Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

Paradoxica · Dec 5, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Find d/dx ( x^2logx) ... = 2xlnx + x

Hence evaluate Integral from 1 to e xlnx dx

I get the answer : (e^2 + 1)/4

Could someone see if they get the same result.

The answer says its (e^2 - e + 1)/ 2

Still can't figure where I have gone wrong.

$\int_1^e x\ln{x} dx = \frac{1}{2}\int_1^e (2x\ln{x} + x - x)dx = \frac{1}{2} \left ( x^2 \ln{x} - \frac{x^2}{2} \right )_1^e$

$= \frac{1}{2}\left ( e^2\ln{e} - \frac{e^2}{2} - 1^2\ln{1} + \frac{1^2}{2} \right ) = \frac{e^2 +1}{4}$

$Wolfram Alpha verifies this result. The answer given is simply incorrect.$

appleibeats · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Consider the curve y = -xlnx + 4

Given that y'' < 0 for all values of x in the natural domain. What is the natural domain and range of the curve??

InteGrand · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Consider the curve y = -xlnx + 4

Given that y'' < 0 for all values of x in the natural domain. What is the natural domain and range of the curve??

$We have $y=f(x)=-x\ln x + 4$. Note that since we have a $\ln x$, we need $x>0$. Hence the natural domain is the positive reals $\mathbb{R}^+$ (i.e. the set of all real numbers $x$ such that $x>0$). Note that as $x \to 0^+$, $-x\ln x \to 0^{+}$, so $f(x) \to 4^+$. As $x\to \infty$, $f(x)\to -\infty$. Since the second derivative is always negative, we can find the global maximum by setting the first derivative to 0.$

$We have $f^\prime (x) = -\left(\ln x + x \cdot \frac{1}{x} \right)=-\left(\ln x + 1 \right)$. So $f^\prime (x) = 0 \Longleftrightarrow \ln x = -1$, i.e. $x = \mathrm{e^{-1}}$. So we have the global max. at $x=\mathrm{e}^{-1}$. At this point, $f(x) = f\left(\mathrm{e}^{-1} \right) = -\mathrm{e}^{-1} \cdot (-1) + 4 = 4 + \mathrm{e}^{-1}$. So the global max. is $ 4 + \mathrm{e}^{-1}$. So the range is $y \leq 4 + \mathrm{e}^{-1} $, i.e. $\{y\in \mathbb{R} : y \leq 4 + \mathrm{e}^{-1} \}$ (this is read as ``the set of real $y$ such that $y$ is less than or equal to $4 + \mathrm{e}^{-1}$'').$

InteGrand · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$So in summary:$

$Natural domain: $x>0$$

$Range: $y\leq 4 + \mathrm{e}^{-1}$.$

appleibeats · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Given that k is an even integer, find Integral from -2 to 2 x(e^x)^k dx

Paradoxica · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
$$Given that $k$ is an even integer, find $\int_{-2}^{2} x(e^x)^k dx$

$Define $k= 2n \; \; \; \forall n \in \mathbb Z$

$\int_{-2}^2 x(e^{x})^{2n} dx = \int_{-2}^2 xe^{2nx} dx \rightarrow $IBP$ \rightarrow \frac{xe^{2nx}}{2n} \big |_{-2}^2 - \int_{-2}^2 \frac{e^{2nx}}{2n} dx = \left ( \frac{xe^{2nx}}{2n} - \frac{e^{2nx}}{4n^2} \right )_{-2}^2$

$= \left ( \frac{(2nx-1)e^{2nx}}{4n^2} \right )_{-2}^2 = \frac{(4n-1)e^{4n}-(-4nx-1)e^{-4n}}{4n^2} = \frac{(4n-1)e^{4n}+(4n+1)e^{-4n}}{4n^2}$

$$Degenerate case: $n= 0$

$\int_{-2}^2 xdx = 0$

leehuan · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Given that k is an even integer, find Integral from -2 to 2 x(e^x)^k dx

Are you sure this is 3U and there was no precursor? Because I would've also used IBP here (except for, as pointed out, the specific case of k=0)

appleibeats · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Yes it is 3U and no precursor question , and what is IBP?

leehuan · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

This is what happens when InteGrand is not here.

InteGrand · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Yes it is 3U and no precursor question , and what is IBP?

IBP stands for Integration by Parts, an integration technique taught in HSC 4U Maths.

leehuan · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
IBP stands for Integration by Parts, an integration technique taught in HSC 4U Maths.

Do you have a 3U method for this question?

Paradoxica · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
This is what happens when InteGrand is not here.

I know. Wolfram Alpha refuses to return a more elementary solution,and I can't think of anything.

leehuan · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Paradoxica said:
I know. Wolfram Alpha refuses to return a more elementary solution,and I can't think of anything.

You accidentally mucked up your degenerate case btw. That integral is 0

Paradoxica · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
You accidentally mucked up your degenerate case btw. That integral is 0

thanks, now we must work on a more elementary solution...

leehuan · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

This is so stupid. No definite integral tricks work here either, and we don't have an odd function to manipulate.

applei, exercise + question number please...

Paradoxica · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
This is so stupid. No definite integral tricks work here either, and we don't have an odd function to manipulate.

applei, exercise + question number please...

The integral itself, when treated as a function, is odd though. However, I fail to see how this will help us prove the formula for the definite integral.

InteGrand · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
Do you have a 3U method for this question?

$No, but possibly appleibeats typed it wrongly and it was meant to be $x\mathrm{e}^{x^k}$ (in which case we can clearly do it using analysis of odd/even functions assuming $k>0$)? Nesting in powers is done by doing the powers from the top downwards.$

Paradoxica · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
$No, but possibly appleibeats typed it wrongly and it was meant to be $x\mathrm{e}^{x^k}$ (in which case we can clearly do it using analysis of odd/even functions assuming $k>0$)? Nesting in powers is done by doing the powers from the top downwards.$

$Yeah, I suspected that might have been the case from the very beginning, it looked a bit off to me when I saw the post, but given the bracket nesting, I assumed he knew what he was doing. And that looks like it isn't the case.$

InteGrand · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Paradoxica said:
$Yeah, I suspected that might have been the case from the very beginning, it looked a bit off to me when I saw the post, but given the bracket nesting, I assumed he knew what he was doing. And that looks like it isn't the case.$

$Yeah, maybe he thought that $\mathrm{e}^{x^k}$ means $\left(\mathrm{e}^x\right)^k$ (which, of course, it doesn't).$

leehuan · Dec 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

xe^(x^k) will go into odd functions by default, yep

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