
The domain of the function \[f\left( x \right)=\dfrac{\left( \sqrt{-\log _{0.3}^{\left( x-1 \right)}} \right)}{\left( \sqrt{-{{x}^{2}}+2x+8} \right)}\]
1.\[\left( 1,4 \right)\]
2.\[\left( -2,4 \right)\]
3.\[\left[ 2,\left. 4 \right) \right.\]
4. None of these
Answer
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Hint: In order to find the domain of the given function \[f\left( x \right)=\dfrac{\left( \sqrt{-\log _{0.3}^{\left( x-1 \right)}} \right)}{\left( \sqrt{-{{x}^{2}}+2x+8} \right)}\], firstly, we will be finding the roots of the given quadratic equation. Then we will be checking for the function at which they are defined for \[f\]. Then upon checking for them, we will be obtaining the domain of the function.
Complete step-by-step solution:
Now let us learn about the domain and range of functions. A domain of a function \[f\left( x \right)\]is the set of all values for which the function is defined and the range of the function is the set of all values that \[f\] takes. We can also define the special functions whose domains are limited. The natural domain of a function is the set of all allowable input values. Natural domain is basically the x values for which the function is defined. 'Domain' or 'restricted domain' is 'man-made' .
Now let us find the domain of the given function i.e. \[f\left( x \right)=\dfrac{\left( \sqrt{-\log _{0.3}^{\left( x-1 \right)}} \right)}{\left( \sqrt{-{{x}^{2}}+2x+8} \right)}\]
Now let us find the roots of the equation \[\left( \sqrt{-{{x}^{2}}+2x+8} \right)\]
\[\begin{align}
& -{{x}^{2}}+2x+8=0 \\
& \Rightarrow {{x}^{2}}-2x-8=0 \\
& \Rightarrow {{x}^{2}}+2x-4x-8=0 \\
& \Rightarrow x\left( x+2 \right)-4\left( x+2 \right)=0 \\
& \Rightarrow \left( x+2 \right)\left( x-4 \right) \\
\end{align}\]
\[\therefore \] The roots are \[\left( x+2 \right),\left( x-4 \right)\].
Now let us start checking for the domain.
\[-\log _{0.3}^{\left( x-1 \right)}\ge 0,\left( x+2 \right)\left( x-4 \right)<0\]
Upon solving,
\[\left( x-1 \right)>0\] and \[x\ne -2,4\]
\[\begin{align}
& \Rightarrow x\ge 2,-2 < x < 4,x >1 ,x\ne -2,4 \\
& \Rightarrow x\in \left[ 2,\left. 4 \right) \right. \\
\end{align}\]
\[\therefore \] The domain of the function is \[\left[ 2,\left. 4 \right) \right.\].
Hence option 3 is correct.
Note:We must note that for quadratic equations, we must be finding the roots in order to find the domain of any function. While finding the domain, the denominator of a fraction cannot be zero. While solving for a domain, the function must be checked for all possible values.
Complete step-by-step solution:
Now let us learn about the domain and range of functions. A domain of a function \[f\left( x \right)\]is the set of all values for which the function is defined and the range of the function is the set of all values that \[f\] takes. We can also define the special functions whose domains are limited. The natural domain of a function is the set of all allowable input values. Natural domain is basically the x values for which the function is defined. 'Domain' or 'restricted domain' is 'man-made' .
Now let us find the domain of the given function i.e. \[f\left( x \right)=\dfrac{\left( \sqrt{-\log _{0.3}^{\left( x-1 \right)}} \right)}{\left( \sqrt{-{{x}^{2}}+2x+8} \right)}\]
Now let us find the roots of the equation \[\left( \sqrt{-{{x}^{2}}+2x+8} \right)\]
\[\begin{align}
& -{{x}^{2}}+2x+8=0 \\
& \Rightarrow {{x}^{2}}-2x-8=0 \\
& \Rightarrow {{x}^{2}}+2x-4x-8=0 \\
& \Rightarrow x\left( x+2 \right)-4\left( x+2 \right)=0 \\
& \Rightarrow \left( x+2 \right)\left( x-4 \right) \\
\end{align}\]
\[\therefore \] The roots are \[\left( x+2 \right),\left( x-4 \right)\].
Now let us start checking for the domain.
\[-\log _{0.3}^{\left( x-1 \right)}\ge 0,\left( x+2 \right)\left( x-4 \right)<0\]
Upon solving,
\[\left( x-1 \right)>0\] and \[x\ne -2,4\]
\[\begin{align}
& \Rightarrow x\ge 2,-2 < x < 4,x >1 ,x\ne -2,4 \\
& \Rightarrow x\in \left[ 2,\left. 4 \right) \right. \\
\end{align}\]
\[\therefore \] The domain of the function is \[\left[ 2,\left. 4 \right) \right.\].
Hence option 3 is correct.
Note:We must note that for quadratic equations, we must be finding the roots in order to find the domain of any function. While finding the domain, the denominator of a fraction cannot be zero. While solving for a domain, the function must be checked for all possible values.
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