Each of the four inequalities given below defines a region in the xy-plane. Let P be the property for any two points (x₁, y₁) and (x₂, y₂) in the region, the point $(\frac{1}{2} (x_{1} + x_{2}), \frac{1}{2} (y_{1} + y_{2}))$ is also in the region. Then which of them does not satisfy property P

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$x^{2} + 2 y^{2} \leq 1$

$max {| x |, | y |} \leq 1$

$x^{2} - y^{2} \leq 1$

$y^{2} - x^{2} \leq 0$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$o p t i o n (C) : G i v e n x^{2} - y^{2} \leq 1 w h i c h r e p r e s e n t s i n s i d e r e g i o n o f h y p e r b o l a . A h y p e r b o l a h a s t w o p a r t s . I f w e t a k e 2 p o i n t s, o n e i n o n e p a r t a n d o t h e r i n o t h e r p a r t, t h e m i d - p o i n t n e e d n o t t o b e i n s i d e t h e h y p e r b o l a . L e t' s t a k e (\frac{- 1}{2} , 0) a n d (\frac{1}{4} , 0) a s t w o p o i n t s . T h e i r m i d - p o i n t i s (- \frac{1}{8} , 0) i s n o t i n t h e r e g i o n o f h y p e r b o l a . H e n c e, o p t i o n C d o e s n o t s a t i s f y p r o p e r t y P .$