$L e t f (x) = \frac{x - [x]}{1 + x - [x]}, x ϵ R, [] d e n t o e s t h e g r e a t e s t i n t e g e r f u n c t i o n . T h e n, t h e r a n g e o f f i s$

(0,1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[0,1]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B

The graph of y = x – [x] is as shown below

When x is an integer, x – [x] = 0

Hence, f(x) = 0 when x is an integer

$x \Rightarrow [x] a s x t e n d s t o a n i n t e g e r . A s \to 1, \frac{x}{1 + x} \to \frac{1}{2} H e n c e, t h e r a n g e o f f (x) i s [0, \frac{1}{2})$

Suggest Corrections

Similar questions

Let f (x) = \frac{x - [x]}{1 + x - [x]}, x ϵ R, [] dentoes the greatest integer function . Then, the range of f is

f (x) = \frac{x - [x]}{1 + x - [x]}, x ϵ R

f (x) = \frac{x - [x]}{1 + x - [x]}, x ϵ R

f (x) = \frac{x - [x]}{1 + x - [x]}, x ϵ R

y = \frac{x - [x]}{1 - [x] + x}

[.]

Join BYJU'S Learning Program