FunctionHard
Question
A function f : R → R satisfies the condition x2 f(x) + f(1 - x) = 2x - x4. Then f(x) is:
Options
A.- x2 - 1
B.- x2 + 1
C.x2 - 1
D.- x4 - 1
Solution
Method 1 : (usual but lengthy)
x2 f(x) + f(1 - x) = 2x - x4 .....(1)
replace x by (1 - x) in equation (1)
(1 - x)2 f(1 - x) + f(x) = 2 (1 - x) - (1 - x)4 .....(2)
eliminate f(1 - x) by equation (1) and (2)
we get
f(x) = 1 - x2
Method 2 :
Since R.H.S. is polynomial of 4th degree and also by options consider f(x) = ax2 + bx + c
x2 f(x) + f(1 - x) = 2x - x4
⇒ x2 (ax2 + bx + c) + a (1 - x)2 + b (1 - x) + c = 2x - x4
by comparing coefficients
a = - 1
b = 0
c = 1
∴ f(x) = - x2 + 1
x2 f(x) + f(1 - x) = 2x - x4 .....(1)
replace x by (1 - x) in equation (1)
(1 - x)2 f(1 - x) + f(x) = 2 (1 - x) - (1 - x)4 .....(2)
eliminate f(1 - x) by equation (1) and (2)
we get
f(x) = 1 - x2
Method 2 :
Since R.H.S. is polynomial of 4th degree and also by options consider f(x) = ax2 + bx + c
x2 f(x) + f(1 - x) = 2x - x4
⇒ x2 (ax2 + bx + c) + a (1 - x)2 + b (1 - x) + c = 2x - x4
by comparing coefficients
a = - 1
b = 0
c = 1
∴ f(x) = - x2 + 1
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