question_answer

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, \[{{(O{{P}_{n-1}})}^{2}}={{(O{{M}_{n-1}})}^{2}}+{{({{P}_{n-1}}{{M}_{n-1}})}^{2}}\]

\[\Rightarrow \]\[{{\alpha }_{n-1}}^{2}=2{{\alpha }_{n}}^{2}+\frac{1}{2}{{\alpha }_{n-1}}^{2}\]

\[\Rightarrow \]\[{{\alpha }_{n}}=\frac{1}{2}{{\alpha }_{n-1}}\]

\[\therefore \]\[O{{P}_{n}}={{\alpha }_{n}}=\frac{1}{2}{{\alpha }_{n-1}}=\frac{1}{{{2}^{2}}}{{\alpha }_{n-2}}\]

\[=.......=\frac{1}{{{2}^{n}}}={{\left( \frac{1}{2} \right)}^{n}}\]