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```				   If y=log e x then for x>1, show that we must have
(a) x-1>y    (b) x2-1 >y    (c) (x-1) / x
```

7 years ago

Share

```										 hi,
if two functions are equal at a point , & after that point if one has always a high derivative than other , it means that it increases  more with the same change in x than other,it clearly emplies that it would always be high than other
(a) x-1>y
at x =1 ,
l.h.s.=r.h.s . = 0
d/dx(l.h.s) = d/dx(x-1) =1
d/dx(r.h.s) = d/dx (log e x) = 1/x
if x >1,
1/x <1
d/dx(l.h.s) > d/dx(r.h.s)
emplies,
x-1>y

(b) x2-1 >y  ,                at x=1
l.h.s = rh.s = 0
d/dx(l.h.s) =d/dx (x2-1)
= 2x
d/dx(r.h.s) =d/dx (log e x ) =  1/x
we know,
for x>1 ,
2x >1/x
hence,
x2-1 >y

(c)  (x-1) / x <y
at x= 1,

l.h.s = rh.s = 0,

d/dx(l.h.s) =d/dx ( (x-1) / x )
= 1/x2
d/dx(l.h.s) =d/dx (log e x )
= 1/x
for x>1 ,
1/x2 <1/x, hence
(x-1) / x <y
```
7 years ago
```										hi tushar,

x = e^y , now for 1st case substitute e^y in place of x and u will get (e^y - 1) > y, here u can see that the function on the left hand side is  a exponential function and for x>1, e^y > 0 ,also for any value of y , the left side fxn will increase with a slope of e^y while right side fxn will increase only with a slope of 1 so, for all value of y ,(e^y - 1)>y........

similarly for 2nd case the eqn is (e^2y - 1)> y ,we can say that it is always valid for x>1......

now in 3rd case , ur eqn wud be (1 - 1/e^y) here u can see its slope is 1/e^y , so its slope is always less than 1 for all x>1....so the given condition (x-1) / x
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best .

Regards,
IIT - Delhi

```
7 years ago

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