Hi friends ! I need your help in clarifying my doubts here in this note. Can you help me in solving the below problems in which I found some difficulty. Any help would be appreciated.

\(\large \underline{\text{Functions}}\)

\(1)\) If \(f(x) = \log(x^{\log x^{\log x}})\), then show that \(f\left(\dfrac{1}{x}\right) = - f(x)\). (cleared)

\(2)\) If \(f(x) + f(x + 2) = x^2\), then find the value of \(f(x + 2) - f(x - 2)\) in terms of \(x\). (cleared)

\(3)\) If \(f(x)\) and \(g(x)\) are periodic functions with periods 7 and 11 respectively then find the period of \(F(x) = f(x)g(\dfrac{x}{5}) - g(x)f(\dfrac{x}{3})\).

\(4)\) Find the domain of the function \(f(x) = \sqrt{\dfrac{|x + 3| + x}{x + 2} - 1}\). (cleared)

\(5)\) If \(a^2 + b^2 + c^2 = 1\) then find the range of \(ab + bc + ca\). (cleared)

\(6)\) Prove that \(f(x) = \dfrac{2x(\sin x + \tan x)}{2 \left \lfloor \dfrac{x + 2\pi}{\pi} \right \rfloor - 3}\) is an odd function where \(\lfloor . \rfloor\) denotes greatest integer function. (cleared)

\(7)\) Let \(f(x) = \begin{cases} x^2 - 4x + 3 & , x < 3 \\ x - 4 & , x \geq 3 \\ \end{cases}\) and \(g(x) = \begin{cases} x - 3 & , x < 4 \\ x^2 + 2x + 2 & , x \geq 4 \\ \end{cases}\)

Describe the function \(fog(x)\) and find its domain.

\(8)\) If \(f(a - x) = f(a + x)\) and \(f(b - x) = f(b + x)\) for all real values of \(x\) where \(a,b ~(a > b)\) are constants. The prove that \(f(x)\) is a periodic function and find it's period.

That's all for now. I will keep posting my doubts here if they further arise. I hope I get solutuions for my doubts.

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## Comments

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TopNewestProblem 5:I assume \(a,b,c\) are real numbers.Let's work on the maximum value first.

By Cauchy-Schwarz inequality, we can see that \((ab + bc + ca)^2 \leq (a^2 + b^2 + c^2)^2 = 1 \implies |ab + bc + ac | \leq 1 \).

\(ab + bc + ac \) can take a maximum value of 1 when \(a=b=c=\pm\frac1{\sqrt3} \).

Now for the minimum value.

Since \(a^2 + b^2 + c^2 = 1\), then \(a,b,c\) are in the range \([-1,1] \). And so \[a + b + c \geq -3 \implies (a+b+c)^2 \geq 0 \implies \underbrace{ (a^2 + b^2 + c^2)}_{=1} + 2 (ab + ac + bc) \geq 0 \implies ab + ac + bc \geq -\frac12 . \]

\(ab + bc + ac \) can take a minimum value of \(-\frac12\) when the unordered solution of \((a,b,c) \) is \(\left(0, \frac1{\sqrt2} , -\frac1{\sqrt2} \right) \).

Hence, the range of \((a,b,c) \) is \( \left [ -\frac12 , 1 \right ] \).

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Problem 1Plug in \( \frac 1x \) and simplify using logarithm and power rules rules

\[ \begin{align} f \left( \frac 1x \right) &= \log \left(\left(\frac 1x\right)^{\log\left(\left(\frac 1x\right)^{\log\left(\frac 1x\right)}\right)}\right) \\ &= \log \left(\left(\frac 1x\right)^{\log\left(\left(\frac 1x\right)^{-\log x}\right)}\right) \\ &= \log \left(\left(\frac 1x\right)^{\color{red}\log\left( x^{\log x}\right)}\right) \\ &= {\color{red}\log\left( x^{\log x}\right)}\cdot \log \left(\frac 1x\right) \\ &= \log\left( x^{\log x}\right)\cdot \left(-\log x\right) \\ &= (-1)\cdot\log\left( x^{\log x}\right)\cdot \left(\log x\right) \\ &= (-1)\cdot\log \left(x^{\log\left(x^{\log x}\right)}\right) \\ &= (-1) \cdot f(x) \\ &= -f(x) \end{align} \]

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Can you explain me how you got the 4th step.

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It's a use of the logarithm rule \( \log_b\left(a^c\right)=c\cdot\log_b(a) \). I've added an intermediate step to make this clear, and the exponent is also marked in red.

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Problem 4Firstly, realize that \(x=-2\) is not in the domain. This will be important later.

