isi 2004

[seema]

f(x)= (x-a)^3 + (x-b)^3 + (x-c)^3, a< b< c. The number of real roots of f(x)=0  ?

a) 3
b) 2
c) 1
d) 0

[anon_econ]

I guess it should be 1. It's a strictly increasing function and it's everywhere continuous..

[deepak]

I believe Vasudha is right.

This represents the equation of a strictly increasing function (first order : 2[(x-a)^2+(x-b)^2+(x-c)^2] - always positive)

Also, the second order condition shows that the function has an inflection point at (a+b+c)/3. So it has 1 real root.