Re: ISI 2014 PEA Answer Key

Posted by JMKeynes on May 03, 2016; 6:31pm
URL: http://discussion-forum.276.s1.nabble.com/ISI-2014-PEA-Answer-Key-tp7587867p7600340.html

For q27 answer is (d).

If the function is of the form f(z)=z,
then from the given equation (let's call it it eq1): f(x+y)*f(x-y)=
􀀀(f(x) + f(y))^2 - 4*x^2*f(y),
we have (x+y)*(x-y) = (x+y)^2 - 4*x^2*y

Solving we get a quadratic equation 2*x^2-x-y=0.
Solving for x, we get,

 x=(1+(1+8y)^.5) or x=(1-(1+8y)^.5)..........(2)

So for a given value of y, eq1 will not hold for any x not satisfying (2). But as per the question eq1 should hold for all x,y that are real.
Contradiction!
So a function of the form f(z)=z cannot hold. Or, f(2)=2 cannot hold.