ISI 2014 DOUBT

[varnika1880]

GUYS PLZ HELP WITH Q 27http://http://www.isical.ac.in/~deanweb/Syll-Sample-MSQE-PEA-2014.pdf

[duck]

Hi.. :)

Given: f(x+y) f(x-y) = (f(x) + f(y))^2 - 4 x^2 f(y) holds for all values of x and y.

→ The above equation should hold when x = y = 0 i.e
2f(0) = 4f(0) ⇒ f(0) = 0 ..... (I)

→ The above equation should hold when x = y i.e
f(2x) f(0) = (2f(x))^2 - 4x^2f(x)
⇒ f(x) [ x^2 - f(x)] = 0
⇒ f(x) = 0 or f(x) = x^2

Therefore, f(x) ≠ 2.
Option (d) is the answer.

[varnika1880]

THANKS A LOT