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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 = ?

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 = ?

Grade:11

2 Answers

SIBASHIS MAHAPATRA
14 Points
10 years ago

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

Jit Mitra
25 Points
10 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. :) ..

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