A real valued function f(x) = |x^2 - 3x + 2| | vedclass

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(D) Given $f(x) = |x^2 - 3x + 2| + 2x - 3$ on $[-2, 1]$.Since $x^2 - 3x + 2 = (x - 1)(x - 2)$, for $x \in [-2, 1]$, $(x - 1) \le 0$ and $(x - 2) < 0$,

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$10$

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(D) Given $f(x) = |x^2 - 3x + 2| + 2x - 3$ on $[-2, 1]$.Since $x^2 - 3x + 2 = (x - 1)(x - 2)$, for $x \in [-2, 1]$, $(x - 1) \le 0$ and $(x - 2) Thus, $|x^2 - 3x + 2| = x^2 - 3x + 2$.Substituting this into $f(x)$, we get $f(x) = x^2 - 3x + 2 + 2x - 3 = x^2 - x - 1$.To find the extrema on $[-2, 1]$, we check the critical points and endpoints.$f'(x) = 2x - 1$. Setting $f'(x) = 0$ gives $x = 1/2$, which is in $[-2, 1]$.Calculate values at critical point and endpoints:$f(-2) = (-2)^2 - (-2) - 1 = 4 + 2 - 1 = 5$.$f(1) = (1)^2 - (1) - 1 = -1$.$f(1/2) = (1/2)^2 - (1/2) - 1 = 1/4 - 1/2 - 1 = -5/4$.Comparing these, the absolute maximum $M = 5$ and the absolute minimum $m = -5/4$.Then $M - 4m = 5 - 4(-5/4) = 5 + 5 = 10$.

Column $I$	Column $II$
$(A)$ If $-1 < x < 1$,then $f(x)$ satisfies	$(p)$ $0 < f(x) < 1$
$(B)$ If $1 < x < 2$,then $f(x)$ satisfies	$(q)$ $f(x) < 0$
$(C)$ If $3 < x < 5$,then $f(x)$ satisfies	$(r)$ $f(x) > 0$
$(D)$ If $x > 5$,then $f(x)$ satisfies	$(s)$ $f(x) < 1$

$A$ real valued function $f(x) = |x^2 - 3x + 2| + 2x - 3$ is defined on $[-2, 1]$. If $m$ and $M$ are absolute minimum and absolute maximum values of $f$ respectively, then $M - 4m =$

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