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If $P$ is a point on the segment $A B$ of length $12 \mathrm{~cm}$, then the position of $P$ for $A P^{2}+B P^{2}$ to be minimum is such that
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$P$ is the midpoint of segment $A B$
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$\begin{aligned} \text { Let } \mathrm{d}(\mathrm{AP}) &=\mathrm{x} \Rightarrow \mathrm{d}(\mathrm{BP})=12-\mathrm{x} \\ \mathrm{f}(\mathrm{x}) &=\mathrm{AP}^{2}+\mathrm{BP}^{2} \\ &=\mathrm{x}^{2}+(12-\mathrm{x})^{2} \\ &=2 \mathrm{x}^{2}-24 \mathrm{x}+144 \\ \therefore \quad \mathrm{f}^{\prime}(\mathrm{x}) &=4 \mathrm{x}-24 \text { and when } \mathrm{f}^{\prime}(\mathrm{x})=0, \text { we get } \mathrm{x}=6 . \\ \mathrm{f}^{\prime \prime}(\mathrm{x}) &=4>0 \\ \text { Hence } \mathrm{f}(\mathrm{x}) \text { is minimum at } \mathrm{x}=6 & \end{aligned}$
$\begin{aligned} \text { Let } \mathrm{d}(\mathrm{AP}) &=\mathrm{x} \Rightarrow \mathrm{d}(\mathrm{BP})=12-\mathrm{x} \\ \mathrm{f}(\mathrm{x}) &=\mathrm{AP}^{2}+\mathrm{BP}^{2} \\ &=\mathrm{x}^{2}+(12-\mathrm{x})^{2} \\ &=2 \mathrm{x}^{2}-24 \mathrm{x}+144 \\ \therefore \quad \mathrm{f}^{\prime}(\mathrm{x}) &=4 \mathrm{x}-24 \text { and when } \mathrm{f}^{\prime}(\mathrm{x})=0, \text { we get } \mathrm{x}=6 . \\ \mathrm{f}^{\prime \prime}(\mathrm{x}) &=4>0 \\ \text { Hence } \mathrm{f}(\mathrm{x}) \text { is minimum at } \mathrm{x}=6 & \end{aligned}$
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