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If $\log_{2}(f(x)+|\sin x|)=\log_{2} x$ then:

A) $f(x)>0$ for each $x \in R$

B) $\lim_{x\to\infty}f(x)= +\infty$

C) the function is strictly increasing

D) $f(\pi)=\pi$

So firstly I define domain $f(x)+|\sin x|>0 \implies f(x)>-|\sin x|$

And we have $f(x)+|\sin x|=x \implies f(x)=x-|\sin x|$ but I'm not sure about next step

$x-|\sin x|>-|\sin x| \implies x>0$

So the answers B, C, D are correct?

egreg
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Mark
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3 Answers3

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B) is correct because $f(x)=x-| \sin x |$ is dominated by $O(x)$ as $x \to \infty$.

D) is correct if you plug in $\pi -|\sin \pi |=0$

EDIT:

C) there's another (slower) way of showing monotonicity of the function without taking the derivative. Consider

$$ f(x+a)-f(x)=a+|\sin x|-|\sin(x+a)| \geq 0 $$ We need to show this difference is always positive for $a>0$. This amounts to showing $$ |\sin (x+a)| -|\sin x| \leq a\\ $$ RHS is always positive by definition of $a$. LHS is negative for $x \in [\frac{\pi}{2}(2k-1), \pi k], k \in \mathbb{Z}^+ $, so the inequality is trivial. For the positive segments of the function ($x \in [\pi k, \frac{\pi}{2}(2k+1)], k \in \mathbb{Z}$) the difference is at most $1$. Due to periodicity it's enough to consider only the first segment, where the function coincides with $\sin x$. What we need to show is that $$ \sin(x+a)- \sin x <a $$ for $0<x<\frac{\pi}{2}$. We star by expanding $\sin (x+a)$: $$ \sin (x+a)- \sin x = \sin x \cos a +\sin a \cos x - \sin x \\ = \sin x(\cos a -1) +\sin a \cos x \\ \leq \sin a \cos x \\ \leq \sin a \\ \leq a $$ The first inequality is due to $\sin x >0$ and $\cos a<1$, the second inequality is due to $\cos x <1$.

Hence $|\sin(x+a)|-|\sin x| \leq a \ \forall \ a>0$ and $f(x)$ is a monotonically increasing function.

Alex
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$\log_{2}(f(x)+|\sin x|)=\log_{2}x$ implies that, $f(x)=x-|\sin x|$. Therefore it's obvious that A) is incorrect (take for example $x=-1$). Can you try out to find whether the remaining statements are correct?

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In order that the condition is satisfied, you must have $$ f(x)+|\sin x| = x,\quad x>0,\quad f(x)+|\sin x|>0 $$ The first gives $f(x)=x-|\sin x|$ that is true for $x>0$, because it's well known that $\sin x<x$ for all $x>0$.

Is $f$ strictly increasing? Well, $f'(x)=1-\cos x\frac{|\sin x|}{\sin x}$ where $\sin x\ne 0$, which is positive where defined (why can't it be zero?). So $f$ is indeed strictly increasing.

Questions A, B and D are easily settled.

egreg
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