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

Which of the following functions are bijective?

A
f:R[0,) defined as f(x)=e sgn(x)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
f:[3,4][4,6] defined by f(x)=|x1|+|x2|+|x+3|+|x4|
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
f:RR defined by f(x)=[x]+cos(π[x]) where [.] denotes the greatest integer function
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
f:RR defined as f(x)=min{x+2,2x+4}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B f:[3,4][4,6] defined by f(x)=|x1|+|x2|+|x+3|+|x4|
sgn(x)=|x|x,x00,x=0

sgn(x)=1, x<00, x=01, x>0

esgn(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪e1=1e,x<0e0=1,x=0e1=e,x>0

Neither one-one nor onto.
Hence, esgn(x) is not bijective.


f:[3,4][4,6]
f(x)=|x1|+|x+2|+|x3|+|x4|
For 3x4
f(x)=x1+x2+x3+4x =2x2
f is one-one.

3x462x842x26 f is onto.
It is bijective.


f:RR
f(x)=[x]+cosπ[x]
If 0x<1,
f(x)=0+cosπ(0)=1
Not one-one, so, not bijective.


f:RR
f(x)=min{x+2,2x+4}


Not one-one
Not onto.
So, f is not bijective.

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