$A variable point P moving inside a sqaure whose coordinates of vertices are (1,1), (-1,-1),(-1,1) and (1,-1)$ $in such a way that P is closer to the diagonals of sqaure$ $compare to the coordinate axes. If A is the area of the region tranverse by P then the value of [A] i s$

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answer is 2.

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Detailed Solution

$G i v e n t h a t t h e p o i n t P i s m o v i n g c l o s e t o d i a g o n a l s w h e n c o m p a r e d w i t h a x e s$

$T o u n d e r s \tan d t h e s i t u a t i o n, c o n s i d e r t h e m o v e m e n t o f t h e p o i n t P i n t h e f i r s t q u a d r a n t$

$T h e p o i n t P i s c l o s e t o d i a g o n a l m e a n s i t w i l l n o t m o v e t h e a r e a b e l o w t h e r a y w h i c h m a k e s 22.5 d e g r e e s a n g l e w i t h x - a x i s$

$a n d t h e p o i n t P d o e s n o t m o v e a b o v e t h e l i n e w h i c h m a k e s a n a n g l e 67.5 d e g r e e s w i t h x - a x i s$

$T h e s h a d e d r e g i o n i s t h e r e g i o n w h e r e t h e p o i n t P m o v e s i n t h e f i r s t q u a d r a n t$

$s i m i l a r l y i n a l l o t h e r q u a d r a n t s t o o$

$\begin{matrix} A = 4 times the above shaded region \\ = 4 (2) (Area of triangle OAB - Area of triangle OAP) \\ = 8 (\frac{1}{2} (1) (1) - \frac{1}{2} (1) (\sqrt{2} - 1)) \\ = 4 (1 - \sqrt{2} + 1) \\ = 4 (2 - \sqrt{2}) \\ = 2.343 \\ [A] = 2 \end{matrix}$