If for $f(x) = 2x - {x^2}$, Lagrange’s theorem satisfies in $[0, 1]$, then the value of $c \in [0,\,1]$ is
If for $f(x) = 2x - {x^2}$, Lagrange’s theorem satisfies in $[0, 1]$, then the value of $c \in [0,\,1]$ is
- A
$c = 0$
- B
$c = \frac{1}{2}$
- C
$c = \frac{1}{4}$
- D
$c = 1$
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