 # Doubt from functions

For 1st am getting 6 roots

Don't have the answer bro...share solution...
@Raunak_2020 try this..

For first try to make graph of function let
f(x) has x_1,x_2,x_3 root
f(f(x))=0
f(x)=x_1 nd two more root imagining then after
x_1 will be constant u know graph of f(x)
No of time it cuts u will get the solutions

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Its correct @akash_2020
For second also try to make graph

@akash_2020, @Sneha_2021 what will we do if the cubic is not getting factorised....like here we knew about one of the factors being 2 so we could easily locate the other 2 roots. ..
Say for example x^3 - 3x +1

Ivt theoram will give u idea of roots (jee domain theoram)

Other another way is descartes theoram of polynomial (rmo type) no need for jee but telling one aspect of telling roots

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@Sneha_2021 for x^3-3x+1=0, number of different real solutions of f (f (x)) =0 should be 7?? Pls check..
One of my roots is between 0 and 1 and the other 2 between (1,infinity) and (-infinity,-1)..

Yes 7 is correct

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