Next, set up the question of domain into an inequality. Notice that \(\frac{|x+3|+x}{x+2}\) must be greater than or equal to 1 in order for the function to be real. Therefore, an inequality can be set up: \[\frac{|x+3|+x}{x+2}\ge 1\]

Solving for a standard inequality:

\[|x+3|+x\ge x+2\] \[|x+3|\ge 2\] \[x\le -5;~ x\ge -1\]

However, it is always good to check one's work. Plugging in -3 for \(x\) checks out, but one can see that all positive values will result in outputs greater than 1. Therefore, one must go back for errors.

Notice that when \(x<-2\), \(x+2\) is negative. While this is a simple and obvious observation, this is key to understanding the error. When solving inequalities, the sign must be flipped when multiplying by a negative number. Therefore, one must solve for \(|x+3|+x\ge x+2\) when \(x>-2\), and \(|x+3|+x\le x+2\) when \(x<-2\).

Solving for \(x<-2\):

\[|x+3|+x\le x+2\] \[|x+3|\le 2\] \[-5\le x\le -1\]

Because \(x<-2\), \[-5\le x<-2\]

To find the other half of the domain, use the original answer and the inequality \(x>-2\):

\[-1\le x\]

To put the final domain in interval notation: \(x\in [-5,~-2),~[-1,~\infty)\)

Sorry for the long explanation! I wanted to put my whole thought process in the explanation so people would understand how they got caught up. I can reduce this if preferred.

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Excellent Explanation. This must be long enough for this explanation for the people to understand. No there is no problem with the length.

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Problem 2If we plug in \( x-2 \) into the equation and rearrange, we get

\[ f(x+2) = x^2 - f(x) \]

\[ f(x-2) = (x-2)^2 - f(x) \]

Since we want to find \( f(x+2)-f(x-2) \) let's replace them by what we've got

\[ \begin{align} f(x+2)-f(x-2) & = [x^2-f(x)]-[(x-2)^2-f(x)] \\ & = x^2-f(x)-(x^2-4x+4-f(x)) \\ & = x^2-f(x)-x^2+4x-4+f(x) \\ & = \boxed{4x-4} \end{align} \]

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Thank you. I did some mistake in the first step itself.

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Problem 6An odd function satisfies \( f(-x)=-f(x) \), so let's plug in \( -x \) and simplify.

Note that \( \sin x \), \( \tan x \) and \( x \) are all odd functions.

\[ \begin{align} f(-x) &= \frac {2(-x)(\sin(-x)+\tan(-x))} {2\left\lfloor \frac {-x+2\pi}{\pi} \right\rfloor -3} \\ &= \frac {(-1)2x(-\sin x -\tan x)} {2\left\lfloor -\frac {x}{\pi}+2 \right\rfloor -3} \\ &= \frac {2x(\sin x +\tan x)} {2\left\lfloor -\frac {x}{\pi}+2 \right\rfloor -3} \\ &= \frac {2x(\sin x +\tan x)} {2\left\lfloor -\frac {x}{\pi}\right\rfloor +2-3} \\ &= \frac {2x(\sin x +\tan x)} {2\left\lfloor -\frac {x}{\pi}\right\rfloor -1} \\ &{\color{red}\stackrel{?}{=}} {\color{red}\frac {2x(\sin x +\tan x)} {2\left(\left\lfloor \frac {x}{\pi}\right\rfloor-1\right)-1}} \\ &= \frac {2x(\sin x +\tan x)} {2\left\lfloor \frac {x}{\pi}\right\rfloor-2-1} \\ &= \frac {2x(\sin x +\tan x)} {2\left\lfloor \frac {x}{\pi}\right\rfloor-3} \\ &= -\frac {2x(\sin x +\tan x)} {3-2\left(\left\lfloor\frac {x}{\pi}+\frac{2\pi}{\pi}-2\right\rfloor\right)} \\ &= -\frac {2x(\sin x +\tan x)} {3-2\left(\left\lfloor\frac {x+2\pi}{\pi}\right\rfloor\right)+4} \\ &= -\frac {2x(\sin x +\tan x)} {7-2\left\lfloor\frac {x+2\pi}{\pi}\right\rfloor} \\ \end{align} \]

Here, I am kind of stuck, because it doesn't look like this function is really odd, maybe someone can go on or correct my steps. I'm not quite sure if the step marked with \( {\color{red}\stackrel{?}{=}} \) is valid.

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Your technique is good but I too think that there is a mistake.

