A- 0-1B- 0-1-C- -0-1-D- 0-1

The correct option is B (0,1] f(x)=ln ( x^2+e/x^2+1 )=ln ( x^2+1 1+e/x^2 + 1 )=ln ( 1+e 1/x^2+1 ) Clearly range is (0,1 ] Hence (B) is correct answer ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Range

A. 0,1B. 0,1]...

Question

\(\if ~f(x) = ln \left ( \frac{x^2 +e}{x^2 + 1} \right ), then range of f(x) is

(0,1)

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

(0,1]

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!

{0,1}

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

Open in App

Solution

The correct option is B
(0,1]

$f (x) = l n (\frac{x^{2} + e}{x^{2} + 1}) = l n (\frac{x^{2} + 1 - 1 + e}{x^{2} + 1}) = l n (1 + \frac{e - 1}{x^{2} + 1}) C l e a r l y r a n g e i s (0, 1] H e n c e (B) i s c o r r e c t a n s w e r .$

Suggest Corrections

Similar questions

Q. Let

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

, where [ x] denotes the greatest integer function. Then, the range of f is

Q. If

x ϵ R

and

m = \frac{x^{2}}{(x^{4} - 2 x^{2} + 4)}

, then m lies in the interval

Q. If the function

f : R \to

A given by

f (x) = \frac{x^{2}}{x^{2} + 1}

is a surjection, then A is

Q. The range of

f (x) = cos (\sqrt{x})

Q. Let

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

, where [ x] denotes the greatest integer function. Then, the range of f is

Join BYJU'S Learning Program

\(\if ~f(x) = ln \left ( \frac{x^2 +e}{x^2 + 1} \right ), then range of f(x) is

The correct option is B (0,1] f(x)=ln(x2+ex2+1)=ln(x2+1−1+ex2+1)=ln(1+e−1x2+1)Clearly range is (0,1]Hence (B)is correct answer.

The correct option is B
(0,1]

$f (x) = l n (\frac{x^{2} + e}{x^{2} + 1}) = l n (\frac{x^{2} + 1 - 1 + e}{x^{2} + 1}) = l n (1 + \frac{e - 1}{x^{2} + 1}) C l e a r l y r a n g e i s (0, 1] H e n c e (B) i s c o r r e c t a n s w e r .$