Question

Let \[f:\left[ {4,\infty } \right) \to \left[ {1,\infty } \right)\] be a function defined by \[f\left( x \right) = {5^{x\left( {x - 4} \right)}}\] . Then what is the value of \[{f^{ - 1}}\left( x \right)\]?
A. \[2 - \sqrt {4 + {{\log }_5}x} \]
B. \[2 + \sqrt {4 + {{\log }_5}x} \]
C. \[{\left( {\dfrac{1}{5}} \right)^{x\left( {x - 4} \right)}}\]s
D. None of these

Answer 1

Hint: Here, an exponential function and its domain and range is given. First, consider the given function as \[y = {5^{x\left( {x - 4} \right)}}\]. Then, take the logarithm to base 5 on both sides and simplify the equation. After that, apply the quadratic formula and find the value of the variable \[x\]. Then, check the equality condition of the variable \[x\]. In the end, find the value of \[{f^{ - 1}}\left( y \right)\] and then substitute \[y = x\] in the equation and solve it to get the required answer.

Formula Used: Quadratic Formula: The solution of the quadratic equation \[a{x^2} + bx + c = 0\] , where \[a \ne 0\] is
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step solution: The given function is \[f\left( x \right) = {5^{x\left( {x - 4} \right)}}\].

Let’s simplify the given function.
Consider, \[y = {5^{x\left( {x - 4} \right)}}\]
Take the logarithm to base 5 on both sides of the above equation.
We get,
\[{\log _5}\left( y \right) = {\log _5}\left( {{5^{x\left( {x - 4} \right)}}} \right)\]
\[ \Rightarrow {\log _5}y = x\left( {x - 4} \right)\]
\[ \Rightarrow {\log _5}y = {x^2} - 4x\]
\[ \Rightarrow {x^2} - 4x - {\log _5}y = 0\]
Now apply the quadratic formula and find the value of the variable \[x\].
Here, \[a = 1,b = - 4\] and \[c = - {\log _5}y\]
\[x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 1 \right)\left( { - {{\log }_5}y} \right)} }}{{2\left( 1 \right)}}\]
\[ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 4{{\log }_5}y} }}{2}\]
\[ \Rightarrow x = \dfrac{{2\left( {2 \pm \sqrt {4 + {{\log }_5}y} } \right)}}{2}\]
\[ \Rightarrow x = 2 \pm \sqrt {4 + {{\log }_5}y} \]
But, the domain of the given function is \[\left[ {4,\infty } \right)\].
So, the value is \[x \ge 4\].
\[x = 2 + \sqrt {4 + {{\log }_5}y} \]
\[ \Rightarrow {f^{ - 1}}\left( y \right) = 2 + \sqrt {4 + {{\log }_5}y} \]

Now substitute \[y = x\] in the above equation.
We get,
\[{f^{ - 1}}\left( x \right) = 2 + \sqrt {4 + {{\log }_5}x} \]

Option ‘B’ is correct

Note: Students often make mistakes in the quadratic formula. They take \[x = \dfrac{{b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] as the quadratic formula. But the correct formula is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].