IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

sneha_91

using lmvt prove

ln(1+x)<x , x>o

jaggy

first cosider ln(1+x)-x as a function and u can see dat its cont. and diff.

now its derivative is 1/1+x -1.thus its -x/1+x which is decreasin 4 every value of x>0.

hence f(x)<f(0)

and ans. x>ln(1+x).

h.p.

give me rates

mukundmadhav

Let f(x)= ln(1+x)
f'(x)= 1/1+x

Lagrange Mean Value theorem states that, if d lies between f'(a) and f'(b), then there exists at least one c between a and b such that f'(c) = d.

f'(x) does not change sign for x > 0
It is = 1 at x = 0.. So y = x is tangent at x = 0..
Now f(x) is continually decreasing for x > 0 and = x at x = 0
So x > f(x)

RyuAmakusa

@mukund i think that is not the statement of lmvt anyway now
applying lmvt beetween (0,x) x is any no. >0
f'(x)= f(x)-f(0)/x f(x)=ln(1+x)-x =>f'(x)=(ln(1+x)/x)-1 now ln function increases at a rate less than x so f'(x)<0=>ln(1+x)/x-1<0
=>ln(1+x)

jaggy

yeah mukund is wrong and my soln. is correct..

at least give me rates

mukundmadhav

It is the correct statement..The statement in most textbooks, which is that f'(c) = 0 is only partially complete..

jaggy

can u give me full explnation.

RyuAmakusa

well mukund f'(c)=0 is not partially complete it is totally wrong

mukundmadhav

Ok I left out d = f(b) - f(a) / (b-a) in the original statement... Happy now?

RyuAmakusa

and 1 more thing it does not say f'(a)<d<f'(b) it says a<c<b.....and.......what is there to be happy in this?????are u angry???if i should not have said that then i am sorry..

mukundmadhav

That thing is right.. Think about it.. And I checked it on the net re.. Nothing wrong with it..

jaggy

good reply from ray

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