117386
The domain of the function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\)
1 \([2,3]\)
2 \([2,3)\)
3 \([1,2]\)
4 \([1,2)\)
Explanation:
B Given, function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) Case - (I) :- Since, \(\sin ^{-1}(\mathrm{x})\) lies between \([-1,1]\), So, \(-1 \leq \mathrm{x}-3 \leq 1\) \(-1+3 \leq \mathrm{x} \leq 1+3\) \(2 \leq \mathrm{x} \leq 4\) Case-II :- Since, the denominator has to be greater than or equal to 0 , Hence, \(9-x^2 \geq 0\) \(x^2 \leq 9\) \(x \leq \pm 3\) \(-3 \leq x \leq 3\) So, \(f(x)\) will be defined in the common portion of \(2 \leq x\) \(\leq 4\) and \(-3 \leq x \leq 3\) And the common portion is \(2 \leq \mathrm{x}\lt 3\) or, \(x \in[2,3)\)
Jamia Millia Islamia-2007
Sets, Relation and Function
117387
Domain of the function \(\log \left|x^2-9\right|\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-[-3,3]\)
3 \(\mathrm{R}-\{-3,3\}\)
4 None of these
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\log \left|\mathrm{x}^2-9\right|\) We know \(\log \left|x^2-9\right|\) is defined when, \(\left|x^2-9\right|>0\), Now \(\left|x^2-9\right|\) is positive for all \(x \in R\). But, \(\log \left|x^2-9\right|\) is not defined when, \(\left|x^2-9\right|=0\) \(|(x-3)(x+3)|=0\) \(x=-3,3\)Hence, domain of the function \(\log \left|\mathrm{x}^2-9\right|\) is \(\mathrm{R}-\{-3,3\}\).
Manipal UGET-2012
Sets, Relation and Function
117388
In which of the following domains \(f(x)\) is continuous, where \(f(x)=\frac{|x+2|}{\tan ^{-1}(x+2)}\) ? (Note : \(\mathbf{R}\) denotes set of real numbers)
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{-2,2\}\)
3 \(\mathrm{R}-\{-2\}\)
4 \(\{-2,2\}\)
Explanation:
C Given, \(f(x)=\frac{|x+2|}{\tan ^{-1}(x+2)}\) For \(\mathrm{f}(\mathrm{x})\) to be defined, \(\mathrm{x}+2 \neq 0\) \(\Rightarrow \mathrm{x} \neq-2\)Thus we can see that \(f(x)\) is defined and continuous for all \(x\) except \(x=-2\) Thus, required domain is \(R-\{-2\}\).
117386
The domain of the function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\)
1 \([2,3]\)
2 \([2,3)\)
3 \([1,2]\)
4 \([1,2)\)
Explanation:
B Given, function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) Case - (I) :- Since, \(\sin ^{-1}(\mathrm{x})\) lies between \([-1,1]\), So, \(-1 \leq \mathrm{x}-3 \leq 1\) \(-1+3 \leq \mathrm{x} \leq 1+3\) \(2 \leq \mathrm{x} \leq 4\) Case-II :- Since, the denominator has to be greater than or equal to 0 , Hence, \(9-x^2 \geq 0\) \(x^2 \leq 9\) \(x \leq \pm 3\) \(-3 \leq x \leq 3\) So, \(f(x)\) will be defined in the common portion of \(2 \leq x\) \(\leq 4\) and \(-3 \leq x \leq 3\) And the common portion is \(2 \leq \mathrm{x}\lt 3\) or, \(x \in[2,3)\)
Jamia Millia Islamia-2007
Sets, Relation and Function
117387
Domain of the function \(\log \left|x^2-9\right|\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-[-3,3]\)
3 \(\mathrm{R}-\{-3,3\}\)
4 None of these
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\log \left|\mathrm{x}^2-9\right|\) We know \(\log \left|x^2-9\right|\) is defined when, \(\left|x^2-9\right|>0\), Now \(\left|x^2-9\right|\) is positive for all \(x \in R\). But, \(\log \left|x^2-9\right|\) is not defined when, \(\left|x^2-9\right|=0\) \(|(x-3)(x+3)|=0\) \(x=-3,3\)Hence, domain of the function \(\log \left|\mathrm{x}^2-9\right|\) is \(\mathrm{R}-\{-3,3\}\).
Manipal UGET-2012
Sets, Relation and Function
117388
In which of the following domains \(f(x)\) is continuous, where \(f(x)=\frac{|x+2|}{\tan ^{-1}(x+2)}\) ? (Note : \(\mathbf{R}\) denotes set of real numbers)
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{-2,2\}\)
3 \(\mathrm{R}-\{-2\}\)
4 \(\{-2,2\}\)
Explanation:
C Given, \(f(x)=\frac{|x+2|}{\tan ^{-1}(x+2)}\) For \(\mathrm{f}(\mathrm{x})\) to be defined, \(\mathrm{x}+2 \neq 0\) \(\Rightarrow \mathrm{x} \neq-2\)Thus we can see that \(f(x)\) is defined and continuous for all \(x\) except \(x=-2\) Thus, required domain is \(R-\{-2\}\).
