Find-the-area-common-to-min-x-y-2-and-max-x-y-4-x-denotes-the-greatest-integer-less-than-or-equal-to-x-

Tinku Tara June 4, 2023 Mensuration 0 Comments

Facebook Tweet Pin

Question Number 38568 by ajfour last updated on 27/Jun/18

Find the area common to min{[x], [y] } =2 and max{[x], [y] } =4 . [x] denotes the greatest integer less than or equal to x.

F i n d t h e a r e a c o m m o n t o

m i n {[x], [y]} = 2 a n d

m a x {[x], [y]} = 4 .

[x] d e n o t e s t h e g r e a t e s t i n t e g e r

l e s s t h a n o r e q u a l t o x .

Commented by ajfour last updated on 27/Jun/18

Right sketch,right answer Sir. Find area bounded by min{∣x∣,∣y∣}=2 max{∣x∣,∣y∣}=4

R i g h t s k e t c h, r i g h t a n s w e r S i r .

F i n d a r e a b o u n d e d b y

m i n {∣ x ∣, ∣ y ∣} = 2

m a x {∣ x ∣, ∣ y ∣} = 4

Commented by MrW3 last updated on 27/Jun/18

Commented by MrW3 last updated on 27/Jun/18

common area = 2

c o m m o n a r e a = 2

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

area=8^2 −4^2 =64−16=48

a r e a = 8^{2} - 4^{2} = 64 - 16 = 48

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

area formed by ∣x∣=4 and∣y∣=4 is8×8=64 area formed by ∣x∣=2 and ∣y∣=2 is=4×4=16 so=area required 64−16=48

a r e a f o r m e d b y ∣ x ∣= 4 a n d ∣ y ∣= 4 i s 8 \times 8 = 64

a r e a f o r m e d b y ∣ x ∣= 2 a n d ∣ y ∣= 2 i s = 4 \times 4 = 16

s o = a r e a r e q u i r e d 64 - 16 = 48

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

Commented by ajfour last updated on 27/Jun/18

min{ } , max{ } has a meaning.

m i n {}, m a x {} h a s a m e a n i n g .

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

you solve the question...

y o u s o l v e t h e q u e s t i o n \dots

Commented by MrW3 last updated on 27/Jun/18

area bounded by min{[x],[y]}=2 max{[x],[y]}=4 is 8. area bounded by min{∣x∣,∣y∣}=2 max{∣x∣,∣y∣}=4 is 16.

a r e a b o u n d e d b y

m i n {[x], [y]} = 2

m a x {[x], [y]} = 4

i s 8 .

a r e a b o u n d e d b y

m i n {∣ x ∣, ∣ y ∣} = 2

m a x {∣ x ∣, ∣ y ∣} = 4

i s 16 .

Commented by ajfour last updated on 27/Jun/18

Area for the second one is 16.

A r e a f o r t h e s e c o n d o n e i s 16 .

Commented by MrW3 last updated on 27/Jun/18

you′r right.

{y o u}^{'} r r i g h t .

Commented by MrW3 last updated on 27/Jun/18

Commented by ajfour last updated on 27/Jun/18

Great, Sir. you are fast.

G r e a t, S i r . y o u a r e f a s t .

Commented by ajfour last updated on 28/Jun/18

Too Good Sir.

T o o G o o d S i r .

Commented by MrW3 last updated on 28/Jun/18

Commented by MrW3 last updated on 28/Jun/18

the pink color area is also bounded area which is 64−16=48.

t h e p i n k c o l o r a r e a i s a l s o b o u n d e d

a r e a w h i c h i s 64 - 16 = 48 .

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Facebook Tweet Pin

Leave a Reply Cancel reply