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Let f(x) = x 3 +x 2 +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)).
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 --- 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, Naga Ramesh IIT Kgp - 2005 batch
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
---
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, Naga Ramesh IIT Kgp - 2005 batch
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