logarithms and functionIf log 2 ⁡ ( f ( x ) + | sin ⁡ x | ) = log 2 ⁡ x then:A) f ( x ) &gt; 0 for each x ∈ RB) lim x → ∞ f ( x ) = + ∞C) the function is strictly increasingD) f ( π ) = πSo firstly I define domain f ( x ) + | sin ⁡ x | &gt; 0 ⟹ f ( x ) &gt; − | sin ⁡ x | And we have f ( x ) + | sin ⁡ x | = x ⟹ f ( x ) = x − | sin ⁡ x | but I'm not sure about next step x − | sin ⁡ x | &gt; − | sin ⁡ x | ⟹ x &gt; 0So the answers B, C, D are correct?

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wasipewelr · Accepted Answer

B) is correct because   f  (  x  )  =  x  −      |    sin  ⁡  x      |   is dominated by   O  (  x  ) as   x  →  ∞D) is correct if you plug in   π  −      |    sin  ⁡  π      |    =  0EDIT:C) there&#039;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  )      |    ≥  0We need to show this difference is always positive for   a  &amp;gt;  0. This amounts to showing      |    sin  ⁡  (  x  +  a  )      |    −      |    sin  ⁡  x      |    ≤  a  RHS is always positive by definition of   a. LHS is negative for   x  ∈  [      π    2    (  2  k  −  1  )  ,  π  k  ]  ,  k  ∈            Z        +  , so the inequality is trivial. For the positive segments of the function (  x  ∈  [  π  k  ,      π    2    (  2  k  +  1  )  ]  ,  k  ∈      Z  ) the difference is at most   1. Due to periodicity it&#039;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  &amp;lt;  afor   0  &amp;lt;  x  &amp;lt;      π    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    ≤  sin  ⁡  a  cos  ⁡  x    ≤  sin  ⁡  a    ≤  aThe first inequality is due to   sin  ⁡  x  &amp;gt;  0 and   cos  ⁡  a  &amp;lt;  1, the second inequality is due to   cos  ⁡  x  &amp;lt;  1.Hence       |    sin  ⁡  (  x  +  a  )      |    −      |    sin  ⁡  x      |    ≤  a     ∀     a  &amp;gt;  0 and   f  (  x  ) is a monotonically increasing function.

Michelle Mendoza · Accepted Answer

log          2        ⁡  (  f  (  x  )  +      |    sin  ⁡  x      |    )  =      log          2        ⁡  x implies that,   f  (  x  )  =  x  −      |    sin  ⁡  x      |  . Therefore it&#039;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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