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I'm still not sure about the marked step, but at least it's almost what I wanted to get in the end.

\[ \begin{align} f(-x) &= \frac {2(-x)(\sin(-x)+\tan(-x))} {2\left\lfloor \frac {-x+2\pi}{\pi} \right\rfloor -3} \\ &= \frac {(-1)2x(-\sin x -\tan x)} {2\left\lfloor -\frac {x}{\pi}+2 \right\rfloor -3} \\ &= \frac {2x(\sin x +\tan x)} {2\left\lfloor -\frac {x}{\pi}+2 \right\rfloor -3} \\ &= \frac {2x(\sin x +\tan x)} {2\left\lfloor -\frac {x}{\pi}\right\rfloor +4-3} \\ &= \frac {2x(\sin x +\tan x)} {2\left\lfloor -\frac {x}{\pi}\right\rfloor +1} \\ &= \frac {2x(\sin x +\tan x)} {2\left(-\left\lfloor \frac {x}{\pi}\right\rfloor-1\right)+1} \\ &= \frac {2x(\sin x +\tan x)} {-2\left\lfloor \frac {x}{\pi}\right\rfloor-2+1} \\ &= \frac {2x(\sin x +\tan x)} {-2\left\lfloor \frac {x}{\pi}\right\rfloor-1} \\ &= -\frac {2x(\sin x +\tan x)} {2\left\lfloor \frac {x}{\pi}\right\rfloor+1} \\ &= -\frac {2x(\sin x +\tan x)} {2\left\lfloor\frac {x}{\pi}+\frac{2\pi}{\pi}-2\right\rfloor+1} \\ &= -\frac {2x(\sin x +\tan x)} {2\left\lfloor\frac {x+2\pi}{\pi}\right\rfloor-4+1} \\ &= -\frac {2x(\sin x +\tan x)} {2\left\lfloor\frac {x+2\pi}{\pi}\right\rfloor-3} \\ \end{align} \]

EDIT: I've found my mistake, it was in line four, but now I corrected it and everything works out in the end.

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It's easy to see that \(f_1(x) \) is an even function. And if \(f_2(x) \) is an odd function, then so is \( f(x) = \frac{f_1(x)}{f_2(x)} \).

What's left to do is to prove that \(f_2(x) \) is an odd function. Or equivalently, we just have to prove that \(f_2(x) + f_2(-x) = 0 \).

Consider the identity, \( \{ x \} + \{ - x \} = 1\), where \( \{ \cdot \} \) denotes the fractional part function.

This identity can be made evident by superimposing two graphs \(g_1(x) = \{x \} \) and \(g_2 (x) = \{-x\} \) together. Similarly, \[ \left\{ \frac x\pi \right \} + \left\{ -\frac x\pi \right \}= 1\quad \implies \quad \left ( \frac x\pi - \left\lfloor \frac x\pi \right \rfloor \right ) + \left ( \frac {-x}\pi - \left\lfloor \frac {-x}\pi \right \rfloor \right ) = 1 \implies \left\lfloor \frac x\pi \right \rfloor + \left\lfloor \frac {-x}\pi \right \rfloor = -1.\] Hence, \[\begin{eqnarray} f_2(x) + f_2(-x) &=& 2 \left \lfloor \frac{x + 2\pi}{\pi} \right \rfloor - 3 + 2 \left \lfloor \frac{-x + 2\pi}{\pi} \right \rfloor - 3 \\ &=& 2 \left( \left \lfloor \frac{x + 2\pi}{\pi} \right \rfloor + \left \lfloor \frac{-x + 2\pi}{\pi} \right \rfloor \right) - 6 \\ &=& 2 \left( \left \lfloor \frac x \pi + 2 \right \rfloor + \left \lfloor \frac {-x} \pi + 2 \right \rfloor \right) - 6 \\ &=& 2 \left( \left \lfloor \frac x \pi \right \rfloor + \left \lfloor \frac {-x} \pi \right \rfloor \right) - 6 + 8 = 2(-1) + 2= 0. \end{eqnarray} \]

And we're done!

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@Agnishom Chattopadhyay, @Chew-Seong Cheong, @Pi Han Goh , @Blan Morrison, @X X

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My goodness! I am honored to be recognized as someone to be reached out to! Thank you for tagging me and showing me these challenges!

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Problem 8.\(f(x)=f(a+(x-a))=f(a-(x-a))=f(2a-x)=f(b-(x+b-2a))=f(b+(x+b-2a))=f(x+(2b-2a))\), so \(f(x)\) is a periodic function.Since \(a>b\), the period is \(|2b-2a|=2a-2b\)

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Can someone here please suggest some good olympiad papers similar to the Indian INMOs?

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