Question

Let $ f\left( x \right) = \dfrac{1}{{\left( {1 - {x^2}} \right)}} $ . Then, $ range\left( f \right) = $ ?
A. $ \left( { - \infty ,1} \right] $
B. $ \left[ {1,\infty } \right) $
C. $ \left[ { - 1,1} \right] $
D.None of these

Answer 1

Hint: Range refers to the complete set of dependent variables of a function. In order to approach the range for the given function, write the value of $ x $ in terms of $ y $ by solving the function step by step. Check all the values that can be satisfied for the condition obtained for $ y $ , which gives the range for the function we need.

Complete step-by-step answer:
Let, $ y = f\left( x \right) $ that implies, $ y = \dfrac{1}{{\left( {1 - {x^2}} \right)}} $ .
Multiplying both the sides by $ \left( {1 - {x^2}} \right) $ :
 $ y \times \left( {1 - {x^2}} \right) = \dfrac{1}{{\left( {1 - {x^2}} \right)}} \times \left( {1 - {x^2}} \right) $
 $ \Rightarrow y\left( {1 - {x^2}} \right) = 1 $
Dividing both the Left-hand and right-hand side by $ y $ , and we get:
 $ \Rightarrow \dfrac{{y\left( {1 - {x^2}} \right)}}{y} = \dfrac{1}{y} $
 $ \Rightarrow 1 - {x^2} = \dfrac{1}{y} $
Adding both the sides by $ {x^2} $ :
\[ \Rightarrow 1 - {x^2} + {x^2} = \dfrac{1}{y} + {x^2}\]
 $ \Rightarrow 1 = \dfrac{1}{y} + {x^2} $
Subtracting both the sides by $ \dfrac{1}{y} $ , we get:
 $ \Rightarrow 1 - \dfrac{1}{y} = \dfrac{1}{y} + {x^2} - \dfrac{1}{y} $
 $ \Rightarrow 1 - \dfrac{1}{y} = {x^2} $
Exchanging the sides, the equation can be written as:
 $ \Rightarrow {x^2} = 1 - \dfrac{1}{y} $
Multiplying and dividing the operand $ 1 $ on the right side by $ y $ , in order to get a common denominator.
 $ \Rightarrow {x^2} = \dfrac{y}{y} - \dfrac{1}{y} $
Taking the denominator common, we get:
 $ \Rightarrow {x^2} = \dfrac{{y - 1}}{y} $
In order to get the value of $ x $ , taking square root both the sides:
 $ \Rightarrow \sqrt {{x^2}} = \sqrt {\dfrac{{y - 1}}{y}} $
Since, we know that the value of $ \sqrt {{x^2}} $ is $ x $ , applying this on the above equation, we get:
 $ \Rightarrow x = \sqrt {\dfrac{{y - 1}}{y}} $
From this equation that is: $ x = \sqrt {\dfrac{{y - 1}}{y}} $ , we need to know the range of the function.
Since, the value is under square root, and we know that the radicand inside the root must be greater than zero, that gives:
  $ \dfrac{{y - 1}}{y} \geqslant 0 $ ……………(1)
And, also we know that the denominator for a fraction cannot be zero as it will give an indefinite value, that implies:
 $ y \ne 0 $ ……………..(2)
Now, multiplying $ y $ on both the sides of the equation $ \dfrac{{y - 1}}{y} \geqslant 0 $ , we get:
 $ \dfrac{{y - 1}}{y} \times y \geqslant 0 \times y $
 $ \Rightarrow y - 1 \geqslant 0 $
Adding both the sides by $ 1 $ :
 $ \Rightarrow y - 1 + 1 \geqslant 0 + 1 $
 $ \Rightarrow y \geqslant 1 $ ………………. (3)
So, from (1), (2) and (3), we get that $ y $ is defined when:
 $ \dfrac{{y - 1}}{y} \geqslant 0 $ , $ y \ne 0 $ and $ y \geqslant 1 $
That means the range for the values of $ y $ that satisfies the above conditions is: $ [1,\infty ) $
Therefore, the range of the function $ f\left( x \right) = \dfrac{1}{{\left( {1 - {x^2}} \right)}} $ is $ [1,\infty ) $ or $ range\left( f \right) = \left[ {1,\infty } \right) $ .
Hence, Option B is correct.
So, the correct answer is “Option B”.

Note: The interval is one side closed and one side opened because the values are greater than equal to that means the value is also included, if it was only greater than that the value was not included and it will be an open bracket.
If in the radicand it’s a fraction then it’s important to check the limitations of the denominator.