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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 => R = 2√2/tanα = 8 units Alternate Solution: We have , from the figure (R + 1)2 = (R - 1)2 + (4√2)2 => R = 8. << Back || Next >>
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
=> R = 2√2/tanα = 8 units
Alternate Solution:
We have , from the figure
(R + 1)2 = (R - 1)2 + (4√2)2
=> R = 8.
<< Back || Next >>
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IIT JEE 2009 Mathematics Paper2 Code 1 Solutions...