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If $\mathrm{A}=\left\{\mathrm{x} / 9 \mathrm{x} \geq \mathrm{x}^2+20\right\}$ and $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{R}$ is defined by $f(x)=2 x^3-15 x^2+36 x-48$, then the maximum value of $f(x)$ is
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Verified Answer
The correct answer is:
7
$$
\begin{aligned}
& \text { Given } \mathrm{A}=\left\{x: 9 \mathrm{x} \geq \mathrm{x}^2+20\right\} \\
& \mathrm{A}=\{x: x \in[4,5]\}=[4,5] \\
& \text { Given } f(x)=2 x^3-15 x^2+36 x-48 \\
& f^{\prime}(x)=6\left(x^2-5 x+6\right)=6(x-3)(x-2)
\end{aligned}
$$
For maxima or minima $f^{\prime}(x)=0$
$\begin{aligned} & \Rightarrow x=3,2 \\ & \text { but } A=[4,5]\end{aligned}$
Hence $f(x)$ is increasing function in $\mathrm{A}=[4,5]$ Therefore maximum value of $f(x)$ will be obtained at $x=5$ $\therefore f(5)=7$
\begin{aligned}
& \text { Given } \mathrm{A}=\left\{x: 9 \mathrm{x} \geq \mathrm{x}^2+20\right\} \\
& \mathrm{A}=\{x: x \in[4,5]\}=[4,5] \\
& \text { Given } f(x)=2 x^3-15 x^2+36 x-48 \\
& f^{\prime}(x)=6\left(x^2-5 x+6\right)=6(x-3)(x-2)
\end{aligned}
$$
For maxima or minima $f^{\prime}(x)=0$
$\begin{aligned} & \Rightarrow x=3,2 \\ & \text { but } A=[4,5]\end{aligned}$
Hence $f(x)$ is increasing function in $\mathrm{A}=[4,5]$ Therefore maximum value of $f(x)$ will be obtained at $x=5$ $\therefore f(5)=7$
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