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Let f (x)= x^3-3x+1. Find the number of ...

Let f (x)= `x^3`-3x+1. Find the number of different real solution of the equation f (f(x) =0

A

2

B

4

C

5

D

7

Text Solution

Verified by Experts

The correct Answer is:
D
`f'(x)=3x^(2)-3=3(x^(2)-1)`
`f(-1)=3,f(1)=-1,f(3)=19gt0`
Graph of f(x) is as shown in the following figure.

`f(0)=0` has root `x_(1),x_(2),x_(3)` such that
`x_(1)lt-1,-1 lt x_(2)lt1, 1 ltx_(3)lt3`
`rArr" "f(x)=x_(1)` has only one real root since, `x_(1)lt-1`
`f(x)=x_(2)` has 3 real roots since `-1lt x_(2)lt1`
`f(x)=x_(3)` has 3 real roots since `1ltx_(3)lt3`
The roots of `f(f(x))=0` only if `f(x)=x, f(x)=x_(2),f(x)=x_(3)`
`therefore " Number real roots of "f(f(x))=0` is 7
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