Mathematics Question Thread

[david.wang28]

okay I got it, thanks for the help

No problem

[RuiAce]

Hi there,
I'm not doing 2u right now but I stumbled across this random challenging 2u question, and I have no idea how to do it. Can anyone please help me out? Thanks

That question certainly has a typo. It should be \( S_n = \frac12 (a+\ell) \left(1+ \frac{\ell-a}{d} \right) \). Note the 1.

Other than that though, it can be fudged. Recall that \(S_n = \frac{n}{2}(a+\ell) \) where \(\ell = T_n = a+ (n-1)d\). Literally rearranging \(\ell = a+(n-1)d\) gives \(n = 1+\frac{\ell-a}{d}\), which you can sub into there. The second part is then easy - just sub \(a=550\), \(\ell=990\) and \(d=11\) in, noting that the last multiple of 11 between 550 and 1000 is actually 990.

[david.wang28]

That question certainly has a typo. It should be \( S_n = \frac12 (a+\ell) \left(1+ \frac{\ell-a}{d} \right) \). Note the 1.

Other than that though, it can be fudged. Recall that \(S_n = \frac{n}{2}(a+\ell) \) where \(\ell = T_n = a+ (n-1)d\). Literally rearranging \(\ell = a+(n-1)d\) gives \(n = 1+\frac{\ell-a}{d}\), which you can sub into there. The second part is then easy - just sub \(a=550\), \(\ell=990\) and \(d=11\) in, noting that the last multiple of 11 between 550 and 1000 is actually 990.

Thanks for the help!

[shaynec19]

Could someone please help me with this.

[jamonwindeyer]

Could someone please help me with this.

Hey! The trick with the sum of the 6th-10th terms is to take \(S_{10}\) and subtract \(S_5\)! the equations you'll form are:



Simplify and solve these simultaneously!! Let us know if you need help with that

[shaynec19]

Thanks Jamon!
They are the equations I already derived, but I cannot for the life of me solve them to get an integer of a or d.
I have tried a couple of methods..
Cheers

[RuiAce]

Thanks Jamon!
They are the equations I already derived, but I cannot for the life of me solve them to get an integer of a or d.
I have tried a couple of methods..
Cheers

\begin{align*}a+19d &= 131\tag{1}\\ \frac{10}{2}(2a+9d) - \frac{5}{2}(2a+4d)&=235\\ 5(2a+9d) - 5(a+2d)&=235\\ a+7d&=47\tag{2}\end{align*}
\[ (2)-(1)\text{ gives }12d = 84 \implies \boxed{d=7}. \]

[shaynec19]

\begin{align*}a+19d &= 131\tag{1}\\ \frac{10}{2}(2a+9d) - \frac{5}{2}(2a+4d)&=235\\ 5(2a+9d) - 5(a+2d)&=235\\ a+7d&=47\tag{2}\end{align*}
\[ (2)-(1)\text{ gives }12d = 84 \implies \boxed{d=7}. \]

Thanks a mill Rui!

[Youssefh_]

Can someone please help me understand how to differentiate these sorts of questions

[david.wang28]

Can someone please help me understand how to differentiate these sorts of questions

Just remember to take the power down, multiply the coefficient, and then minus the power by one. Also, can you take a clearer screenshot please? Thanks

[fun_jirachi]

Just remember to take the power down, multiply the coefficient, and then minus the power by one. Also, can you take a clearer screenshot please? Thanks

Note that you can't actually do this. Think a) is asking for

Basically you want to be rearranging in into the form e^(ln (whatever you have)) which is true by log laws (if you don't understand, ask again ). Then you follow your regular thing for differentiating the natural exponential (basically the chain rule, and it's on your reference sheet), and simplify where necessary.

Hope this helps

[spnmox]

Sketch y=(x^2+1)/e^x

[fun_jirachi]

From this we can see we have a stationary point at (1, 2/e), and that the function is decreasing for all real x.
Note also that the domain is all real x, and that since x^2+1>=1 for all real x and e^x>0 for all real x, then f(x) > 0 for all real x. Also note that this will result in a horizontal asymptote at y=0. Do your usual thing with limits as x approaches -infinity and nature around the stationary point to finish up, then sketch. You should get something relatively similar to e^-x for the most part.

Also try and show some working next time, so instead of me telling you basically everything, I can properly help out with where you're going wrong in particular!

Hope this helps

[spnmox]

From this we can see we have a stationary point at (1, 2/e), and that the function is decreasing for all real x.
Note also that the domain is all real x, and that since x^2+1>=1 for all real x and e^x>0 for all real x, then f(x) > 0 for all real x. Also note that this will result in a horizontal asymptote at y=0. Do your usual thing with limits as x approaches -infinity and nature around the stationary point to finish up, then sketch. You should get something relatively similar to e^-x for the most part.

Also try and show some working next time, so instead of me telling you basically everything, I can properly help out with where you're going wrong in particular!

Hope this helps

Hi! Sorry, I'll be more specific.

I found the stat pt and tried to find slope around it, except I got that the gradient is the same ? so is it an inflexion point?
as for limits, I got as x--> infinity, y--> - infinity and as x--> - infinity, y--> infinity. is there a way to figure these out without subbing in very large and very small numbers?

[spnmox]

Sketch y=xlnx.

so i differentiated and got (1/e, -1/e) as a min TP, and no inflexion points; x--> infinity and y--> infinity, and x can't approach negative infinity because x can't be equal to 0 therefore there's an asymptote at x=0, is that right? I have looked at the answers and i just don't get the open circle thing at (0,0). There's an asymptote at x=0 and also an x-intercept at 0, why