Author Topic: 3U Maths Question Thread (Read 1239776 times) Tweet Share

Calley123 · « **Reply #3465 on:** June 05, 2018, 09:07:26 pm »

Quote from: Calley123 on June 05, 2018, 09:02:17 pm

Thank you!!

Please help me with Q13 !
Inverse functions

Opengangs · « **Reply #3466 on:** June 05, 2018, 09:29:33 pm »

-1

Quote from: Calley123 on June 05, 2018, 09:07:26 pm

Please help me with Q13 !
Inverse functions

$ \begin{align*}\frac{d}{dx}\left(x\cos^{-1}x - \sqrt{1 - x^2}\right) &= \cos^{-1}x - \frac{x}{\sqrt{1 - x^2}} - \frac{-2x}{2\sqrt{1 - x^2}} \\ &= \cos^{-1}x - \frac{x}{\sqrt{1 - x^2}} + \frac{x}{\sqrt{1 - x^2}} \\ &= \cos^{-1}x \end{align*}$

$\text{Because we found that } \frac{d}{dx}\left(x\cos^{-1}x - \sqrt{1 - x^2}\right) = \cos^{-1}x \\ \text{it follows that: } \\ \\ \int \cos^{-1}x\,dx = \int \frac{d}{dx}\left(x\cos^{-1}x - \sqrt{1 - x^2}\right)\,dx .$
$\begin{align*} \text{So, } \int_0^{\frac{1}{2}} \cos^{-1}x\,dx &= \int_0^{\frac{1}{2}} \frac{d}{dx}\left(x\cos^{-1}x - \sqrt{1 - x^2}\right)\,dx \\ &= \frac{1}{2}\cdot \cos^{-1}\left(\frac{1}{2}\right) - \sqrt{1 - \left(\frac{1}{2}\right)^2} - \left(0\cdot \cos^{-1}(0) - \sqrt{1}\right) \\ &= \frac{\pi}{6} - \sqrt{\frac{4 - 1}{4}} + 1 \\ &= \frac{\pi}{6} - \frac{\sqrt{3}}{2} + 1 \\ &= \left(\frac{\pi}{6} + 1 - \frac{\sqrt{3}}{2}\right)\text{units}^2 \end{align*}$

arii · « **Reply #3467 on:** June 08, 2018, 10:49:21 pm »

Find all possible values of "a" for (x+2)(x-3)²(x+a)≥0.

RuiAce · « **Reply #3468 on:** June 08, 2018, 11:46:37 pm »

Quote from: arii on June 08, 2018, 10:49:21 pm

Find all possible values of "a" for (x+2)(x-3)²(x+a)≥0.

This is missing information. Do we want to find the possible values of $a$ that make $ (x+2)(x-3)^2(x+a)\geq 0$ for all $x$ or something?

Or are we just trying to solve an inequality?

arii · « **Reply #3469 on:** June 09, 2018, 06:24:22 pm »

Quote from: RuiAce on June 08, 2018, 11:46:37 pm

This is missing information. Do we want to find the possible values of $a$ that make $ (x+2)(x-3)^2(x+a)\geq 0$ for all $x$ or something?

Or are we just trying to solve an inequality?

They want all the possible values of "a" that make the inequality true for all real values of x.

RuiAce · « **Reply #3470 on:** June 09, 2018, 06:28:20 pm »

Quote from: arii on June 09, 2018, 06:24:22 pm

They want all the possible values of "a" that make the inequality true for all real values of x.

In that case the only possible value is $a = 2$, or else the graph of $ y = (x+2)(x-3)^2(x+a) $ always dips below the $x$-axis at some point.

Edit: That minus sign should not have been there

arii · « **Reply #3471 on:** June 09, 2018, 09:10:02 pm »

When I was doing the simultaneous equations for the R co-ordinates, stuff didn't quite cancel out due to the sign and that made the rest of it quite complex... Not sure where I went wrong so could you help me with this?

RuiAce · « **Reply #3472 on:** June 10, 2018, 09:52:23 am »

Quote from: arii on June 09, 2018, 09:10:02 pm

When I was doing the simultaneous equations for the R co-ordinates, stuff didn't quite cancel out due to the sign and that made the rest of it quite complex... Not sure where I went wrong so could you help me with this?

$OA\text{ clearly has equation }\boxed{y=\frac{p}{2}x}$
$m_{BC} = \frac{\frac{a}{p^2}+a}{-\frac{2a}{p} - 0} = -\frac{\frac1{p^2}=1}{\frac2p} = -\frac{p^2+1}{2p}\\ \text{so }BC\text{ has equation }\boxed{y+a = -\left(\frac{p^2+1}{2p}\right)x}$
$\text{For the coordinates of }R:\\ \begin{align*}\frac{p}{2}x + a &= -\left( \frac{p^2+1}{2p} \right)x\\ \left( \frac{p^2+1}{2p} \right) x + \frac{p^2}{2p}x &=-a\\ \left( \frac{2p^2+1}{2p} \right)x&=-a\\ \therefore x &= -\frac{2ap}{2p^2+1}\\ \implies y&= -\frac{ap^2}{2p^2+1}\end{align*}$
$\text{Upon dividing the coordinates of }R,\\ \frac{y}{x} = \frac{p}{2} \implies \boxed{p = \frac{2y}{x}}.\\ \text{So subbing into the second we have}\\ \begin{align*}y &= -\frac{a\left( \frac{4y^2}{x^2} \right)}{2\left( \frac{4y^2}{x^2} \right) + 1}\\ y &= -\frac{4ay^2}{8y^2+x^2}\\ 1 &= -\frac{4ay}{8y^2+x^2}\\ 8y^2+x^2&=-4ay\\ x^2+8y^2+4ay&=0\end{align*}$
Note: To be fully accurate, one should check the case of $p = 0$, i.e. $R = (0,0) $ separately.

