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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
9 years ago

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

Jit Mitra
25 Points
9 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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