Edexcel A2 C4 Mathematics June 2016 - Official Thread

@Zacken @Euclidean

Any more hard/interesting questions then? :wink:

Reply 461

XxKingSniprxX

When I see a person helping out someone with their Maths problem...

Reply 462

Original post by BBeyond

@Zacken @Euclidean

Any more hard/interesting questions then? :wink:

I'll think up some when I get home. :-)

Reply 463

Original post by BBeyond

@Zacken @Euclidean

Any more hard/interesting questions then? :wink:

Try solving this:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} + 4xe^{-x^2}\sqrt{y+3} = 0\end{equation*}

subject to the conditions

x \geq 0

and that

y=6

when

x=0

. Express your solution in the form

y = f(x)

and find

\lim_{x \to \infty} y

Reply 464

Original post by Zacken

Try solving this:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} + 4xe^{-x^2}\sqrt{y+3} = 0\end{equation*}

subject to the conditions

x \geq 0

and that

y=6

when

x=0

. Express your solution in the form

y = f(x)

and find

\lim_{x \to \infty} y

I'll try after my dinner, cheers!

Reply 465

Original post by BBeyond

I'll try after my dinner, cheers!

Bon appetit.

Reply 466

Original post by Zacken

Try solving this:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} + 4xe^{-x^2}\sqrt{y+3} = 0\end{equation*}

subject to the conditions

x \geq 0

and that

y=6

when

x=0

. Express your solution in the form

y = f(x)

and find

\lim_{x \to \infty} y

Kinda rushed this as I have a few things to do, so there may be a mistake somewhere :rofl:

Spoiler

(edited 8 years ago)

Reply 467

Original post by edothero

Kinda rushed this as I have a few things to do, so there may be a mistake somewhere :rofl:

Yeaaah. It's

y+3

but you seem to have switched to

y-3

half way through which gets you ugly ass square roots. :tongue:

Reply 468

Original post by Zacken

Yeaaah. It's

y+3

but you seem to have switched to

y-3

half way through which gets you ugly ass square roots. :tongue:

Oh ffs

There we go, I've changed it :colondollar:

(edited 8 years ago)

Reply 469

Original post by Zacken

Try solving this:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} + 4xe^{-x^2}\sqrt{y+3} = 0\end{equation*}

subject to the conditions

x \geq 0

and that

y=6

when

x=0

. Express your solution in the form

y = f(x)

and find

\lim_{x \to \infty} y

Spoiler

Posted from TSR Mobile

Reply 470

Original post by edothero

[Awesome work]

All correct. Now what's

\lim_{x \to \infty} y

Reply 471

Original post by BBeyond

...

Original post by edothero

...

All correct, both of you. Good work, too bad we're only warming up. Let's get to the meaty bit: Let

y

be any solution to:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{\mathrm{d}y}{\mathrm{d}x} - xe^{6x^2}(y+3)^{1-k} = 0 \quad (x\geq 0)\end{equation(}

Find a

k

such that we have

e^{-3x^2}y

tending to a finite non-zero limit that you should determine as

x \to \infty

Reply 472

Original post by Zacken

All correct. Now what's

\lim_{x \to \infty} y

As x tends to infinity, e^x tends to 0, as we end up with

e^{-2x^{2}} + 4e^{-x^{2}} + 1

then

\lim_{x \to \infty} y

would = 1 ?
Is this the right way of going about that?

Reply 473

Original post by edothero

As x tends to infinity, e^x tends to 0, as we end up with

e^{-2x^{2}} + 4e^{-x^{2}} + 1

then

\lim_{x \to \infty} y

would = 1 ?
Is this the right way of going about that?

Yep, you needn't expand though:

x \to \infty \Rightarrow e^{-x^2} \to 0 \Rightarrow y \to (0+2)^2 - 3 = 2^2 - 3 = 4-3=1

Reply 474

Original post by Zacken

Yep, you needn't expand though:

x \to \infty \Rightarrow e^{-x^2} \to 0 \Rightarrow y \to (0+2)^2 - 3 = 2^2 - 3 = 4-3=1

Ah yes makes more sense :biggrin:

I'll attempt that second question later tonight after revision. Seems tough!

Reply 475

Original post by Zacken

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{\mathrm{d}y}{\mathrm{d}x} - xe^{6x^2}(y+3)^{1-k} = 0 \quad (x\geq 0)\end{equation(}

Find a

k

such that we have

e^{-3x^2}y

tending to a finite non-zero limit that you should determine as

x \to \infty

I think it's time you give me a hint :biggrin:

Spoiler

Reply 476

Original post by edothero

I think it's time you give me a hint :biggrin:

Spoiler

Do the obvious thing: isolate, multiply both sides by what you want, simplify. Look at limits.

Reply 477

Original post by edothero

I think it's time you give me a hint :biggrin:

Spoiler

That's about as far as I made it as well ahahah

Reply 478

Original post by Zacken

Do the obvious thing: isolate, multiply both sides by what you want, simplify. Look at limits.

Not getting anywhere I'm afraid :colonhash:

Reply 479