The function f(x) = x^{3} - 6x^{2} + ax + b i | vedclass

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(A) Given $f(x) = x^{3} - 6x^{2} + ax + b$. Since $f(2) = 0$,$8 - 24 + 2a + b = 0 \Rightarrow 2a + b = 16$.Since $f(4) = 0$,$64 - 96 + 4a + b = 0 \Rig

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both $(S_1)$ and $(S_2)$ are true

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(A) Given $f(x) = x^{3} - 6x^{2} + ax + b$. Since $f(2) = 0$,$8 - 24 + 2a + b = 0 \Rightarrow 2a + b = 16$.Since $f(4) = 0$,$64 - 96 + 4a + b = 0 \Rightarrow 4a + b = 32$.Solving these,$a = 8, b = 0$. Thus $f(x) = x^{3} - 6x^{2} + 8x$.$f^{\prime}(x) = 3x^{2} - 12x + 8$. The roots of $f^{\prime}(x) = 0$ are $x = \frac{12 \pm \sqrt{144 - 96}}{6} = 2 \pm \frac{\sqrt{48}}{6} = 2 \pm \frac{2}{\sqrt{3}}$.For $(S_1)$: $f^{\prime}(2) = -4$ and $f^{\prime}(4) = 8$. Since $f^{\prime}(x)$ is continuous,by the Intermediate Value Theorem,there exists $x_{1} \in (2, 4)$ such that $f^{\prime}(x_{1}) = -1$ (as $-1 \in (-4, 8)$). Also,$f^{\prime}(x)$ has a root $x_{2} = 2 + \frac{2}{\sqrt{3}} \approx 3.15 \in (2, 4)$. Since $f^{\prime}(3) = 27 - 36 + 8 = -1$,we have $x_{1} = 3 For $(S_2)$: $f$ is decreasing on $(2, x_{4})$ and increasing on $(x_{4}, 4)$ where $x_{4} = 2 + \frac{2}{\sqrt{3}}$. $f(x_{4}) = (2 + \frac{2}{\sqrt{3}})^{3} - 6(2 + \frac{2}{\sqrt{3}})^{2} + 8(2 + \frac{2}{\sqrt{3}}) = -\frac{16}{3\sqrt{3}}$.$2f^{\prime}(x_{3}) = \sqrt{3}f(x_{4}) = \sqrt{3}(-\frac{16}{3\sqrt{3}}) = -\frac{16}{3} \approx -5.33$. So $f^{\prime}(x_{3}) = -\frac{8}{3} \approx -2.67$. Since $f^{\prime}(x)$ takes all values in $[-4, 8]$ on $(2, 4)$,such $x_{3}$ exists. Thus $(S_2)$ is true.

List-$I$	List-$II$
$(P)$ The minimum value of $n$ for which the function $f(x)=\left[\frac{10 x^3-45 x^2+60 x+35}{n}\right]$ is continuous on the interval $[1,2]$,is	$(1)$ $8$
$(Q)$ The minimum value of $n$ for which $g(x)=\left(2 n^2-13 n-15\right)\left(x^3+3 x\right), x \in R$,is an increasing function on $R$,is	$(2)$ $9$
$(R)$ The smallest natural number $n$ which is greater than $5$,such that $x=3$ is a point of local minima of $h(x)=\left(x^2-9\right)^{n}\left(x^2+2 x+3\right)$,is	$(3)$ $5$
$(S)$ Number of $x_0 \in R$ such that $l(x)=\sum_{k=0}^4\left(\sin \|x-k\|+\cos \left\|x-k+\frac{1}{2}\right\|\right), x \in R$ is not differentiable at $x_0$,is	$(4)$ $6$
	$(5)$ $10$

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