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)).

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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