Let f -0- -3- -0- -3-log e 2- def ined f x -log e - x 2-1-tan -1 x then f x isA- one - one and ontoB- onto but not one - oneC- neither one - one nor ontoD- one - one but not onto

The correct option is A one – one and onto f'(x)=x+1/x^2+1 ⇒ f(x) is increasing in[0,√(3)], So it is one one function. and since range is equal to codomain, i ...

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Byju's Answer

Standard XII

Mathematics

Bijective Function

Let f :[0, √3...

Question

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

one – one and onto

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one – one but not onto

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onto but not one – one

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neither one – one nor onto

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Solution

The correct option is A
one – one and onto

$f^{'} (x) = \frac{x + 1}{x^{2} + 1} \Rightarrow f (x) i s i n c r e a s i n g i n [0, \sqrt{3}]$ , So it is one one function.
and since range is equal to codomain, it is onto.

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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

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

Q. The function

f : R \to R

defined by

f (x) = (x - 1) (x - 2) (x - 3)

is
(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto

Q. The function f : R

\to

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

$f : R \to R, f (x) = x | x | i s$

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

The correct option is A one – one and onto f′(x)=x+1x2+1 ⇒ f(x) is increasing in[0,√3], So it is one one function. and since range is equal to codomain, it is onto.

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

The correct option is A
one – one and onto

$f^{'} (x) = \frac{x + 1}{x^{2} + 1} \Rightarrow f (x) i s i n c r e a s i n g i n [0, \sqrt{3}]$ , So it is one one function.
and since range is equal to codomain, it is onto.