Functions Doubt

[prisma holly]

<@&1227988399579730072>

Note for OP

+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.

[hushed perch]

Also if you don't mind, what is the source of the question?

[hazy parcel]

This is a pretty good question

[hushed perch]

Ye exactly.

[lavish flax]

Our sir gave the ques by their

maybe its a original ques created by him

or taken from some book

[hushed perch]

Ah I see.

[lavish flax]

Please explain the first step

how f(x) < x

[lavish flax]

fr but I am right now dumb for it

I subtly imporiving it will take time

[hazy parcel]

f(x+1) - f(x) ≤ (x+1) - (x) for all x.

[lavish flax]

oh

how do you get to know how to transform it

like intead of writing 1

[hazy parcel]

Practice.

[lavish flax]

you did

[lavish flax]

I do still

thinkinh of approach takes too much time

[hazy parcel]

It does yeah

[lavish flax]

[hushed perch]

Oh sorry mb.
Consider the following inequalities:
f(x+1)-f(x)<=1
f(x)-f(x-1)<=1
and so forth till f(2)-f(1)<=1
Add all these and you get f(x+1)<=x+f(1)=x+1
So f(x+1)<x+1 or f(x)<x

Or just look at it like this: difference between f(x+1) and f(x) will be at max 1. So the max. value of f(x) can be attained for x in natural when f(1),f(2)... are terms of an A.P. with common difference 1.