4U Maths Question Thread

[spnmox]

Hi, please could I have some help with proving this absolute value inequality:
l (a+b+c) l = lal + lbl + lcl

[RuiAce]

Hi, please could I have some help with proving this absolute value inequality:
l (a+b+c) l = lal + lbl + lcl

Were you trying to use \( \leq \) there?

(Note: If so, then this is just the usual triangle inequality. First prove that \( |z_1|+|z_2| \geq |z_1+z_2| \) using the usual geometric approach, and then prove by induction that \( |z_1|+|z_2|+\dots + |z_n| \geq |z_1+z_2+\dots+z_n| \). And then sub \(n=3\), with \(z_1 = a\), \(z_2 = b\), \(z_2 = c\).)

[david.wang28]

The thing that completely horrifies me with this question is how they expect you to know how to sketch \(x^{2/3} = y^{2/3} + 1\), and the awkwardness of the wording. Sketching that curve should, in theory, be a completely separate question on its own, guided using stuff like implicit differentiation and identifying the domain of the curve.

The wording of the question is also ambiguous, as it could represent the region below. They should have specified that they specifically want the region in the first quadrant.

Nevertheless, the question becomes doable with those assumptions put into place (albeit with some awkwardness). The idea is that they want you to find the green area shown below, but with the aid of the blue area. I will start it off.
(Image removed from quote.)
\[ \text{Note that when }x=2\sqrt{2}, \, y = 1.\\ \text{The combined area of the blue and green region is therefore just }2\sqrt{2}.\]
\[ \text{Note that the equation given rearranges to }x = \left( y^{2/3}+1\right)^{3/2}.\\ \text{The area of the blue region is therefore equal to}\\ \int_0^1 \left( y^{2/3}+1\right)^{3/2}\,dy\]
\[ \text{Let }y = tan^3\theta\implies dy = 3\tan^2\sec^2\theta.\text{ Then}\\ \begin{align*}\int_0^1 \left( y^{2/3} +1\right)^{3/2}\,dy &= \int_0^{\pi/4} \left( \tan^2\theta + 1 \right)^{3/2} (3\tan^3\theta \sec^2\theta)\,d\theta\\ &= 3\int_0^{\pi/4} \sec^2\theta (\sec^2\theta-1)\sec^2\theta\,d\theta\\ &= 3\left( \int_0^{\pi/4} \sec^6\theta\,d\theta + \int_0^{\pi/4} \sec^4\theta\,d\theta \right) \\ &= 3\left( \left[I_6\right]_0^{\pi/4} + \left[I_4\right]_0^{\pi/4} \right)\\ &= 3\left( \frac{\sec^4\frac\pi4 \tan \frac\pi4}{5} - \frac{\sec^4 0 \tan 0}5 - \frac45 \left[I_4\right]_0^{\pi/4} + \left[ I_4 \right]_0^{\frac\pi4}\right)\\ &= 3\left( \frac45 + \frac15 \left[ I_4\right]_0^{\pi/4}\right)\end{align*}\]
The remaining use of the reduction formulae and the final subtraction at the end is left as your exercise.

Subject to minor computational error.

Thank you Rui! But surely this question won't be in an Extension 2 exam?

[RuiAce]

Thank you Rui! But surely this question won't be in an Extension 2 exam?

If it were, it would be chopped up into multiple parts.

[spnmox]

Show that sinx<x for x>0.

Here's my working so far:
Let f(x)=sinx-x
f'(x)=cosx-1 =0 for stat pts
cosx=1
x=0,360
x>0 so x=360 degrees
f''(x)=-sinx
f''(360)=0

so apparently there's a horizontal point of inflexion at x=360
but don't you have to show that it's the absolute max turning point?

[spnmox]

Consider the function f(x) =e^x(1-x/10)^10.  Sketch the curve and from your graph, deduce that e^x <= (1-x/10)^-10 for x<10

I'm having trouble finding the asymptotes and sketching the graph. I've already found that turning points (),1) and (10,0), and I think that as x-->infinity, y--> infinity and x--> -infinite y--> -infinity, hopefully that's right?

I'm not sure how to deduce that inequality either.

[RuiAce]

Show that sinx<x for x>0.

Here's my working so far:
Let f(x)=sinx-x
f'(x)=cosx-1 =0 for stat pts
cosx=1
x=0,360
x>0 so x=360 degrees
f''(x)=-sinx
f''(360)=0

so apparently there's a horizontal point of inflexion at x=360
but don't you have to show that it's the absolute max turning point?

Firstly, for the sake of calculus, never use degrees.

Secondly, \(f^{\prime\prime}(x) = 0\) doesn't always imply that \(x\) is a point of inflexion. The theorem says that if \(x\) is a point of inflexion, then \(f^{\prime\prime}(x) = 0\). But the converse (i.e. what you get when you flip that statement around) does not hold. An easy example is \( h(x) = x^4\) - we can prove that \( h^{\prime\prime}(0) = 0\), but \(x=0\) is certainly not a point of inflexion.
To test if it is actually a point of inflexion, you need to do test both sides for concavity changes. So since you've computed that \(h^{\prime\prime}(2\pi) = 0\), you could plug \(2\pi - 0.01\) and \(2\pi + 0.01\) into \(f(x)\), to determine if there is a concavity change. You'll see that there is not, so the stationary point you found will not be a horizontal point of inflexion.

