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we r given that the curves y =integation from -infinity to x f(t)dt through the point (0,1/2) and y=f(x),where f(x)>0 and f(x) is differntiable for x belongs to R through (0,1).If tangents dranw to both the curves at the point having equal abscissae intersect on the same point on x axis then 


no. of solutions f(x)=2ex = ?

4 years ago

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Answers : (2)

										

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4 years ago
										

From the first relation,


 


dy/dx = f(x)


Equation of tangent at (0,1/2):


 


(y-0.5)/(x-0) = f(x)


or, x.f(x) = y-1/2


 


From the second relation,


 


dy/dx = d(f(x))


 


Equation of tangent at (0,1):


 


(y-1)/(x) = d(f(x))


 


x.d(f(x)) = y-1


 


Note (0,1/2) and (0,1) has same abscissae, so,


On the x-axis, let the common point be (h,0)


Both the equations should satisfy this point.


 


h.f(h) = -0.5          ............(i)


h.d(f(h)) = -1           ...........(ii)


 


dividing (i) and (ii),


 


d(f(h))/f(h) = 2


Integrating both sides,


 


ln (f(h)) = 2h + c


f(h) = e^(2h+c)


 


the function is f(x)=e^(2x+c)


 


Given the y=f(x) passes through (0,1), putting the values, in the above relation,


 


1 = e^(c)


or, c = 0


 


therefore the funtion is, f(x) = e^(2x)


 


 


Now,


 


f(x) = 2e^(x)


or, e^(2x) = 2.e^(x)


or, e^(x) = 2


or, x = ln2


 


so, i get just one solution. And i m really curious to know the answer. :) ..

4 years ago

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