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`         Let f(x) = x3+x2+100x+7sinx  then show that the eqn  1 /  [y-f(!)]   +  2 / [y-f(2)]   + 3 / [y-f(3)] = 0 has exactly one root lying in (f((1) , f(2)).`
8 years ago

Ramesh V
70 Points
```										let g(y) = 1 / [y-f(1)] + 2/[y-f(2)] +3/[y-f(3)]
g' (y) = -(1/[y-f(1)]2 + 2/[ y-f(2)]2 + 3/ [y-f(3)]2)
since g '(y) is negative summ of squares , its always negative
so, g '(y) < 0
g(y) is strictly decreasing function
here f(1) = 102 + 7 sin1
and f(2) = 212 + 7 sin2
clearly f(2) > f(1)
g(  f(1) + 0 ) ---> infinity
g(  f(2) - 0 ) ---> - infinity
So, the mean value theorem
between ( -infinity,infinity) as 0 will lie btn y=f(1) & y=f(2)
hence proved
---
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Regards,
Naga Ramesh
IIT Kgp - 2005 batch
```
8 years ago
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