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

Let a function f defined from RR as f(x)={2mx,x14mx+1,x>1
If function is onto on R, then range of m is

A
[1,)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
[1,0)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
{-1}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(0,)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B [1,0)
Case I: m=k,where k>0
f(x)={2kx,x14kx+1,x>1
If<x11x<2k12kx<2k1f(x)< x1
If 1<x<x<1<4kx<4k<4kx+1<4k+1<f(x)<4k+1 x>1
Since, f(x) is an onto function.
4k+12k14m+12m12m2m11m<0 [As m is negative]

Case II: m=k,where k>0
f(x)={2kx,x14kx+1,x>1
If <x11x<2k12kx<2k1f(x)< x1 ...[1]
If 1<x<4k<4kx<4k+1<4kx+1<4k+1<f(x)< x>1 ...[2]

Since, f(x) is an onto function, the union of [1] and [2] should be R. This is not possible as k is positive.
Therefore, m cannot be positive. Case II is not possible.

Range of f is [1,0)



flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Types of Functions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon