IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ashish_banga

the set of values of all x for which

log ( 1 + x ) $\le$ x is

spideyunlimited

x E ( -inf, 0]

:)

the equality holds at x = 0
plot the graphs of e^x and x+ 1
and u get the desired region.

ashish_banga

answer is ( - 1 , infinity )

shreyasnivas

ashish is right

ashish_banga

are i have asked that question,
i have only provided answer
i want explanation
and u have rated me - what a mistake

ashish_banga

someone here

spideyunlimited

But look at this

:\

:?

.

shreyasnivas

whoops.. what a silly mistake.. well take the rate anyway.. lol.. spidey is right!!

allamraju

Alter method;

Let F(x)=ln(1+x)-x then F'(x)=-x/1+x $\le$ 0 iff x $\ge$ 0 or x<-1. But the fun. is not defined for x<-1

$\ge$ 0 iff -1<x<0

Now,x $\ge$ 0 $\Rightarrow$ F(x) $\le$ F(0)=0[since F is decreasing in this interval] $\Rightarrow$ ln(1+x) $\le$ x

-1<x<0 $\Rightarrow$ F(x) $\le$ F(0)=0[since F is increasing in this interval] $\Rightarrow$ ln(1+x) $\le$ x

animal

hey man it is easy just diff. both sudes as allamraju has done

a shortcut method

also keep in mind the domain of the original function.

we get 1/(1+x)<=1 {rolles theorem}

which is always posiible in the interval (-1,infinity).

hope u got it.

spideyunlimited

But see the graph

.in [-1, 0], log (x+1) >= x

:S

animal

yes spidey u r right in [-1,0) log(1+x)>x which is also seen graphically