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 = ?

2 years ago


Answers : (2)


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. :) ..

2 years ago

hey,this question is in the GRAND MASTERS PACKAGE.Have you completed all the questions of GMP?

2 years ago

Post Your Answer

More Questions On Integral Calculus

Ask Experts

Have any Question? Ask Experts
Post Question
Answer ‘n’ Earn
Attractive Gift
To Win!!!
Click Here for details

More Questions On Integral Calculus

View all Questions »