If P is a point on the segment AB of length 1 | vedclass

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(C) Let $d(AP) = x$. Then $d(BP) = 12 - x$. Define the function $f(x) = AP^{2} + BP^{2} = x^{2} + (12 - x)^{2}$. Expanding this,we get $f(x) = x^{2} +

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$P$ is the midpoint of segment $AB$

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(C) Let $d(AP) = x$. Then $d(BP) = 12 - x$. Define the function $f(x) = AP^{2} + BP^{2} = x^{2} + (12 - x)^{2}$. Expanding this,we get $f(x) = x^{2} + 144 - 24x + x^{2} = 2x^{2} - 24x + 144$. To find the minimum,we calculate the first derivative: $f'(x) = 4x - 24$. Setting $f'(x) = 0$,we get $4x = 24$,which implies $x = 6$. Checking the second derivative,$f''(x) = 4$. Since $f''(x) > 0$,the function $f(x)$ attains a minimum at $x = 6$. Since $x = 6$ is exactly half of the total length $12 	ext{ cm}$,$P$ is the midpoint of segment $AB$.

If $P$ is a point on the segment $AB$ of length $12 \text{ cm}$,then the position of $P$ for $AP^{2} + BP^{2}$ to be minimum is such that

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