FunctionHard
Question
A real valued function f satisfying the equation f(x) + 2f(-x) = x2 - x, then sum of real roots of equation f(f(x)) = 0 is -
Options
A.- 3
B.- 6
C.- 9
D.- 10
Solution
f(x) + 2f(-x) = x2 - x ......(i)
Replace x → - x
f(-x) + 2f(x) = x2 + x
solving (i) & (ii) we get f(x) =
f(f(x)) =
= 0
⇒ f(x) = 0 or f(x) = - 3
⇒
= 0 or = - 3
⇒ x = 0 ; -3 or x2 + 3x + 9 = 0 (No real roots)
∴ sum = - 3
Replace x → - x
f(-x) + 2f(x) = x2 + x
solving (i) & (ii) we get f(x) =
f(f(x)) =
= 0⇒ f(x) = 0 or f(x) = - 3
⇒
= 0 or = - 3⇒ x = 0 ; -3 or x2 + 3x + 9 = 0 (No real roots)
∴ sum = - 3
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