Let f(x)=(x-a)(x-b)- ( a+b 2 ) . If f(x)=0 has both non-negative roots, then the minimum value of f(x) .

Step-by-step solution

Let f(x)=(x-a)(x-b)- ( a+b 2 ) Now, f^ (x)=(x-b)+(x-a)=2 x-b-a Now, f^ (x)=0 aligned x-b+x-a & =0 2 x=a+b x & = a+b 2 aligned Now, f^ (x)=2>0 So, at x= a+b 2 , f(x) has minimum value. Now, at x= a+b 2 aligned f(x) & = ( a+b 2 -a ) ( a+b 2 -b )- ( a+b 2 ) & = -(a-b)^2 4 - ( a+b 2 ) & =- 1 2 a^2+b^2-2 a b+2 a+2 b 2 ] & =- 1 2 a^2+b^2+2 a b-4 a b+2(a+b) 2 ] aligned aligned & Given, (x-a)(x-b)= a+b 2 & x^2-(a+b) x+ ( 2 a b-a-b 2 )=0 aligned Now, roots are non-negative, so (a+b)>0 and 2 a b-(a+b) 2 >0 4 a b-2(a+b) 4 >0