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IIT JEE 2009 Mathematics Paper2 Code 1 Solutions

12.    Let f:R --> R be a continuous function which satisfies f(x) = ∫0 x f(t) dt  Then the value of f(ln5) is

Sol.   0

        f(x) = ∫0 x f(t) dt     =>  f(0) = 0

        also, f'(x) = f(x), x > 0

        => f'(x) = kex, x > 0

         f(0) = 0 and f(x) is continuous => f(x) = 0 for all x > 0

         f(ln5) = 0.

13.    The centres of two circles C1 and C2 each of unit radius are at a distance of 6 units from each other. Let P be the mid point of the line segment joining the centres of C1 and C2 and C be a circle touching circles C1 and C2 externally. If a common tangent to C1 and C passing through P is also a common tangent to C2 and C, then the radius of the circle C is

Sol.   8

        cosa = 2√2/3

        sina = 1/3

        tana = 2√2/R

       tangants

        => R = 2√2/tanα = 8 units

          Alternate Solution:

         We have , from the figure

        (R + 1)2 = (R - 1)2 + (4√2)2

        => R = 8.

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