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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)

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)