Question

A function f ( x ) given by ;

$\quad \qquad { \bigstar { \underline { \boxed { \red { \underbrace { \bf \pmb { f ( x ) = log \bigg\{ \sqrt{x² + 1} - ( x ) \bigg\} }}}}}}}{\bigstar}$

Then f ( x ) is :-

1. An even function
2. An odd function
3. Both even and odd function.
4. Neither Even nor odd function.​

Answer 1

Answer:

f(x)=x2−2x+6 is a surjection.

So the range of f(x) will be equal to its codomain.

f(x)=x2−2x+6

f1(x)=2x−2

=2(x−1)

f(x) will be increasing when x⩾1.

∴f(1)=1−2+6

=5

∴B=[5,∞)

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = log[ \sqrt{ {x}^{2} + 1} - x]$

Now, Consider

$\rm :\longmapsto\:f( - x) = log[ \sqrt{ {( - x)}^{2} + 1} -( - x)]$

$\rm :\longmapsto\:f( - x) = log[ \sqrt{ {x}^{2} + 1} + x]$

can be further rewritten as

$\rm :\longmapsto\:f( - x) = log\bigg[ \sqrt{ {x}^{2} + 1} + x \times \dfrac{ \sqrt{ {x}^{2} + 1 } - x}{ \sqrt{ {x}^{2} + 1} - x } \bigg]$

We know,

$\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this identity, we get

$\rm :\longmapsto\:f( - x) = log\bigg[\dfrac{ {( \sqrt{ {x}^{2} + 1}) }^{2} - {x}^{2} }{ \sqrt{ {x}^{2} + 1} - x } \bigg]$

$\rm :\longmapsto\:f( - x) = log\bigg[\dfrac{{x}^{2} + 1 - {x}^{2} }{ \sqrt{ {x}^{2} + 1} - x } \bigg]$

$\rm :\longmapsto\:f( - x) = log\bigg[\dfrac{1 }{ \sqrt{ {x}^{2} + 1} - x } \bigg]$

$\rm :\longmapsto\:f( - x) = log {[ \sqrt{ {x}^{2} + 1 } - x]}^{ - 1}$

$\rm :\longmapsto\:f( - x) = - \: log {[ \sqrt{ {x}^{2} + 1 } - x]}$

$\bf\implies \:f( - x) \: = - \: f(x)$

$\bf\implies \:f(x) \: is \: odd \: function$

$\bf :\longmapsto\:f(x) = log[ \sqrt{ {x}^{2} + 1} - x] \: is \:an \: odd \: function$

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Basic Definition

Even Function :- A function f(x) is said to be even function iff f(- x) = f(x).

Odd Function :- A function f(x) is said to be odd function iff f(- x) = - f(x).

Periodic Function :- A function f(x) is said to be periodic function of period T iff f(x + T) = f(x).