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If x is real, the minimum value of $f(x)=x^2-8x+20 $ is : |
4 1 2 3 |
4 |
The correct answer is Option (1) → 4 $f(x)=x^2-8x+20$ $f(x)=(x^2-8x+16)+4=(x-4)^2+4$ $(x-4)^2≥0$ Since $(x-4)^2≥0\,∀\,x$, minimum value $(x-4)^2=0$, i.e. at $x=4$ $f(4)=0+4=4$ |
