MHT CET · Maths · Application of Derivatives
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
- A \(P\) divides \(A B\) in the ratio \(2: 3\) internally
- B \(P\) divides \(A B\) in the ratio \(4: 3\) internally
- C \(P\) is the midpoint of segment \(A B\)
- D \(P\) divides \(B A\) in the ratio \(2: 1\) internally
Answer & Solution
Correct Answer
(C) \(P\) is the midpoint of segment \(A B\)
Step-by-step Solution
Detailed explanation
A
\(\text { Let } \mathrm{d}(\mathrm{AP}) =\mathrm{x} \Rightarrow \mathrm{d}(\mathrm{BP})=12-\mathrm{x} \)
\( \mathrm{f}(\mathrm{x}) =\mathrm{AP}^{2}+\mathrm{BP}^{2} \)
\( =\mathrm{x}^{2}+(12-\mathrm{x})^{2} \)
\( =2 \mathrm{x}^{2}-24 \mathrm{x}+144 \)
\( \therefore \mathrm{f}^{\prime}(\mathrm{x}) =4 \mathrm{x}-24 \text { and when } \mathrm{f}^{\prime}(\mathrm{x})=0, \text {we get } \mathrm{x}=6 . \)
\( \mathrm{f}^{\prime \prime}(\mathrm{x}) =4>0 \)
\( \text { Hence } \mathrm{f}(\mathrm{x}) \text { is minimum at } \mathrm{x}=6\)
\(\text { Let } \mathrm{d}(\mathrm{AP}) =\mathrm{x} \Rightarrow \mathrm{d}(\mathrm{BP})=12-\mathrm{x} \)
\( \mathrm{f}(\mathrm{x}) =\mathrm{AP}^{2}+\mathrm{BP}^{2} \)
\( =\mathrm{x}^{2}+(12-\mathrm{x})^{2} \)
\( =2 \mathrm{x}^{2}-24 \mathrm{x}+144 \)
\( \therefore \mathrm{f}^{\prime}(\mathrm{x}) =4 \mathrm{x}-24 \text { and when } \mathrm{f}^{\prime}(\mathrm{x})=0, \text {we get } \mathrm{x}=6 . \)
\( \mathrm{f}^{\prime \prime}(\mathrm{x}) =4>0 \)
\( \text { Hence } \mathrm{f}(\mathrm{x}) \text { is minimum at } \mathrm{x}=6\)
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