Finally, as for the original question itself, hint - you've overcomplicated it. Verifying that \(f^{\prime}(x) \leq 0\) for all real \(x\) is the starting point. Recall that if \(f^\prime(x) \leq 0\) for all real \(x\), i.e. the curve is decreasing (albeit not strictly decreasing), then we have the conclusion
\[ \text{If }x_1 \geq x_2\\ \text{Then }f(x_1) \leq f(x_2) \]

[david.wang28]

Hello,
I am stuck on an integration question. The question and my working out are attached below. Can anyone please help me out? Thanks

[AlphaZero]

Hello,
I am stuck on an integration question. The question and my working out are attached below. Can anyone please help me out? Thanks

\begin{align*}\int_{-1}^1 \frac{x^2}{e^x+1}dx&=\int_0^1\left(\frac{x^2}{e^x+1}+\frac{x^2}{\frac{1}{e^x}+1}\right)dx\\
&=\int_0^1\left(\frac{x^2}{e^x+1}+\frac{x^2e^x}{1+e^x}\right)dx\\
&=\int_0^1\frac{x^2(e^x+1)}{e^x+1}dx\\
&=\qquad \vdots \end{align*}

[david.wang28]

Ahhh k, got the answer. Thank you! 

[wlam]

im lost on part b...
thanks

[fun_jirachi]

From here it should be evident that since (x1,y1) or (x2,y2) lie on the chord, we can just sub them in. x1y1 or x2y2 should cancel out and you should have the given equation.

Hope this helps

[spnmox]

If f(x) = x/sqr(1+x^2), prove by mathematical induction for n>=2 that f(f(...(f (x))...)) = x/sqr(1+nx^2), if there are n number of letter f's on the LHS.

Please help! I don't understand what f of f(x) means. What exactly am I subbing in when I am trying to prove true for n=2 and assuming true for n=k?

Thank you.

[RuiAce]

If f(x) = x/sqr(1+x^2), prove by mathematical induction for n>=2 that f(f(...(f (x))...)) = x/sqr(1+nx^2), if there are n number of letter f's on the LHS.

Please help! I don't understand what f of f(x) means. What exactly am I subbing in when I am trying to prove true for n=2 and assuming true for n=k?

Thank you.

To help you get started, say you have \(h(x) = x-1\). Then, \(h(h(x)) = h(x-1)\). Of course, clearly this simplifies to \( h(h(x)) = x-2\), but it illustrates the idea - you sub the expression the function is equal to, back into itself.

The notation is standard - it is just composition of functions \(g(f(x))\), except \(g(x)\) and \(f(x)\) are taken to be the same thing here. Note we also learnt to use the chain rule in 2U differentiation for composite functions - this let us establish that \( \frac{d}{dx} g(f(x)) = f^\prime(x) \cdot g^\prime(f(x)) \).
\[ \text{So for your question, just for the sake of starting you off,}\\ \text{If }f(x) = \frac{x}{\sqrt{1+x^2}} \]
\[\text{Then}\\ \begin{align*} f(f(x)) &= f\left( \frac{x}{\sqrt{1+x^2}} \right)\\ &= \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1 + \left( \frac{x}{\sqrt{1+x^2}}\right)^2}}\end{align*} \]
For the base case, you wish to prove that that expression equals to \(\frac{x}{\sqrt{1+2x^2}} \).

[spnmox]

To help you get started, say you have \(h(x) = x-1\). Then, \(h(h(x)) = h(x-1)\). Of course, clearly this simplifies to \( h(h(x)) = x-2\), but it illustrates the idea - you sub the expression the function is equal to, back into itself.

The notation is standard - it is just composition of functions \(g(f(x))\), except \(g(x)\) and \(f(x)\) are taken to be the same thing here. Note we also learnt to use the chain rule in 2U differentiation for composite functions - this let us establish that \( \frac{d}{dx} g(f(x)) = f^\prime(x) \cdot g^\prime(f(x)) \).
\[ \text{So for your question, just for the sake of starting you off,}\\ \text{If }f(x) = \frac{x}{\sqrt{1+x^2}} \]
\[\text{Then}\\ \begin{align*} f(f(x)) &= f\left( \frac{x}{\sqrt{1+x^2}} \right)\\ &= \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1 + \left( \frac{x}{\sqrt{1+x^2}}\right)^2}}\end{align*} \]
For the base case, you wish to prove that that expression equals to \(\frac{x}{\sqrt{1+2x^2}} \).

Thank you so much you actual genius! I figured it out and I'm so proud.

I also have another question:

If the sums of the first three terms of an AP and a GP are equal and non-zero, the common difference of the AP and the common ratio of the GP are equal and the ratio of the AM to the GM is 1:-2, find the common ratio of the GP. Also, find the relationship between their first terms.

So far I've just defined all the terms and apparently you need to solve them simultaneously but not sure how to do that.