
If \[f(x) = \dfrac{{\left[ {2x - 1} \right]}}{{\left[ {x + 5} \right]}},x \ne \dfrac{1}{2}\] then what is the value of \[{f^{ - 1}}(x)\] ?
A.\[\dfrac{{\left[ {x + 5} \right]}}{{\left[ {2x - 1} \right]}},x \ne \dfrac{1}{2}\]
B.\[\dfrac{{\left[ {5x + 1} \right]}}{{\left[ {2 - x} \right]}},x \ne 2\]
C.\[\dfrac{{\left[ {x - 5} \right]}}{{\left[ {2x + 1} \right]}},x \ne - \dfrac{1}{2}\]
D.\[\dfrac{{\left[ {5x + 1} \right]}}{{\left[ {2 - x} \right]}},x \ne 2\]
Answer
162.3k+ views
Hint: We will evaluate the inverse of the function by first writing the given function in the form of \[y = f(x)\]. Then we will interchange \[x\] with \[y\] and \[y\] with \[x\]. Finally we will use the formula of inverse \[{f^{ - 1}}(x) = y\] to get the inverse of the function .
Formula used: If \[y = f(x)\] is a function then \[{f^{ - 1}}(y) = x\].
Complete step by step solution: Given- \[f(x) = \dfrac{{\left[ {2x - 1} \right]}}{{\left[ {x + 5} \right]}},x \ne \dfrac{1}{2}\]
Writing the equation in the form of \[y = f(x)\]
∴ \[y = \dfrac{{\left[ {2x - 1} \right]}}{{\left[ {x + 5} \right]}},x \ne \dfrac{1}{2}\]
Replacing \[\,x\] with \[y\]
\[x = \dfrac{{\left[ {2y - 1} \right]}}{{\left[ {y + 5} \right]}},y \ne \dfrac{1}{2}\]
On cross multiplying the terms
\[x\left[ {y + 5} \right] = \left[ {2y - 1} \right]\]
Solving for \[y\] from the equation
\[yx + 5x = 2y - 1\]
Rearranging the terms
\[ \Rightarrow 5x + 1 = y\left[ {2 - x} \right]\]
Dividing by \[\left[ {2 - x} \right]\] both sides
\[ \Rightarrow y = \dfrac{{5x + 1}}{{2 - x}}\] (1.1)
We know by the formula of inverse that
\[{f^{ - 1}}(x) = y\] (1.2)
By comparing equation (1.1) and equation (1.2)
\[\begin{array}{l}{f^{ - 1}}(x) = \dfrac{{\left[ {5x + 1} \right]}}{{\left[ {2 - x} \right]}}\\\end{array}\]
Also, the denominator in the function cannot be equal to zero for inverse to be defined so
\[2 - x \ne 0\]
\[ \Rightarrow x \ne 2\]
Thus
\[\begin{array}{l}{f^{ - 1}}(x) = \dfrac{{\left[ {5x + 1} \right]}}{{\left[ {2 - x} \right]}},x \ne 2\\\end{array}\]
So, Option ‘B’ is correct
Note: In many cases students make mistakes in the inverse calculation part while not interchanging x with y in equation \[y = f(x)\] and thus the calculated inverse is wrong. To avoid this mistake, students must make sure to interchange \[x\] with \[y\] in all places after writing the equation in the form of \[y = f(x)\].
Formula used: If \[y = f(x)\] is a function then \[{f^{ - 1}}(y) = x\].
Complete step by step solution: Given- \[f(x) = \dfrac{{\left[ {2x - 1} \right]}}{{\left[ {x + 5} \right]}},x \ne \dfrac{1}{2}\]
Writing the equation in the form of \[y = f(x)\]
∴ \[y = \dfrac{{\left[ {2x - 1} \right]}}{{\left[ {x + 5} \right]}},x \ne \dfrac{1}{2}\]
Replacing \[\,x\] with \[y\]
\[x = \dfrac{{\left[ {2y - 1} \right]}}{{\left[ {y + 5} \right]}},y \ne \dfrac{1}{2}\]
On cross multiplying the terms
\[x\left[ {y + 5} \right] = \left[ {2y - 1} \right]\]
Solving for \[y\] from the equation
\[yx + 5x = 2y - 1\]
Rearranging the terms
\[ \Rightarrow 5x + 1 = y\left[ {2 - x} \right]\]
Dividing by \[\left[ {2 - x} \right]\] both sides
\[ \Rightarrow y = \dfrac{{5x + 1}}{{2 - x}}\] (1.1)
We know by the formula of inverse that
\[{f^{ - 1}}(x) = y\] (1.2)
By comparing equation (1.1) and equation (1.2)
\[\begin{array}{l}{f^{ - 1}}(x) = \dfrac{{\left[ {5x + 1} \right]}}{{\left[ {2 - x} \right]}}\\\end{array}\]
Also, the denominator in the function cannot be equal to zero for inverse to be defined so
\[2 - x \ne 0\]
\[ \Rightarrow x \ne 2\]
Thus
\[\begin{array}{l}{f^{ - 1}}(x) = \dfrac{{\left[ {5x + 1} \right]}}{{\left[ {2 - x} \right]}},x \ne 2\\\end{array}\]
So, Option ‘B’ is correct
Note: In many cases students make mistakes in the inverse calculation part while not interchanging x with y in equation \[y = f(x)\] and thus the calculated inverse is wrong. To avoid this mistake, students must make sure to interchange \[x\] with \[y\] in all places after writing the equation in the form of \[y = f(x)\].
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