117386
The domain of the function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\)
1 \([2,3]\)
2 \([2,3)\)
3 \([1,2]\)
4 \([1,2)\)
Explanation:
B Given, function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) Case - (I) :- Since, \(\sin ^{-1}(\mathrm{x})\) lies between \([-1,1]\), So, \(-1 \leq \mathrm{x}-3 \leq 1\) \(-1+3 \leq \mathrm{x} \leq 1+3\) \(2 \leq \mathrm{x} \leq 4\) Case-II :- Since, the denominator has to be greater than or equal to 0 , Hence, \(9-x^2 \geq 0\) \(x^2 \leq 9\) \(x \leq \pm 3\) \(-3 \leq x \leq 3\) So, \(f(x)\) will be defined in the common portion of \(2 \leq x\) \(\leq 4\) and \(-3 \leq x \leq 3\) And the common portion is \(2 \leq \mathrm{x}\lt 3\) or, \(x \in[2,3)\)
Jamia Millia Islamia-2007
Sets, Relation and Function
117387
Domain of the function \(\log \left|x^2-9\right|\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-[-3,3]\)
3 \(\mathrm{R}-\{-3,3\}\)
4 None of these
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\log \left|\mathrm{x}^2-9\right|\) We know \(\log \left|x^2-9\right|\) is defined when, \(\left|x^2-9\right|>0\), Now \(\left|x^2-9\right|\) is positive for all \(x \in R\). But, \(\log \left|x^2-9\right|\) is not defined when, \(\left|x^2-9\right|=0\) \(|(x-3)(x+3)|=0\) \(x=-3,3\)Hence, domain of the function \(\log \left|\mathrm{x}^2-9\right|\) is \(\mathrm{R}-\{-3,3\}\).
Manipal UGET-2012
Sets, Relation and Function
117388
In which of the following domains \(f(x)\) is continuous, where \(f(x)=\frac{|x+2|}{\tan ^{-1}(x+2)}\) ? (Note : \(\mathbf{R}\) denotes set of real numbers)
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{-2,2\}\)
3 \(\mathrm{R}-\{-2\}\)
4 \(\{-2,2\}\)
Explanation:
C Given, \(f(x)=\frac{|x+2|}{\tan ^{-1}(x+2)}\) For \(\mathrm{f}(\mathrm{x})\) to be defined, \(\mathrm{x}+2 \neq 0\) \(\Rightarrow \mathrm{x} \neq-2\)Thus we can see that \(f(x)\) is defined and continuous for all \(x\) except \(x=-2\) Thus, required domain is \(R-\{-2\}\).
117386
The domain of the function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\)
1 \([2,3]\)
2 \([2,3)\)
3 \([1,2]\)
4 \([1,2)\)
Explanation:
B Given, function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) Case - (I) :- Since, \(\sin ^{-1}(\mathrm{x})\) lies between \([-1,1]\), So, \(-1 \leq \mathrm{x}-3 \leq 1\) \(-1+3 \leq \mathrm{x} \leq 1+3\) \(2 \leq \mathrm{x} \leq 4\) Case-II :- Since, the denominator has to be greater than or equal to 0 , Hence, \(9-x^2 \geq 0\) \(x^2 \leq 9\) \(x \leq \pm 3\) \(-3 \leq x \leq 3\) So, \(f(x)\) will be defined in the common portion of \(2 \leq x\) \(\leq 4\) and \(-3 \leq x \leq 3\) And the common portion is \(2 \leq \mathrm{x}\lt 3\) or, \(x \in[2,3)\)
Jamia Millia Islamia-2007
Sets, Relation and Function
117387
Domain of the function \(\log \left|x^2-9\right|\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-[-3,3]\)
3 \(\mathrm{R}-\{-3,3\}\)
4 None of these
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\log \left|\mathrm{x}^2-9\right|\) We know \(\log \left|x^2-9\right|\) is defined when, \(\left|x^2-9\right|>0\), Now \(\left|x^2-9\right|\) is positive for all \(x \in R\). But, \(\log \left|x^2-9\right|\) is not defined when, \(\left|x^2-9\right|=0\) \(|(x-3)(x+3)|=0\) \(x=-3,3\)Hence, domain of the function \(\log \left|\mathrm{x}^2-9\right|\) is \(\mathrm{R}-\{-3,3\}\).
Manipal UGET-2012
Sets, Relation and Function
117388
In which of the following domains \(f(x)\) is continuous, where \(f(x)=\frac{|x+2|}{\tan ^{-1}(x+2)}\) ? (Note : \(\mathbf{R}\) denotes set of real numbers)
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{-2,2\}\)
3 \(\mathrm{R}-\{-2\}\)
4 \(\{-2,2\}\)
Explanation:
C Given, \(f(x)=\frac{|x+2|}{\tan ^{-1}(x+2)}\) For \(\mathrm{f}(\mathrm{x})\) to be defined, \(\mathrm{x}+2 \neq 0\) \(\Rightarrow \mathrm{x} \neq-2\)Thus we can see that \(f(x)\) is defined and continuous for all \(x\) except \(x=-2\) Thus, required domain is \(R-\{-2\}\).
