stuck !

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if the function ₀âˆ«^x f(t) dt->5 as |x|->1,then the value of (a) so that the equation 2x + ₀âˆ«^x f(t) dt =a has atleast 2 roots of opposite sign in (-1,1) is
(a) a Îµ (0,1) (b) a Îµ (0,3) (c) a Îµ (-1,âˆž) (d) a Îµ (3,âˆž)
.....may be i am missing something vital ,here ![2]

#Mathematics #Calculus

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2 Answers

62

Lokesh Verma ·2009-12-09 20:38:32

at x=-1, the function 2x+0âˆ«x f(t) dt tends to 3

while at x=1, the function 0âˆ«x f(t) dt tends to 7

at x=0, the function is zero..

so for a root, one -ve and one +ve a should lie between 0 and 3

think in terms of the graph.

I am assuming that the function is continuous which is somehow not given here!

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Debotosh.. ·2009-12-09 20:46:00

yes, the book also says (b) is the answer ! great work ! hurrah ![1]

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