wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let R= the set of real numbers, Z= the set of integers, N= the set of natural numbers. If S be the solution set of the equation (x)2+[x]2=(x1)2+[x+1]2, where (x)= the least integer greater than or equal to x and [x]= the greatest integer less than or equal to x, then

A
S=R
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
S=RZ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
S=RN
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
S=ϕ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C S=RZ
Here (x) = least integer x

and [x] = greatest integer x

For example (2.3)=3,[2,3]=2

(x)[x]=1 if x is not integer and [x]=(x) if xZ

Then (x1)=(x)1,[x+1]=[x]+1

(x)2+[x]2=(x1)2+[x+1]2

(x)2+[x]2=(x)22(x)+1(x)2+2[x]+1

[x](x)+1=0

1+1=0 and xZ

and 0+10 if xZ

Therefore solution set S=RZ

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon