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Year 13 Maths Help Thread

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Original post by RDKGames
What are you doing to the LHS to get 3x3x2\displaystyle \frac{3}{x} \Rightarrow \frac{3}{x^2}?? Do the same to the RHS.


3lnx^2 ?
Original post by kiiten
3lnx^2 ?


Not really. 3ln(x2)=6ln(x)3\ln(x^2) = 6\ln(x)

Just divide RHS by x. Not sure where this question really comes from or why because it doesn't simplify to anything nice.
Original post by RDKGames
Not really. 3ln(x2)=6ln(x)3\ln(x^2) = 6\ln(x)

Just divide RHS by x. Not sure where this question really comes from or why because it doesn't simplify to anything nice.


I wanted to convert that to lnx (3/x^2)

Ill attach the ques....
Original post by kiiten
I wanted to convert that to lnx (3/x^2)

Ill attach the ques....


8.a) ii)

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Where??
Original post by RDKGames
Where??


Sorry i was having trouble attaching it

Posted from TSR Mobile
Original post by kiiten
Sorry i was having trouble attaching it

Posted from TSR Mobile


The proof for decreasing function?? I'm not quite sure what you were trying to do up there.

f(x)=2e2x3x2\displaystyle f'(x)=-\frac{2}{e^{2x}}-\frac{3}{x^2}

Can you see why this differential will always be negative, hence the function is always decreasing?
Original post by RDKGames
The proof for decreasing function?? I'm not quite sure what you were trying to do up there.

f(x)=2e2x3x2\displaystyle f'(x)=-\frac{2}{e^{2x}}-\frac{3}{x^2}

Can you see why this differential will always be negative, hence the function is always decreasing?


Yes - i thought you had to show it by doing dy/dx < 0 (or is it the other way round)
Original post by kiiten
Yes - i thought you had to show it by doing dy/dx < 0 (or is it the other way round)


That is the way to show it, what would be the "other way round"? You do NOT have to solve f(x)<0f'(x)<0 if that's what you're asking, it is a transcendental equation so you wouldn't get any exact answers anyway. You simply have to factor out a -1 then show how the inside of the bracket is always positive therefore f(x)<0f'(x)<0 in the given domain of xR,x>0x \in \mathbb{R},x>0
(edited 7 years ago)
Original post by RDKGames
That is the way to show it, what would be the "other way round"? You do NOT have to solve f(x)<0f'(x)<0 if that's what you're asking, it is a transcendental equation so you wouldn't get any exact answers anyway. You simply have to factor out a -1 then show how the inside of the bracket is always positive therefore f(x)<0f'(x)<0 in the given domain of xR,x>0x \in \mathbb{R},x>0


How could i show inside the bracket is always +ve - just input some values?
Original post by kiiten
How could i show inside the bracket is always +ve - just input some values?


Not really, that wouldn't prove anything for ALL x>0x>0. Since the numerator's are already positive, consider how e2xe^{2x} and x2x^2 are always positive for all real xx
(edited 7 years ago)
Original post by kiiten
How could i show inside the bracket is always +ve - just input some values?

x^2>0 so 3/x^2>0
Similarly e^2x>0 for xER so same follows hence -of them is negative


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Original post by physicsmaths
x^2>0 so 3/x^2>0
Similarly e^2x>0 for xER so same follows hence -of them is negative


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Thanks :smile:

If this was a question in an exam do you know what the mark scheme would say? e.g. certain phrases you must include?
Original post by kiiten
Thanks :smile:

If this was a question in an exam do you know what the mark scheme would say? e.g. certain phrases you must include?


It would say ex>0e2x>0e^{x}>0 \Rightarrow e^{2x}>0 and x2>03x2>0x^2>0 \Rightarrow \frac{3}{x^2}>0 where x>0x>0 therefore f(x)<0f'(x)<0 hence ff is decreasing over the interval but mark schemes never really mention all of it. Would touch up on the fact that the two terms are greater than 0, though.

More inequalities than 'certain phrases' to be honest. This would be like 2 or 3 marks max.
(edited 7 years ago)
Reply 794
Finally, I have a normal A Level question which I'm stuck on (I probably have annoyed many here by posting stuff which should go in the STEP prep thread):

Find the volume of revolution of the shape modelled by the equation x=sint,y=sin2t,0<t<π2x=\sin t,y = \sin 2t, 0<t<\frac{\pi}{2}.

V=π0π2y2dxdtdtV=\pi \int^{\frac{\pi}{2}}_0 y^2 \frac{dx}{dt} dt so I used the fact that y2=sin22t=4sin2tcos2ty^2={\sin}^2 2t=4{\sin}^2 t {\cos}^2 t and dxdt=cost\frac{dx}{dt}= \cos t to get me to 4π0π2sin2tcos3tdt=4π0π2cost3cost5dt4\pi \int^{\frac{\pi}{2}}_0 {\sin}^2 t {\cos}^3 t dt= 4\pi \int^{\frac{\pi}{2}}_0 {\cos t}^3- {\cos t}^5 dt. What do I do next?

Am I bizarrely overcomplicating this?
(edited 7 years ago)
Original post by Palette
Finally, I have a normal A Level question which I'm stuck on (I probably have annoyed many here by posting stuff which should go in the STEP prep thread):

Find the volume of revolution of the shape modelled by the equation x=sint,y=sin2t,0<t<π2x=\sin t,y = \sin 2t, 0<t<\frac{\pi}{2}.

V=π0π2y2dxdtdtV=\pi \int^{\frac{\pi}{2}}_0 y^2 \frac{dx}{dt} dt so I used the fact that y2=sin22t=4sin2tcos2ty^2={\sin}^2 2t=4{\sin}^2 t {\cos}^2 t and dxdt=cost\frac{dx}{dt}= \cos t to get me to 4π0π2sin2tcos3tdt=4π0π2cost3cost5dt4\pi \int^{\frac{\pi}{2}}_0 {\sin}^2 t {\cos}^3 t dt= 4\pi \int^{\frac{\pi}{2}}_0 {\cos t}^3- {\cos t}^5 dt. What do I do next?

Am I bizarrely overcomplicating this?


You can use a reduction formula to integrate cos3t \cos^3 t and cos5t \cos^5 t but there's an easier way of doing this.
If you notice that sin2tcos3tsin2tcost(1sin2t) \sin^2 t\cos^3 t \equiv \sin^2 t\cos t (1-\sin^2 t ) , it should be much easier to integrate.
Area of a circle sector = 30
Arc Length = 10
Find theta

someone pls help I've been trying for a while and can't figure it out, also it's in radians.
Reply 797
Original post by medhelp
Area of a circle sector = 30
Arc Length = 10
Find theta

someone pls help I've been trying for a while and can't figure it out, also it's in radians.


Arc length is rθr\theta and sector area is r2θ2\frac{r^2{\theta}}{2}.
Original post by Palette
Arc length is rθr\theta and sector area is r2θ2\frac{r^2{\theta}}{2}.


Yeah I know the formulae... I just dk how to do it without knowing r or theta
Original post by medhelp
Yeah I know the formulae... I just dk how to do it without knowing r or theta


Use a substitution for rr from the arc length. Get rr in terms of θ\theta.

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