1000 questions! Looks nice

Geometry Level 4

f ( x ) = 2 x ( sin ( x ) + tan ( x ) ) 2 2 + x π 3 \large f(x)=\frac{2x(\sin (x)+ \tan( x))}{2\left \lfloor 2+\frac{x}{\pi} \right \rfloor -3}

For all real x x , which of these answer choices is the property of f ( x ) f(x) ?

f ( x ) f(x) is even function f ( x ) f(x) is odd function f ( x ) f(x) is neither odd nor even function f ( x ) f(x) is even odd simultaneously

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1 solution

Chew-Seong Cheong
Apr 24, 2015

This is a tricky problem.

It is given that f ( x ) = 2 x ( sin x + tan x ) 2 2 + x π 3 = g ( x ) h ( x ) f(x) = \dfrac {2x(\sin{x}+\tan{x})}{2 \lfloor 2+\frac{x}{\pi}\rfloor -3} = \dfrac {g(x)}{h(x)}

We note that:

  • sin x \sin{x} is odd, tan x \tan{x} is odd and the sum of two odd functions is odd ( sin x + tan x ) \Rightarrow (\sin{x}+\tan{x}) is odd.
  • 2 x 2x is odd, ( sin x + tan x ) (\sin{x}+\tan{x}) is odd and the product of two odd functions is even g ( x ) = 2 x ( sin x + tan x ) \Rightarrow g(x) = 2x(\sin{x}+\tan{x}) is even.
  • We note that h ( x ) = 2 2 + x π 3 h(x) = 2 \lfloor 2+\frac{x}{\pi}\rfloor -3 is strictly not an odd function because h ( n π ) = h ( n π ) 2 h (-n\pi) = -h(n\pi)-2 , where n n is an integer but for x n π x \ne n\pi , h ( x ) = h ( x ) h(-x) = -h(x) , it is an odd function.
  • But we find that g ( n π ) = 0 g(n \pi)=0 , therefore f ( x ) f(x) is, in effect, a quotient of an even function g ( x ) g(x) and an odd function h ( x n π ) h(x\ne n \pi) , which is an odd function.

Therefore, f ( x ) is an odd function \boxed{f(x) \text{ is an odd function}} .

This is how it looks like. Isn't it odd?!!!

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