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Yeah, this is how i used to do it during my HSC. just multiply the 3 into the square root. So; dy/dx=4/sqrt(9-16x^2)

Yeah, only way there'd be a nice solution is if x E Q. And there's isn't one for this cubic.

Nah, it isn't lol. I don't think they'll ask such crap questions. They're more likely to ask tan(15)=tan(45-30)

There's probably a shorter way than using tan(105/3) but find the expansion of tan3@ and then let @=x/3. You'll get x^3-3Kx^2-3x+K=0 where K=tan(105). And then just use the cubic formula lol. And sub in a hell lot of values.

Your answer is right but your upper limit is meant to be 1/2.ln(3)

tan((60+45)/3)

Terms of odd powers cancel, so the middle term of the quadratic has to be minus in this case.

I=S[3,8] (x+1-1)/sqrt(x+1)(sqrt(x+1)-1) dx Let u^2=x+1, 2u du=dx: I=S[2,3] (u^2-1)/u(u-1) 2u du =2 S[2,3] (u+1) du =2 [1/2u^2+u] {2,3} =2 (9/2+3-2-2) =2(1/2+3) =7

I=S dx/(4+x^2)^3/2 Let x=2tan@, dx=2sec^2@ d@ I=S 2sec^2 @ d@ / 4^(3/2).sec ^3 @ =1/4 S cos@ d@ =1/4.sin@ +C = 1/4.sin(atan(x/2)) +C = 1/4.x/sqrt(4+x^2) +C

Just make the top into 1+x^2-x^2 and you'll get: I= S [1,sqrt(3)] sqrt(1+x^2)/x^2 dx-S[1,sqrt(3)] 1/sqrt(1+x^2) dx then let x=tanu. I think you can do the rest

Apparently not.

Yeah, think so.

K=a/(b^3*c^2) =(2/5)^4/[(-1/3)^3(3/5)^4] =[(2/5)^4/(3/5)^4]*1/(-1/27) =-27*[2/5/3/5]^4 =-27*[2/3]^4 =[-(3)^3/(3)^4].2^4 =-1/3.2^4 =-(2^4)/3 =-(2^4)*3^(-1)

Yeah, nw.

Yeah probably, but it's a pretty funky question.

I don't think the question is correct as y(1)=c.e^-b=0 but e^-b=/=0

What does iii) say? I can't read it. May think about it later if someone decides to decipher what iii) says.

Certain uni's offer bonus points depending on the marks you get in relevent courses eg E4 in 3u maths could be worth 2 bonus pts. These bonus pts only decrease the cut-off mark for a uni course and doesn't actually boost your ATAR.

Yeah, nw.

Thats incorrect. Divide by e^2x. I=S dx/e^2x+e^x = -S e^-x.-e^-x/(1+e^-x) dx = -S u du/(1+u) = S 1/(u+1) -1 du = ln(u+1)-u +C = ln(e^-x+1)-e^-x+C