1-If we have:

**[x]** **≤ (x/2)**

** **

Then which one is right?

1- **{x: x>1}**

2- **{x: x<1}**

3- **{x: -1<x<1}**

4- **{x: -2<x<0}**

******* ******* ******* ******* *******

2-Which one is **odd**?

1- **Ln {x+**√**(1+x ^{2})}**

2- **e^sinx**

3- **(e^|x|)/(x+1)**

4- **sechx**

## Replies

And also we can find it from their diagram:

A function is "even" when it is symmetry about the y-axis

A function is "odd" when we get origin symmetry.

http://www.mathsisfun.com/algebra/functions-odd-even.html

That's the sinh

^{-1}curve.http://www.ucl.ac.uk/Mathematics/geomath/level2/hyper/hy8b.html

Second problem ;

An odd number is an

integeranyway. But none of the choices give an integer for all values of x (in real numbers).I would say none is odd!

Dear Zahra, we must consider their function, and we have:

f(x) is

oddif we have:f(-x) = - f(x)And it is

evenif we have:f(-x) = f(x)** If

f_{1},fare_{2}evenandg_{1},gare_{2}oddand nonzero, and "a" , "b" be 2 real number and nonzero then we will have:ag_{1}+ bg= Odd_{2}af_{1}+ bf= Even_{2}af_{1}+ bg= Not even and not odd_{1}f_{1}f=_{2}g_{1}g= Even_{2}f_{1}/f=_{2}g_{1}/g= Even_{2}f_{1}g=_{1}f_{1}/g= Odd_{1}If

is an arbitrary function thenfis alwaysf(f_{1})even.Oopssss! How careless I am !!!! So sorry. I know about odd and even functions. But ...Soo Sorry! I feel embarrassed! Thank God my professors aren't here and can not see this stupid mistake of mine!Ohh...

Hmm...Full info about oddity and evenness of functions. Thankssss so much dear.

OK, now let me think.

1-

Ln {x+√(1+x^{2})} >>> correct answerThe statement above is another form for sinh

^{-1}x which is an odd function.2-

e^sinxSince sin(-x) is equal to -sinx , we have : e^(-sinx)=1/(e^sinx)not odd not even3-

(e^|x|)/(x+1)we have |-x|=|x|, so replacing x by -x gives: (e^|x|)/(-x+1) which isneither odd nor even4-

sechxsech x = 1/cosh x

cosh x= [e^(-x) + e^x]/2 . Replacing x by -x gives the same statement. So it is an even function . So sech x = sech (-x)

EVENLet's prove that the first choice is an odd function:

We replace x by -x: Ln {-x+√(1+x

^{2})} =f1If our function is odd, we should have f(-x)= -f(x)

We consider -f(x) =f2. We prove that f1 is equal to f2 which shows that our function is odd : f2= - Ln {x+√(1+x

^{2})} = Ln {[x+√(1+x^{2})]^(-1)}= Ln {1/(x+√(1+x^{2}))}We multiply the numerator and denominator of the fraction by the conjugate of the denominator which is {-x+√(1+x

^{2})}. So we get :f2= Ln {(-x+√(1+x

^{2}))/(-x^{2}+1+x^{2})}= Ln{-x+√(1+x^{2})^{ }} which is equal to f1. PROVED!W0W!!

Zahra!

You are a God in math!!

Very nice dear!

Very nice...

Well done!

I enjoyed reading your precise solution!

Very good, my dear.

Excellent!

Thanks again for your interesting discussions here. Math is my passion.

OK. As for the fist one, I don't know how to prove the total answer. But as it is a multiple choice, we can find a special case to cross out wrong choices.

We consider that x is an integer ( a special case) ; so [x] reduces to x; since integer part of an integer is the integer itself. Now we have : x≤ x/2 which gives: x≤ 0. So the first choice is wrong. [For rejecting this choice we could also pick a number which doesn't satisfy our condition; for example 3/2 .]

Also for 0 < x < 1 which is a positive interval but with

no positive integer; [x] = 0 , so the condition is satisfied. Adding this interval to x≤0, we have x<1.So I think

CHOICE 2is the general interval for our condition. Am I right?Well done!

Excellent dear!!

That's right.