Let f: R rightarrow R be given by f(x)=(x-1)( | vedclass

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(C) Given $f(x) = (x-1)(x-2)(x-5)$.By the Fundamental Theorem of Calculus,$F'(x) = f(x) = (x-1)(x-2)(x-5)$.To find critical points,set $F'(x) = 0$,whi

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$1, 2, 3$

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(C) Given $f(x) = (x-1)(x-2)(x-5)$.By the Fundamental Theorem of Calculus,$F'(x) = f(x) = (x-1)(x-2)(x-5)$.To find critical points,set $F'(x) = 0$,which gives $x = 1, 2, 5$.Using the first derivative test:- For $x - For $1  0$ (increasing).- For $2 - For $x > 5$,$F'(x) > 0$ (increasing).Thus,$F$ has local minima at $x=1$ and $x=5$,and a local maximum at $x=2$.Statement $(1)$ is correct.Statement $(2)$ is correct.Statement $(4)$ is incorrect because $F$ has two local minima and one local maximum.For statement $(3)$,$F(x) = \int_0^x (t^3 - 8t^2 + 17t - 10) dt = \frac{x^4}{4} - \frac{8x^3}{3} + \frac{17x^2}{2} - 10x$. Evaluating $F(x)$ at $x=1, 2, 5$ shows $F(x) Therefore,options $(1), (2), (3)$ are correct.

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