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The line \mathrm{x}+\mathrm{y}=1 meets \mathrm{x}-axis at \mathrm{A} and \mathrm{y}-axis at \mathrm{B}. \mathrm{P} is the mid-point of \mathrm{AB}. \mathrm{P}_{1} is the foot of the perpendicular from \mathrm{P} to \mathrm{O A ; M_{1}} is that from \mathrm{P_{1}} to OP; \mathrm{P_{2}} is that from \mathrm{M_{1}} to \mathrm{O A} and so on. If \mathrm{P_{n}} denotes the \mathrm{n} th foot of the perpendicular on OA from \mathrm{M_{n-1}},then \mathrm{\mathrm{OP}_{n}=} 

Option: 1

1 / 2


Option: 2

1 / 2^{\mathrm{n}}


Option: 3

1 / 2^{\mathrm{n} / 2}


Option: 4

1 / \sqrt{2}


Answers (1)

best_answer

\mathrm{\mathrm{x}+\mathrm{y}=1\: meets \: \mathrm{x}-axis \: at \: \mathrm{A}(1,0)\: and \: \mathrm{y}-axis \: at \: \mathrm{B}(0,1)}.
The ordinates of \mathrm{\mathrm{P}\: are\:(1 / 2,1 / 2)\: and\: \mathrm{PP}_{1}\:\text{ is perpendicular to} \: \mathrm{OA} }.

\mathrm{\Rightarrow \mathrm{OP}_{1}=\mathrm{P}_{1} \mathrm{P}=1 / 2 }
Equation of the line OP is \mathrm{y=x }.

We have
\mathrm{\left(\mathrm{OM}_{\mathrm{n}-1}\right)^{2}=\left(\mathrm{OP}_{\mathrm{n}}\right)^{2}+\left(\mathrm{P}_{\mathrm{n}} \mathrm{M}_{\mathrm{n}-1}\right)^{2} }
\mathrm{=2\left(\mathrm{OP}_{\mathrm{n}}\right)^{2}=2 \mathrm{p}_{\mathrm{n}}^{2}\: (say) }


\mathrm{Also, \: \left(\mathrm{OP}_{\mathrm{n}-1}\right)^{2}=\left(\mathrm{OM}_{\mathrm{n}-1}\right)^{2}+\left(\mathrm{P}_{\mathrm{n}-1} \mathrm{M}_{\mathrm{n}-1}\right)^{2}=2 \mathrm{p}_{\mathrm{n}}^{2}+2 \mathrm{p}_{\mathrm{n}}^{2}}
\mathrm{\Rightarrow \mathrm{p}_{\mathrm{n}}^{2}=\frac{1}{4} \mathrm{p}_{\mathrm{n}-1}^{2} \Rightarrow \mathrm{p}_{\mathrm{n}}=\frac{1}{2} \mathrm{p}_{\mathrm{n}-1}}
\mathrm{\therefore \mathrm{OP}_{\mathrm{n}}=\mathrm{p}_{\mathrm{n}}=\frac{1}{2} \mathrm{p}_{\mathrm{n}-1}=\frac{1}{2^{2}} \mathrm{p}_{\mathrm{n}-2}=\ldots . .=\frac{1}{2^{\mathrm{n}-1}} \mathrm{p}_{1}=\frac{1}{2^{\mathrm{n}}}}

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Gaurav

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