Putting y = x we have,

$2f(x) + {\left[ {f(x)} \right]^2} = 1$

$2f(x) + {\left[ {f(x)} \right]^2} + 1 = 1 + 1$

${[f(x) + 1]^2} = 2$

$f(x) = - 1 \pm \sqrt 2 $

Since f(x) is a constant function, $f'(x) = 0$.

JEE Main, JEE Advanced & NEET Online & Bhopal Coaching