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can someone please help with finding an example satisfying the condition, if possible. i tried doing it. bt couldnt find.

shiya , 6 Years ago
Grade 12th pass
anser 3 Answers
Arun

Last Activity: 6 Years ago

Dear Chitra
 
I am getting my answer as option D i.e. f(x) has no real root.
Please let me know if this is cprrect then I will share my solution

shiya

Last Activity: 6 Years ago

@ Arun you are right. can you please state an exmple. i thought of f(x)=x^2 and g(x)=x^2. but this is not satisfying the condition. 

Arun

Last Activity: 6 Years ago

Let xo be a. real root of f(x)
 
then you have f(x02 + 1) = f(x0) g(x0) = 0
 
that is xo2 +1 is a root of f(x) as well
 
since h(x) = x2 + 1 > x for x belongs to R,it is clear that any hk (x0) is rpeal root of f(x), even with k> m
 
But f(x) has a maximum of m roots. Hence contradiction, f(x) has no realroot

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