Method 1 : (usual but lengthy)

x 2 f(x) + f(1 – x) = 2x – x 4 .....(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 – x 2

Method 2 :

Since R.H.S. is polynomial of 4 th degree and also by options consider f(x) = ax 2 + bx + c

x 2 f(x) + f(1 – x) = 2x – x 4

⇒ x 2 (ax 2 + bx + c) + a (1 – x) 2 + b (1 – x) + c = 2x – x 4

by comparing coefficients

a = – 1

b = 0

c = 1

∴ f(x) = – x 2 + 1

──────────────────────────────────────────────────────────────────────────────────────────