
If the domain of the function $f(x) = {\left[ {{{\log }_{10}}\left( {\dfrac{{5x - {x^2}}}{4}} \right)} \right]^{\dfrac{1}{2}}}$ is $a \leqslant x \leqslant b$, then $a.b$ is
Answer
163.5k+ views
Hint: Check the domain of the logarithmic function and the square root function. Write two inequalities based on the earlier two domains. Find the range from these two inequalities and then find the intersection of the two ranges obtained to get the final answer.
Formula used: If ${\log _a}b = n$, then $b = {a^n}$
Complete step-by-step solution:
We know that for ${\log _b}y$ to exist $y$ needs to be greater than 0. In mathematical terms, we can say that the domain of ${\log _b}y$ is $y \in (0,\infty )$.
Therefore, $\dfrac{{5x - {x^2}}}{4} > 0$
$5x - {x^2} > 0$
$x(5 - x) > 0$
$x \in (0,5)$
We also know that for $\sqrt y $ to exist, y needs to be greater than or equal to 0. In mathematical terms, we can say that the domain of $\sqrt y $ is $y \in [0,\infty )$.
Therefore, ${\log _{10}}\left( {\dfrac{{5x - {x^2}}}{4}} \right) \geqslant 0$
$\dfrac{{5x - {x^2}}}{4} \geqslant {10^0}$
$\dfrac{{5x - {x^2}}}{4} \geqslant 1$
Solving further, we get
$5x - {x^2} \geqslant 4$
${x^2} - 5x + 4 \leqslant 0$
$\left( {x - 4} \right)\left( {x - 1} \right) \leqslant 0$
$x \in \left[ {1,4} \right]$
Taking the intersection of $(0,5)$ and $\left[ {1,4} \right]$ we get $x \in \left[ {1,4} \right]$
Therefore, $a = 1$ and $b = 4$.
Therefore, $a.b$ is $4$.
Note: We can directly check the domain of the logarithmic part of the question to get the range and then the answer. However, that might not always work as the range obtained from a given domain of a square root function might be smaller than the range of the logarithmic part. Therefore, it is safer to get the range of all the functions in the given question.
Formula used: If ${\log _a}b = n$, then $b = {a^n}$
Complete step-by-step solution:
We know that for ${\log _b}y$ to exist $y$ needs to be greater than 0. In mathematical terms, we can say that the domain of ${\log _b}y$ is $y \in (0,\infty )$.
Therefore, $\dfrac{{5x - {x^2}}}{4} > 0$
$5x - {x^2} > 0$
$x(5 - x) > 0$
$x \in (0,5)$
We also know that for $\sqrt y $ to exist, y needs to be greater than or equal to 0. In mathematical terms, we can say that the domain of $\sqrt y $ is $y \in [0,\infty )$.
Therefore, ${\log _{10}}\left( {\dfrac{{5x - {x^2}}}{4}} \right) \geqslant 0$
$\dfrac{{5x - {x^2}}}{4} \geqslant {10^0}$
$\dfrac{{5x - {x^2}}}{4} \geqslant 1$
Solving further, we get
$5x - {x^2} \geqslant 4$
${x^2} - 5x + 4 \leqslant 0$
$\left( {x - 4} \right)\left( {x - 1} \right) \leqslant 0$
$x \in \left[ {1,4} \right]$
Taking the intersection of $(0,5)$ and $\left[ {1,4} \right]$ we get $x \in \left[ {1,4} \right]$
Therefore, $a = 1$ and $b = 4$.
Therefore, $a.b$ is $4$.
Note: We can directly check the domain of the logarithmic part of the question to get the range and then the answer. However, that might not always work as the range obtained from a given domain of a square root function might be smaller than the range of the logarithmic part. Therefore, it is safer to get the range of all the functions in the given question.
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