QJEEAdv22Ma2efc: JEE Advanced 2022 Mathematics - Class 11 JEEAdv22Ma2efc Solution: Let be a circle of radius . Let be circles of equal radius .…

Here if we add center of circles G₁, G₂, G₃ ....... G_n, then we get a polygon of n sides.

From figure you can see one side of polygon makes angle $\theta$ with the center.

$\therefore$ n sides make angle = n $\theta$

We know, $n\theta = 2\pi$

$\Rightarrow \theta = {{2\pi } \over n}$

Here triangle OMN is an isosceles triangle. Line joining of point O and midpoint O of MN (point A) is perpendicular to line MN and perpendicular bisector of angle $\theta$ .

$\therefore$ $\angle MOA = {\theta \over 2} = {\pi \over n}$

From right angle triangle OMA,

$\sin {\pi \over n} = {r \over {R + r}}$

$\Rightarrow R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over n}$

Option A:

When $n = 4$ ,

$R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over 4}$

$\Rightarrow R + r = \sqrt 2 r$

$\Rightarrow R = \left( {\sqrt 2 - 1} \right)r$

$\therefore$ Option (A) is wrong.

Option B :

When $n = 5$ ,

$R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over 5} = r{\mathop{\rm cosec}\nolimits} 36^\circ$

We know, in 0 to ${\pi \over 2}$ , ${\mathop{\rm cosec}\nolimits} \theta$ is decreasing.

$\therefore$ {\mathop{\rm cosec}\nolimits} 45^\circ < {\mathop{\rm cosec}\nolimits} 36^\circ < {\mathop{\rm cosec}\nolimits} 30^\circ

\Rightarrow \sqrt 2 < {\mathop{\rm cosec}\nolimits} 36^\circ < 2

\Rightarrow \sqrt 2 r < r{\mathop{\rm cosec}\nolimits} 36^\circ < 2r

\Rightarrow \sqrt 2 r - r < r{\mathop{\rm cosec}\nolimits} 36^\circ - r < 2r - r

\Rightarrow \left( {\sqrt 2 - 1} \right)r < R < r

$\therefore$ R < r , when $n = 5$

$\therefore$ Option (B) is wrong.

Option C :

When $n = 8$ ,

$R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over 8}$

${\mathop{\rm cosec}\nolimits} \theta$ is decreasing function in 0 to ${\pi \over 2}$ .

$\therefore$ {\mathop{\rm cosec}\nolimits} {\pi \over 8} > {\mathop{\rm cosec}\nolimits} {\pi \over 4}

\Rightarrow r{\mathop{\rm cosec}\nolimits} {\pi \over 8} > \sqrt 2 r

\Rightarrow R + r > \sqrt 2 r

\Rightarrow R > \left( {\sqrt 2 - 1} \right)r

$\therefore$ Option (C) is correct.

Option D :

When $n = 12$ , then

$R + r = r\left( {\cos ec{\pi \over {12}}} \right)$

$\Rightarrow R + r = r \times {{2\sqrt 2 } \over {\sqrt 3 - 1}}$ [as $\cos ec{\pi \over {12}} = {{2\sqrt 2 } \over {\sqrt 3 - 1}}$ ]

$\Rightarrow R + r = r \times {{2\sqrt 2 \left( {\sqrt 3 + 1} \right)} \over 2}$

$\Rightarrow R + r = \sqrt 2 r\left( {\sqrt 3 + 1} \right)$

$\Rightarrow R = \left[ {\sqrt 2 \left( {\sqrt 3 + 1} \right) - 1} \right]r$

\Rightarrow R < \sqrt 2 \left( {\sqrt 3 + 1} \right)r

$\therefore$ Option (D) is correct.