Let f -0- -3- -0- -3-log e 2- defined by f x -log e - x 2-1-tan -1 x then f x isA- One - one and ontoB- One - one but not ontoC- Onto but not one - oneD- Neither one - one nor onto

The correct option is A One – one and ontof'(x) =x+1/x^2+1 gt; 0⇒ f(x) is strictly increasing in [0,√(3)] Also the range of the f(x) = codomain ⇒ f is ...

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

Byju's Answer

Standard IX

Mathematics

Bijective Function

Let f :[0, √3...

Question

Let $f : [0, \sqrt{3}] \to [0, \frac{π}{3} + l o g_{e} 2]$ defined by $f (x) = l o g_{e} \sqrt{x^{2} + 1} + t a n^{- 1} x$ then f(x)is

One – one and onto

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

One – one but not onto

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

Onto but not one – one

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

Neither one – one nor onto

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

Open in App

Solution

The correct option is A One – one and onto
$f^{'} (x) = \frac{x + 1}{x^{2} + 1} > 0 \Rightarrow f (x) i s s t r i c t l y i n c r e a s i n g i n [0, \sqrt{3}]$
Also the range of the f(x) = codomain
$\Rightarrow f i s b i j e c t i v e$

Suggest Corrections

Similar questions

Q. Let

f : [0, \sqrt{3}] \to [0, \frac{π}{3} + l o g_{e} 2]

defined by

f (x) = l o g_{e} \sqrt{x^{2} + 1} + t a n^{- 1} x

then f(x)is

$L e t f : [0, \sqrt{3}] \to [0, \frac{π}{3} + l o g_{e} 2] d e f i n e d f (x) = l o g_{e} \sqrt{x^{2} + 1} + t a n^{- 1} x t h e n f (x) i s$

Q. Let

f : R - {0} \to R

be a function defined by

f (x) = x - \frac{1}{x} .

Then

f

Q. Let

f : R \to R

be a function defined by

f (x) = \frac{x^{2} - 8}{x^{2} + 2}

. Then, f is
(a) one-one but not onto
(b) one-one and onto
(c) onto but not one-one
(d) neither one-one nor onto

Q. The function f : R

\to

Rdefined by f(x) =(x - 1)(x - 2)(x - 3) is

Join BYJU'S Learning Program

Let f:[0,√3]→[0,π3+loge2] defined by f(x)=loge√x2+1+tan−1x then f(x)is

The correct option is A One – one and ontof′(x)=x+1x2+1>0⇒f(x) is strictly increasing in [0,√3] Also the range of the f(x) = codomain ⇒ f is bijective

Let $f : [0, \sqrt{3}] \to [0, \frac{π}{3} + l o g_{e} 2]$ defined by $f (x) = l o g_{e} \sqrt{x^{2} + 1} + t a n^{- 1} x$ then f(x)is

The correct option is A One – one and onto
$f^{'} (x) = \frac{x + 1}{x^{2} + 1} > 0 \Rightarrow f (x) i s s t r i c t l y i n c r e a s i n g i n [0, \sqrt{3}]$
Also the range of the f(x) = codomain
$\Rightarrow f i s b i j e c t i v e$