Question

If f(x) = $ {\cos ^{ - 1}}\left( {x - {x^2}} \right) + \sqrt {1 + \dfrac{1}{{\left| x \right|}}} + \dfrac{1}{{{x^2} - 1}} $ , then find the domain of f(x)
A. $ \left[ {\sqrt 2 ,\left( {\dfrac{{1 - \sqrt 5 }}{2}} \right)} \right] $
B. $ \left[ {\sqrt 2 ,\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)} \right] $
C. $ \left[ { - \sqrt 2 ,\left( {\dfrac{{1 \pm \sqrt 5 }}{2}} \right)} \right] $
D.None of these

Answer 1

Hint: First, we will divide this function in 3 parts and then will solve them separately and find their domain. Then, we will get 3 different domains. So, we will find the intersection of all the three domains. That value of intersection will be the final answer.

Complete step-by-step answer:
Divide this function in three parts.
 $ f\left( x \right) = {f_1}\left( x \right) + {f_2}\left( x \right) + {f_3}\left( x \right) $
Now, we will find the domain of each part of the divided section.
First take $ {f_1}\left( x \right) $
 $ {f_1}\left( x \right) = {\cos ^{ - 1}}\left( {x - {x^2}} \right) $
Domain of $ {\cos ^{ - 1}} $ is
 $ - 1 \leqslant {\cos ^{ - 1}} \leqslant 1 $
From this, we can also get
 $ - 1 \leqslant x - {x^2} $
 $ {x^2} - x - 1 \leqslant 0 $
Now $ {x^2} - x - 1 $ is a quadratic equation. We will find its roots using the formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
In this quadratic equation value a is 1, b is -1 and c is also -1.
 $ x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{2} $
 $ x = \dfrac{{1 \pm \sqrt 5 }}{2} $
So, the domain of $ {f_1}\left( x \right) $ is
 $ \dfrac{{1 - \sqrt 5 }}{2} \leqslant x \leqslant \dfrac{{1 + \sqrt 5 }}{2} $
Now, we will find the domain of $ {f_2}\left( x \right) $
 $ {f_2}\left( x \right) = \sqrt {1 - \dfrac{1}{{\left| x \right|}}} $
Any term inside the root is equal to 0 or greater than 0. So,
 $ 1 - \dfrac{1}{{\left| x \right|}} \geqslant 0 $
 $ 1 \geqslant \dfrac{1}{{\left| x \right|}} $
 $ \left| x \right| \geqslant 1 $
The domain of $ {f_2}\left( x \right) $ is $ \left[ {x \leqslant - 1,x \geqslant 1} \right] $
Now, we will find the domain of $ {f_3}\left( x \right) $
 $ {f_3}\left( x \right) = \dfrac{1}{{{x^2} - 1}} $
X2 can not be equal to 1 as if it will be equal to 1, then the denominator will become 0. So,
The domain of $ {f_3}\left( x \right) $ is:
 $ \left[ {{x^2} < 1,{x^2} \geqslant 2} \right] $
By combining all the three domains we get,
 $ \left[ {\sqrt 2 ,\dfrac{{1 + \sqrt 5 }}{2}} \right] $
The final domain of the function is $ \left[ {\sqrt 2 ,\dfrac{{1 + \sqrt 5 }}{2}} \right] $ .
So, option (B) is the correct answer.
So, the correct answer is “Option B”.

Note: The domain of a function corresponds to the possible values of the independent variable which is x in this case. For this, the entire function is defined. The different values of x define the function.