let quadratic equastion f(x) = ax^2+bx+c,
given f(1) = 3, f(0)= 1
at x=0 f(0)=1 so a.0+b.0+c=1 so c=1
x=1 f(1)= 3 so a+b+c= 3
a+b=2.... (1)
now f(10)= 100a+10b+1 ....(2)
And here given .. at x=1 f(x) is maximum so df(x)/dx=0
which is 2ax+b= 0 ,
puting x=1 2a=-b----(3)
using 1 and 3 a=-2, b=4
now using (2) we get f(10)= -159
so answer is (2)