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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 = ?

3 years ago


Answers : (2)


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3 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)





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. :) ..

3 years ago

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