times \frac {7 \times 6}{1 \times 2} k^2 = \frac {7\times 6 \times 5}{1 \times 2 \times 3} k^3+ 7k \\ \\ \implies 5k^3 -6k^2 + k = 0 \\ \\ \implies...

High school graduates or University graduates??

Congratulations to Bored of Studies on its 20th Birthday. And thanks to its founders. How time passes; I've been a member of the Bored of Studies for over 13 years now, enjoying making whatever little contribution I could along the way. And sadly, I'm now an octo!

Yes, I still take students, but only a few.

3. \frac {dy}{dx} = e^{2y}\\ \\ \therefore \frac {dy}{e^{2y}} = dx \\ \\ \therefore \int dx = \int e^{-2y} dy \\ \\ \therefore x = -\frac {1}{2}e^{-2y} + C

2. \frac {dy}{dx} = 1 - \frac {y}{3} = \frac {3 - y}{3}\\ \\ \therefore \frac {dy}{3-y} = \frac {dx}{3}\\ \\ \therefore \int \cdots dy = \int \cdots dx \\ \\ \therefore -ln |3-y| = \frac {x}{3} + C \\ \\ x = 2 \implies y = 4 \implies -ln|3 - 4| = \frac {2}{3} + C \implies C = -\frac {2}{3} \\...

Question too bloody long!

You can use the "Terry Lee" method to find the square root of -60-32i. Here's how: \sqrt{-32i - 60} = 2\sqrt{-8i -15}\\ \\ \sqrt{-8i - 15} = \sqrt{1 - 8i -16} = \sqrt{1^2 + 2\times 1 \times ({-4i}) +(-4i)^2} \\ \\ \\ $Remember: $ (a + b)^2 = a^2 + 2 \times a \times b + b^2 = (a + b)^2 \\ \\...

Maybe I'll show you how to do RCR later.

\int ^1 _0 \frac {\sqrt x}{1 + x} dx = \int ^1 _0 \frac {\sqrt x}{1+(\sqrt x)^2} \times 2\sqrt x d\sqrt x\\ \\ =2\int^1 _0 (1 - \frac {1}{1 + (\sqrt x)^2} )d \sqrt x \\ \\ =2 \left[\sqrt x - tan^{-1} \sqrt x \right ]^1 _0 \\ \\ = 2[(1 - \frac {\pi}{4} ) -(0 - 0)] = 2 - \frac {\pi}{2} \\ \\...

No.

Integrands which are derivatives(or constant multiples thereof) of composite functions are the ones usually handled by "integration by substitution". They often refer to reverse chain rule - but I've seldom seen how this reversal is achieved. Using my method, I can do the reversal of the chain...

Motion, in general, does not need to be in a straight line. In your Adv & Ext 1 Maths, you are usually dealing with motion in a straight line, because this is the easiest to deal with. Motion in a straight line is often referred to as "rectilinear motion".

\frac {dy}{dx} = -3\frac {d}{dx}(x^4-2x^2-1)^{-1} = -3\frac{d(x^4-2x^2-1)^{-1}}{d(x^4-2x^2-1)} \times \frac {d(x^4-2x^2-1)}{dx}\\ \\ = -3\times {-1}(x^4-2x^2-1)^{-2} \times (4x^3-4x) \\ \\ = \frac{3\times 4x(x-1)(x+1)}{(x^4-x^2-1)^2}\\ \\ = 0 $ at x $=0, -1,1\\ \\ y(-1) = \frac {-3}{1-2-1} =...

Again, if not required to use the substitution: \int ^0 _{-3} \frac {x}{\sqrt {1-x} }dx = -\int ^0 _{-3}\frac {1-x - 1}{\sqrt {1-x} }dx\\ \\ = \int ^0 _{-3} (1-x)^{0.5} d(1-x)-\int ^0 _{-3}(1-x)^{-0.5}d(1-x)\\ \\ = \left[ \frac{2}{3}(1-x)^{1.5} -2(1-x)^{0.5} \right]^0 _{-3} \\ \\ =-(\frac...

Avoiding formal substitution: \int ^1 _0 x\sqrt{1-x^2} dx = -\frac {1}{2}\int ^1 _0 (1-x^2)^{0.5} d(1-x^2) \\ \\ =-\frac{1}{2} \left[\frac {(1-x^2)^{1.5}}{1.5} \right]^1 _0 = -\frac {1}{2}\left [ (\frac{2}{3} \times(1-1)^{1.5}) -(\frac {2}{3}(1-0)^{1.5})\right ]\\ \\ = -\frac{1}{2} \times...

a) \int y\sqrt{y+1}]dy = \int (y+1 - 1)\sqrt {y+1} d(y+1) \\ \\ = \int ((y+1)^{\frac {3}{2}} - (y+1)^\frac{1}{2}) d(y+1) = \frac{2}{5}(y+1)^\frac{5}{2} - \frac{2}{3}(y+1)^\frac {3}{2} + C c) \int \frac {x}{\sqrt{2x-1}} dx = \frac {1}{2}\times \frac {1}{2}\int \frac {(2x-1 + 1)}{\sqrt{2x-1}}...

Try to find out why you were not doing better? What areas were you having difficulties in your MX1; can you do better with more effort? Put in extra effort and learn from any "mistakes" you may have made in your 1st test. Now see how you go in your next test and see how you go, before deciding...

You need to provide more details of what happened. Which new topics were you experiencing problems with. When you got lower marks, why were you getting lower marks? etc etc. Without more background info, we can only guess what or where your difficulty lies.

Above from an old Signpost Year 8 Book. Say we are given a parallelogram ABCD. e start with the definition of the parallelogram, in this case only that it is a quadrilateral with opposite sides, viz AB & DC and AD & BC being parallel. That's all we are given. How do you prove that: i) AB = DC...