Let f(x)=x^2+a x+b . If x R , there exist a real value of y such that f(y)=f(x)+y , then find the maximum value of 100 a

Step-by-step solution

Given, f(y)=f(x)+y aligned & y^2+a y+b=x^2+a x+b+y & y^2+y(a-1)-x^2-a x=0 aligned As, y R , so D 0 x R aligned & (a-1)^2+4 (x^2+a x ) 0 x R & 4 x^2+4 a x+a^2-2 a+1 0 x R aligned Now, D 0 aligned & 16 a^2-16 (a^2-2 a+1 ) 0 & 2 a-1 0 a 1 2 aligned So, a_ . = 1 2 Hence, maximum value of 100 a =50