clovvy · « **Reply #3473 on:** June 10, 2018, 07:30:26 pm »

With binomial theorem when it ask for greatest term, is it asking for coefficient? Because I always get the number that have the highest coefficient when solving with $\frac{T_{k+1}}{T_k}>1$ but the book is telling me the answer being the term higher than that
e.g. from old fitz $\text{Find which are the numerically greatest terms of the following (i)}\, (3+4x)^{12}\, \text{if}\, x=\frac{1}{2}\\ \text{I get k=5 but why is the answer telling me}\, T_6$

RuiAce · « **Reply #3474 on:** June 10, 2018, 07:37:06 pm »

Quote from: clovvy on June 10, 2018, 07:30:26 pm

With binomial theorem when it ask for greatest term, is it asking for coefficient? Because I always get the number that have the highest coefficient when solving with $\frac{T_{k+1}}{T_k}>1$ but the book is telling me the answer being the term higher than that
e.g. from old fitz $\text{Find which are the numerically greatest terms of the following (i)}\, (3+4x)^{12}\, \text{if}\, x=\frac{1}{2}\\ \text{I get k=5 but why is the answer telling me}\, T_6$

Once you sub in $ x = \frac12$ things change. The greatest coefficient is used in the case where $x$ is unknown, and we just care about the coefficients. The greatest term, however, becomes a thing once we actually sub $x$ in for something.

For that question, if we just wanted the greatest coefficient, we'd take $T_k $ to just be the coefficient, and nothing else. i.e. $T_{k+1}= \binom{12}{k} 3^{12-k} 4^k $.

But if we wanted the greatest term after subbing in $ x = \frac12$, we'd have to consider the actual term $T_{k+1} = \binom{12}{k}3^{12-k}4^k x^k$, but after subbing in $x = \frac12$. That is, we'd be considering $T_{k+1} = \binom{12}{k} 3^{12-k} 4^k \left(\frac12 \right)^k $

clovvy · « **Reply #3475 on:** June 10, 2018, 08:06:07 pm »

Quote from: RuiAce on June 10, 2018, 07:37:06 pm

Once you sub in $ x = \frac12$ things change. The greatest coefficient is used in the case where $x$ is unknown, and we just care about the coefficients. The greatest term, however, becomes a thing once we actually sub $x$ in for something.

For that question, if we just wanted the greatest coefficient, we'd take $T_k $ to just be the coefficient, and nothing else. i.e. $T_k= \binom{12}{k} 3^{12-k} 4^k $.

But if we wanted the greatest term after subbing in $ x = \frac12$, we'd have to consider the actual term $T_k = \binom{12}{k}3^{12-k}4^k x^k$, but after subbing in $x = \frac12$. That is, we'd be considering $T_k = \binom{12}{k} 3^{12-k} 4^k \left(\frac12 \right)^k $

I did check with the calculator and k=5 gives a bigger answer so I don't really know what is going on

RuiAce · « **Reply #3476 on:** June 10, 2018, 08:07:20 pm »

There's a few typo's in my post above. I'll fix it.
\[ \text{The formula is actually }\boxed{T_{k+1} = \binom{n}{k} a^{n-k}b^k}\text{ in the expansion of }(a+b)^n\\ \text{Not }T_k. \]
So $k=5$ is fine. Because when $k=5$, we have $T_6$.

clovvy · « **Reply #3477 on:** June 10, 2018, 08:12:46 pm »

Quote from: RuiAce on June 10, 2018, 08:07:20 pm

There's a few typo's in my post above. I'll fix it.
\[ \text{The formula is actually }\boxed{T_{k+1} = \binom{n}{k} a^{n-k}b^k}\text{ in the expansion of }(a+b)^n\\ \text{Not }T_k. \]
So $k=5$ is fine. Because when $k=5$, we have $T_6$.

ah dammit that's right, so I was on the right track altogether just forgetting that k+1 lol

Calley123 · « **Reply #3478 on:** June 10, 2018, 09:29:36 pm »

Please helo!
Applications of calculus to the physical world

Questions below.

Thanks in advance!

RuiAce · « **Reply #3479 on:** June 10, 2018, 09:36:48 pm »

Quote from: Calley123 on June 10, 2018, 09:29:36 pm

Please helo!
Applications of calculus to the physical world

Questions below.

Thanks in advance!

This is a very common 2U-type question and you're expected to know how to do it. $\text{Assume that the mass is given }M=M_0e^{-kt}$
$\text{The half-life is 1600, so when }t=1600, M=\frac{M_0}{2}\\ \text{So we have}\\ \begin{align*}\frac{M_0}{2}&= M e^{-1600k}\\ \frac12 &= e^{-1600k}\\ \ln \frac12 &= -1600k\\ k &= -\frac{\ln \frac12}{1600}\\ &= \frac{\ln 2}{1600}\end{align*}$
$\text{So }\boxed{M = M_0 e^{-\frac{\ln 2}{1600}t}}$
_____________________
$\text{When }t=500\\ \begin{align*}M&=M_0e^{-\frac{500\ln 2}{1600}}\\ &= 0.805245\dots M_0\end{align*}\\ \text{So approximately 80.5 percent}$
_____________________
$\text{When }M=0.25M_0\\ \begin{align*}\frac14 M_0 &=M_0 e^{-\frac{\ln 2}{1600}t}\\ \ln \frac14 &= -\frac{\ln 2}{1600}t \end{align*}\\ \text{You should be able to finish it off from here.}$

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