Search results

Re: MX2 Integration Marathon u=-x I=\int_{-\pi/2}^{\pi /2} \frac{\cos x}{1+e^x} \ dx = \int_{\pi /2}^{-\pi /2} \frac{-\cos(-u)}{1+e^{-u}} \ du= \int_{-\pi /2}^{\pi /2} \frac{\cos u}{1+e^{-u}} \ du \therefore \ 2I = \int_{-\pi /2}^{\pi /2} \frac{\cos x}{1+e^{x}} + \frac{\cos x}{1+e^{-x}} \...

Re: HSC 2013 4U Marathon The numbers in pascal's triangle can be represented by binomial coefficients, the nth row and kth spot number is given by: \binom{n}{k} Thus, if there exists an AP in Pascal's triangle, for some 0\leq k \leq n \binom{n}{k} , \binom{n}{k+1} , \binom{n}{k+2} ...

Re: MX2 Integration Marathon Yep that's it. === \int_0^1 x^m(1-x)^n \ dx

Re: HSC 2013 3U Marathon Thread The symmetry argument is really elegant, but the more obvious way which is kinda what I was leaning to, involved calculus and tangents (i.e. not very elegant)

Re: HSC 2013 4U Marathon $For a triangle of side lengths$ \ \ a, b, c $Show that the area of such a triangle is$ \ \ A = \sqrt{s(s-a)(s-b)(s-c)} $Where$ \ \ s= \frac{a+b+c}{2} $Further show that the radius of a circle inscribed inside the triangle has a radius$...

Re: HSC 2013 4U Marathon Dividing using long division, we get that: \frac{x^3 + ax^2 + bx + c}{x - \alpha} = x^2+(a+\alpha)x + (b+\alpha(a+\alpha)) Discriminant is greater than zero: (a+\alpha)^2 - 4(b + \alpha(a+\alpha)) > 0 Expanding: a^2-2\alpha a - 3\alpha^2 -4b > 0 0> 3\alpha^2...

Re: MX2 Integration Marathon \int \sin^{-1} \sqrt{ \frac{x}{1+x}} \ dx

Re: MX2 Integration Marathon Yep. $First note that$ \cos^2(x-\pi/2) = \cos^2(x+ \pi / 2) \therefore $for$ \ \ 0\leq x \leq pi \ \ $graph is symetrical about$ \ \ \pi / 2 \therefore \ \int_0^{\pi} \frac{1}{1+\cos^2 x} \ dx = 2 \int_0^{\pi /2}\frac{1}{1+\cos^2 x} \ dx \int_0^{\pi /2}...

Re: HSC 2013 4U Marathon Not quite, that might be its locus, but we don't know what alpha is (i.e. Re(z-i)/(z+1) = 0 ) === Also twinklegal, my explanation is wrong, the variable square is above the fixed square of units y, but still using similar triangles as we wake a triangle to the top of...

Re: MX2 Integration Marathon \int_0^1 \frac{x}{1+ \frac{x}{1+ \frac{x}{1+ \frac{x}{1 + \dots}}}} \ dx EDIT: Thanks asianese

Re: HSC 2013 3U Marathon Thread $Prove that when the range is a maximum, the angle the projectile hits the plane, is perpendicular to the angle of projection$

Re: HSC 2013 3U Marathon Thread Through drawing a diagram, we can see that the side of a square, s is given by: s=2r-2y Hence area of one square is: 4(r-y)^2 Volume thus by integrating from r to -r, i.e. from the top to the bottom, we get 32/3 r^3 Hopefully that's right. ====...

Re: MX2 Integration Marathon =\int \frac{\sin x + \cos x}{\sin^2 x + \cos^2x (1-\sin^2 x} \ dx = \int \frac{\sin x + \cos x}{1-\sin^2x \cos^2x} \ dx u=\sin x - \cos x u^2 = 1-2\sin x \cos x \therefore \ \sin x \cos x = \frac{1}{2}(1-u^2) I = \int \frac{du}{1 - \frac{1}{4}(1-u^2)^2} =...

Re: MX2 Integration Marathon = \int \frac{\sec^2 x \ dx}{\cos^2 x + \sec^2 x } u=\tan x =\int \frac{du}{1+u^2 + \frac{1}{1+u^2}}= \frac{u^2+1}{u^4+2u^2+2} \ du $Next we find some$ \ \ a, b \ $so that$ u^4+2u^2+2=(u^2+au+\sqrt{2})(u^2+ bu + \sqrt{2}) We can find through expansion or...

Re: HSC 2013 4U Marathon a, b, c \ $are positive integers such that$ a+b+c+ab+ac+bc+abc= 7\cdot 11 \cdot 13 - 1 $Find$ \ \ a,b,c

You have assumed that B, C, K are all collinear (second step) it is not mentioned in the question and is in fact proven in the last part. To prove it is a cyclic quadrilateral, first we notice that angle ABD is equal to ACD (angle same segment) Moreover, angle ACD is equivalent to angle XCD...

Re: HSC 2013 4U Marathon Yep, $By considering the cases whereby the positive numbers$ \ \ a, b, c \ \ $are sides of a triangle and when they are not sides of a triangle, prove the inequality$ (-a+b+c)(a-b+c)(a+b-c) \leq abc

If the function is simple enough we can simply integrate a/v after a change in variable for x. EDIT: Nvm this will result us in inding v in terms of x anyway. Sometimes v cannot be a function of x valid for all positions x. i.e. in SHM, we have x=cos t, v= -sin t, we have 2 velocities for...

\frac{dv}{dx} = \frac{dv}{dt} \times \frac{dt}{dx} = \frac{1}{v} \times a So if we have v=f(t) then a=f'(t)

That makes much more sense, nice.