QJEEAdv21Maae54: JEE Advanced 2021 Mathematics - Class 11 JEEAdv21Maae54 Solution: Consider M with . The number of all those circles D n that are…

Question

Verified Solution • 2025–26 Session moderate

Consider M with

r = {{({2^{199}} - 1)\sqrt 2 } \over {{2^{198}}}}

. The number of all those circles D_n that are inside M is

Accepted Answer

\because

r = {{({2^{199}} - 1)\sqrt 2 } \over {{2^{198}}}}

Now,

\sqrt 2 {s_{n - 1}} + {a_n} &lt; \left( {{{{2^{199}} - 1} \over {{2^{198}}}}} \right)\sqrt 2

2\sqrt 2 \left( {1 - {1 \over {{2^{n - 1}}}}} \right) + {1 \over {{2^{n - 1}}}} &lt; \left( {{{{2^{199}} - 1} \over {{2^{198}}}}} \right)\sqrt 2

\therefore

2\sqrt 2 - {{\sqrt 2 } \over {{2^{n - 2}}}} + {1 \over {{2^{n - 1}}}} &lt; 2\sqrt 2 - {{\sqrt 2 } \over {{2^{198}}}}

{1 \over {{2^{n - 2}}}}\left( {{1 \over 2} - \sqrt 2 } \right) &lt; - {{\sqrt 2 } \over {{2^{198}}}} \Rightarrow {{2\sqrt 2 - 1} \over {2\,.\,{2^{n - 2}}}} &gt; {{\sqrt 2 } \over {{2^{198}}}}

\Rightarrow {2^{n - 2}} &lt; \left( {2 - {1 \over {\sqrt 2 }}} \right)\,.\,{2^{197}}

\therefore

n

\le

199

So, number of circles = 199

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