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Re: HSC 4U Integration Marathon 2017 $\noindent let $u=\frac{\pi}{2}-x \Rightarrow $ d$x = -$d$u.$ When $x=\frac{\pi}{2}, u=0.$ When $x=0, u=\frac{\pi}{2}.\\\therefore I= \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{\cos x+\sin x}$ d$x=-\int_{\frac{\pi}{2}}^{0}\frac{\cos (\frac{\pi}{2}-u)}{\cos...

Re: HSC 4U Integration Marathon 2017 $\noindent$ I= \int \frac{x^2-1 }{(x^2+1)\sqrt{1+x^4}}$ d$x=\int \frac{x^2-1}{x(x^2+1)\sqrt{x^2+\frac{1}{x^2}}}$ d$x \\ $ dividing the denominator and the numerator by $x$ we get $\\\\ I =\int \frac{x-\frac{1}{x}}{x(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}}$...

$\noindent$ $ It is known that if $ f'(x)\geq 0$ and $f(0)=0$, then $f(x)\geq 0$ for $x> 0.\\\\$ Show that $\sin x-x+\frac{x^3}{6}\geq 0$ for $x> 0

$\noindent cis($a$)+cis($b$)+cis($c$)=0$\\\Rightarrow (\cos a +\cos b+ \cos c)+i(\sin a+\sin b+ \sin c)=0\\ \Rightarrow \cos a +\cos b+ \cos c=0 $ and $\sin a+\sin b+ \sin c=0\\ \\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{cis(a)}+\frac{1}{cis(b)}+\frac{1}{cis(c)}\\ \\ =...

$\noindent Note that $(\sin \theta + \cos \theta )^2=1+2\sin \theta\cos \theta \\\Rightarrow \sin \theta \cos \theta = \frac{1}{2}(r^2-1)\\\\ $a) $\csc \theta + \sec \theta = \frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta }=\frac{2r}{r^2-1}\\\\$b) $\tan \theta + \cot \theta =...

Re: HSC 4U Integration Marathon 2017 $\noindent Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{1}{3+5\cos x}$ d$x

Re: HSC 4U Integration Marathon 2017 \int \frac{1}{x^2-3x+2}$ d$x = \int \frac{1}{(x-2)(x-1)} $ d$x \\ =\int( \frac{1}{x-2}-\frac{1}{x-1}) $ d$x \\\\= $ln$|x-2|-$ln$|x-1|+$C$

Re: HSC 4U Integration Marathon 2017 I_{m,n}=\int_{0}^{1}x^m(1-x)^n$ d$x \\$ Using IBP, $\\ I_{m,n}= \frac{(1-x)^nx^{m+1}}{m+1}|_{0}^{1}+\int_{0}^{1}\frac{x^{m+1}}{m+1}\cdot n(1-x)^{n-1}$ d$x \\$ Using IBP again, and again, etc, it is clear that $\\\\I_{m,n}=...

Re: HSC 4U Integration Marathon 2017 Lol i think this method is a complete unintended tedious one use the substitution x=sqrt(3)sin(theta). the new limits are now 0 to sin^-1(1/sqrt(3)). and dx=sqrt(3)cos(theta). Sub it in, the integrand becomes 9sin^2(theta)cos^2(theta). Which can be written...

You're in the right track. The question says that the entire frame has the height h, and width y. like this: . So doing what u basically did, you'll get 3h+4y=12. Making y the subject, you get 1/4(12-3h).

Re: HSC 2018 4U Marathon An alternate method (also a more tedious one, esp at the end): $\noindent$ z\neq 1\Rightarrow z-1\neq 0 \\\therefore $ by dividing all terms by $ (z-1)^6,$ we get $ 1+\frac{(z+1)^6}{(z-1)^6}=0\\\Rightarrow \left ( \frac{z+1}{z-1} \right )^{6}=-1 \\ \Rightarrow...

If any data is given in the question, always round up your answers to the lowest significant figures given in the question. For example "if 0.20 g of something reacts with 0.7255 g of something else" then your final answer should be in 2 sig figs, cause of 0.20 which has the least sig figs...

$\noindent$a_{n}=a_{n-1}+a_{n-3},\\ $ $a_{1}=1\\a_{2}=1\\ a_{3}=2 not 100% sure though

arithmetic progression; https://en.wikipedia.org/wiki/Arithmetic_progression

No worries! Hope that helped. And yeah it does take a tiny bit longer than typing it normally

hopefully this is clearer than before $\noindent a) Since all the cooefficients are real, due to the conjugate root theorem, $3+2i$ is also a root. Now the sum of roots is $-b=3+2i+3-2i=6\Rightarrow b=-6\\ $ The product of the roots is $c=(3+2i)(3-2i)=13\\ \\ $b) Since all the coefficients are...

bruh my school required us to do a ~3 minute skit for prelim chem and now we have to make a brochure (A3 laminated). Can't be bothered spending my time on that, would much prefer theories and pracs all year. and nah im doing forensics

m888888888 you lucky kid

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread You made a small silly mistake, the point B has coordinates (0,-4). So PB^2 would be x^2+(y+4)^2

Re: HSC Chemistry Marathon 2017 lmao