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JEE MAIN & ADVANCE 12th PCM Maths -Relation & Function Module-1 Demo Video

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Hello, students, so continuing with examples on even and odd functions, next we have this, we need to prove two statements.

Now number one, we will be starting with g of minus x, which is equal to this. Try to see, it is equal to g of x, so g must be an even function. Proceeding in a similar function, second, g of minus x is equal to this which is nothing of minus of g x, so g must be an odd function. Remember f of x plus f of minus x is an even function for any f. And f of x minus f of minus x is an odd function for any f.

Now we have an important remark, try to see, “Every function can be represented as a sum of an even function and an odd function”. How, let us see, try to see.  We just proved that this is an even function and this is an odd function, try to see if we add them we get f of x, so I wrote f of x as a sum of even part and odd part. So that’s what this remark is all about. So whenever you are asked to write a function as a sum of even and odd function, you have to do it like this, okay.

Moving ahead we have next example that is this, we need to check if it is an even or odd function. First step is clear, you have to start with f of minus x which is this. Now writing e par minus x as 1 by e par x and taking LCM we have this. Now we add and subtract x in the numerator. Why? I will tell you. Try to see the first term will be broken into two terms, this. Advantage is 1 minus e par x cancels. Now why I did that, because I want to prove that f of minus x looks like fx or minus fx. So in numerator I do not need e par x as in f of x as there is no e par x in the numerator. So I wanted to cancel it, and hence I added and subtracted x. So moving like this equal to f of x so this f must be an even function.

At last we have this, I want you to prove this question at home, I give you two hints, I am your friend. Number one is this, that integer comes out of gint under addition and number 2 is gint of minus x is minus 1 minus gint of x for k not belonging to  integer. In the example try to see x plus pi divide by pi. It will be x by pi plus 1, so 1 will come out, using Hint number 1. Then now try to see, you will have to start with f of minus x. So obviously in the denominator inside gint you will get minus x by pi. So this minus will be tackled using Hint number 2. Further you can solve it, I have full confidence on you.

So best of luck, take care, god bless.

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