Pratham Ashish
Last Activity: 15 Years ago
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