The line \[x+y=1\] meets x-axis at A and y-axis B, P is the mid-point of AB |
\[{{P}_{1}}\] is the foot of the perpendicular from P to OA; |
\[{{M}_{1}}\] is that of \[{{P}_{1}}\] to \[OP;\]\[{{P}_{2}}\] is that of \[{{M}_{1}}\] to OA; |
\[{{M}_{2}}\] is that of \[{{P}_{2}}\] to \[OP;\]\[{{P}_{3}}\] is that of \[{{M}_{2}}\] to OA; and so on. |
If \[{{P}_{n}}\] denotes the \[{{n}^{th}}\] foot of the perpendicular on OA; then \[O{{P}_{n}}\] is |
A) \[{{\left( \frac{1}{2} \right)}^{n-1}}\]
B) \[{{\left( \frac{1}{2} \right)}^{n}}\]
C) \[{{\left( \frac{1}{2} \right)}^{n+1}}\]
D) none of these
Correct Answer: B
Solution :
we have \[{{(O{{M}_{n-1}})}^{2}}={{(O{{P}_{n}})}^{2}}+{{({{P}_{n}}{{M}_{n-1}})}^{2}}\] |
\[=2{{(O{{P}_{n}})}^{2}}=2{{\alpha }_{n}}^{2}\] (say) |
Also, \[{{(O{{P}_{n-1}})}^{2}}={{(O{{M}_{n-1}})}^{2}}+{{({{P}_{n-1}}{{M}_{n-1}})}^{2}}\] |
\[\Rightarrow \]\[{{\alpha }_{n-1}}^{2}=2{{\alpha }_{n}}^{2}+\frac{1}{2}{{\alpha }_{n-1}}^{2}\] |
\[\Rightarrow \]\[{{\alpha }_{n}}=\frac{1}{2}{{\alpha }_{n-1}}\] |
\[\therefore \]\[O{{P}_{n}}={{\alpha }_{n}}=\frac{1}{2}{{\alpha }_{n-1}}=\frac{1}{{{2}^{2}}}{{\alpha }_{n-2}}\] |
\[=.......=\frac{1}{{{2}^{n}}}={{\left( \frac{1}{2} \right)}^{n}}\] |
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