3
1
2
4
Given ydx−(x+2y²)dy=0⇒ydx−xdy=2y²dy⇒ydx−xdy/y²=2dy⇒d(x/y)=2dyIntegrating we get,⇒(x/y)=2y+c⇒2y²+cy=x=f(y)Given f(√2)=1⇒2(√2)²−c=1⇒c=1⇒f(y)=2y²+y∴f(1)=2(1)²+1=3