Question

Find the domain of definition of the function \[f\left( x \right) = \sqrt {{{\log }_{10}}\dfrac{{\left[ {5x - {x^2}} \right]}}{4}} \] is
A. \[\left[ {1,4} \right]\]
B. \[\left[ {1,0} \right]\]
C. \[\left[ {0,5} \right]\]
D. \[\left[ {5,0} \right]\]

Answer 1

Hint: We have to find the domain of the given function \[f\left( x \right) = \sqrt {{{\log }_{10}}\dfrac{{\left[ {5x - {x^2}} \right]}}{4}} \]. We will find the value of \[x\] for which numerator are defined and we will apply the standard condition of logarithm function to be defined and square root function to be defined. So, by applying all these facts we can get the values of \[x\].

Formula Used:
The basic logarithmic expansion formula are given as
1. \[\begin{array}{l}{\log _b}m = x\\ \Rightarrow m = {b^x}\end{array}\]
2. \[{a^0} = 1\] where \[a \in R\]

Complete step-by-step solution:
Given that the function is\[f\left( x \right) = \sqrt {{{\log }_{10}}\dfrac{{\left[ {5x - {x^2}} \right]}}{4}} \]…(1)
We will find the domain of the given function by making two cases.
As we know that the quantity of logarithmic function is always greater then zero.
So, the first case is \[\dfrac{{5x - {x^2}}}{4} > 0\].
Now, we will simplify the above expression and find a quadratic equation.
\[\begin{array}{l}5x - {x^2} > 0 \times 4\\5x - {x^2} > 0\end{array}\]
Further, we will multiply the quadratic function with minus sign and we know that if any equation is multiply with minus sign then the inequality sign changes, we get
\[\begin{array}{l} - \left( {5x - {x^2}} \right) < 0\\{x^2} - 5x < 0\end{array}\]
Furthermore, we will simplify the above quadratic equation with factorization by taking out the common factor, we get
\[x\left( {x - 5} \right) < 0\]
So, there are two values of \[x\], we get
\[x < 0\] and \[x < 5\]
From above the domain of function lies in open bracket because the inequality is not with the equal sign.
Thus, the domain is \[x \in \left( {0,5} \right)\].
As we also know that the value under square root will never be negative.
So, the second case is \[{\log _{10}}\left( {\dfrac{{5x - {x^2}}}{4}} \right) \ge 0\].
Now, we will apply the logarithmic function expansion that is \[{\log _b}m = x\] then
\[m = {b^x}\], we get
\[\dfrac{{5x - {x^2}}}{4} \ge {\left( {10} \right)^0}\]
Further, we will apply the formula \[{a^0} = 1\] where \[a \in R\], we get
\[\dfrac{{5x - {x^2}}}{4} \ge 1\]
Furthermore, we will simplify the above expression to find quadratic equation, we get
\[\begin{array}{l}5x - {x^2} \ge 4\\5x - {x^2} - 4 \ge 0\end{array}\]
Now, we will multiply the quadratic function with minus sign and we know that if any equation is multiply with minus sign then the inequality sign changes, we get
\[{x^2} - 5x + 4 \le 0\]
Further, we will apply the factorisation method to find the roots of \[x\], we get
\[\begin{array}{l}{x^2} - 4x - x + 4 \le 0\\x\left( {x - 4} \right) - 1\left( {x - 4} \right) \le 0\end{array}\]
Furthermore, we will simplify the above quadratic equation with factorization by taking out the common factor, we get
\[\left( {x - 4} \right)\left( {x - 1} \right) \le 0\]
So, there are two values of \[x\], we get
\[x \le 1\] and \[x \le 4\]
From above the domain of function lies in a close bracket because the inequality is with the equal sign.
Thus, the domain is \[x \in \left[ {1,4} \right]\].

Hence, the option (A) is correct

Note: In these types of questions always try to satisfy the given function. The limitations of all functions like square root, logarithm function, trigonometric function, etc must satisfy some conditions to be defined.