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Let f(x) = ax 2 +bx+c, where a,b,c belongs to R and a is not equals to zero. If f(x) = x has non real roots, then show that the eqn f(f(x)) = x has all non real roots.

Let f(x) = ax2+bx+c, where a,b,c belongs to R and a is not equals to zero. If f(x) = x has non real roots, then show that the eqn f(f(x)) = x has all non real roots.

Grade:12

3 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question below

f(x) = ax^2 + bx + c
f(x) =x
ax^2 + bx + c = x
ax^2 + (b-1)x + c = 0
This equation have non real roots.
(b – 1)2< 4ac
f(f(x)) = x
a(f(x))^2 + 2bf(x) + c = x
a(ax^2 + bx + c)^2 + 2b(ax^2 + bx + c) + c = x
a(ax^4 + b^2x^2 + c^2 + 2abx^3 + 2bcx + 2acx^2) + 2b(ax^2 + bx + c) + c = xa^2x^4 + (b^2 + 2ac + 2ab)x^2 + (c^2 + 2bc + c) + 2abx^3 + (2bc-2b ^2-1)x = 0

a^2x^4 + 2abx^3 + (b^2 + 2ac + 2ab)x^2 + (2bc-2b^2-1)x + (c^2 + 2bc + c) = 0
So this equation also have non real roots.
mycroft holmes
272 Points
6 years ago
Without Loss of Generality we can assume that a>0.
 
Then f(x) = x has no real roots implies that f(x)> x for all real x.
 
This also implies that f(f(x))>x for all real x.
 
This means f(f(x)) – x = 0 has non-real roots.
mycroft holmes
272 Points
6 years ago
I need to add an explanation:
 
f(x)> x for all x implies f(f(x))>f(x)>x. That’s how we get the inequality f(f(x))>x for all x

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