117294
The domain and range of the function \(f\) given by \(f(x)=2-|x-5|\) is
1 Domain \(=\mathrm{R}^{+}\), Range \(=(-\infty, 1]\)
2 Domain \(=\mathrm{R}\), Range \(=(-\infty, 2]\)
3 Domain \(=\mathrm{R}\), Range \(=(-\infty, 2)\)
4 Domain \(=\mathrm{R}^{+}\), Range \(=(-\infty, 2]\)
Explanation:
B Given, \(f(x)=2-|x-5|\) Domain of \(f(x)\) is defined for all real values of \(x\). Since, \(|x-5| \geq 0 \Rightarrow-|x-5| \leq 0\) \(\Rightarrow 2-|\mathrm{x}-5| \leq 2\) \(\Rightarrow \mathrm{f}(\mathrm{x}) \leq 2\) Hence, range of \(f(x)\) is \((-\infty, 2]\).
VITEEE-2016
Sets, Relation and Function
117295
The domain of the function \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2} \text { is }\)
1 \(]-3,-2.5[\cap]-2.5,-2[\)
2 \([-2,0[\cup] 0,1[\)
3 \(] 0,1[\)
4 None of the above
Explanation:
B We have, \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) For this function to be define - \((1-x)>0, \neq 1 \text { and }(x+2) \geq 0\) \(\Rightarrow \quad \mathrm{x}\lt 1, \neq 0\) and \(\mathrm{x} \geq-2\) So, the domain of \(f(x)\) is - \(\mathrm{x} \in[-2,0[\cup] 0,1[\)
VITEEE-2013
Sets, Relation and Function
117296
The domain of the function \(f(x)=\frac{\sqrt{4-x^2}}{\sin ^{-1}(2-x)}\) is
1 \([0,2]\)
2 \([0,2)\)
3 \([1,2)\)
4 \([1,2]\)
Explanation:
C Case (I) - \(4-x^2 \geq 0\) \(x^2-4 \leq 0\) \((x-2)(x+2) \leq 0\) \(x \in[2,2]\) Case (II) - \(\sin ^{-1}(2-x) \neq 0 \text { and } 2-x \neq 0\) \(-1 \leq \sin ^{-1}(2-x) \leq 1 \text { and } x \neq 2\) \(-1 \leq(2-x) \leq 1\) \(-1-2 \leq-x \leq 1-2\) \(-3 \leq-x \leq-1\) \(3 \geq x \geq 1\) \(x \in[1,3]-\{2\}\) \({[1,2] \cup(2,3]}\)So, the domain of the given function is \([1,2)\).
VITEEE-2012
Sets, Relation and Function
117297
If \(D\) is the set of all real \(x\) such that \(1-e^{(1 / x)-1}\) is positive, then \(D\) is equal to
117294
The domain and range of the function \(f\) given by \(f(x)=2-|x-5|\) is
1 Domain \(=\mathrm{R}^{+}\), Range \(=(-\infty, 1]\)
2 Domain \(=\mathrm{R}\), Range \(=(-\infty, 2]\)
3 Domain \(=\mathrm{R}\), Range \(=(-\infty, 2)\)
4 Domain \(=\mathrm{R}^{+}\), Range \(=(-\infty, 2]\)
Explanation:
B Given, \(f(x)=2-|x-5|\) Domain of \(f(x)\) is defined for all real values of \(x\). Since, \(|x-5| \geq 0 \Rightarrow-|x-5| \leq 0\) \(\Rightarrow 2-|\mathrm{x}-5| \leq 2\) \(\Rightarrow \mathrm{f}(\mathrm{x}) \leq 2\) Hence, range of \(f(x)\) is \((-\infty, 2]\).
VITEEE-2016
Sets, Relation and Function
117295
The domain of the function \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2} \text { is }\)
1 \(]-3,-2.5[\cap]-2.5,-2[\)
2 \([-2,0[\cup] 0,1[\)
3 \(] 0,1[\)
4 None of the above
Explanation:
B We have, \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) For this function to be define - \((1-x)>0, \neq 1 \text { and }(x+2) \geq 0\) \(\Rightarrow \quad \mathrm{x}\lt 1, \neq 0\) and \(\mathrm{x} \geq-2\) So, the domain of \(f(x)\) is - \(\mathrm{x} \in[-2,0[\cup] 0,1[\)
VITEEE-2013
Sets, Relation and Function
117296
The domain of the function \(f(x)=\frac{\sqrt{4-x^2}}{\sin ^{-1}(2-x)}\) is
1 \([0,2]\)
2 \([0,2)\)
3 \([1,2)\)
4 \([1,2]\)
Explanation:
C Case (I) - \(4-x^2 \geq 0\) \(x^2-4 \leq 0\) \((x-2)(x+2) \leq 0\) \(x \in[2,2]\) Case (II) - \(\sin ^{-1}(2-x) \neq 0 \text { and } 2-x \neq 0\) \(-1 \leq \sin ^{-1}(2-x) \leq 1 \text { and } x \neq 2\) \(-1 \leq(2-x) \leq 1\) \(-1-2 \leq-x \leq 1-2\) \(-3 \leq-x \leq-1\) \(3 \geq x \geq 1\) \(x \in[1,3]-\{2\}\) \({[1,2] \cup(2,3]}\)So, the domain of the given function is \([1,2)\).
VITEEE-2012
Sets, Relation and Function
117297
If \(D\) is the set of all real \(x\) such that \(1-e^{(1 / x)-1}\) is positive, then \(D\) is equal to
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117294
The domain and range of the function \(f\) given by \(f(x)=2-|x-5|\) is
1 Domain \(=\mathrm{R}^{+}\), Range \(=(-\infty, 1]\)
2 Domain \(=\mathrm{R}\), Range \(=(-\infty, 2]\)
3 Domain \(=\mathrm{R}\), Range \(=(-\infty, 2)\)
4 Domain \(=\mathrm{R}^{+}\), Range \(=(-\infty, 2]\)
Explanation:
B Given, \(f(x)=2-|x-5|\) Domain of \(f(x)\) is defined for all real values of \(x\). Since, \(|x-5| \geq 0 \Rightarrow-|x-5| \leq 0\) \(\Rightarrow 2-|\mathrm{x}-5| \leq 2\) \(\Rightarrow \mathrm{f}(\mathrm{x}) \leq 2\) Hence, range of \(f(x)\) is \((-\infty, 2]\).
VITEEE-2016
Sets, Relation and Function
117295
The domain of the function \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2} \text { is }\)
1 \(]-3,-2.5[\cap]-2.5,-2[\)
2 \([-2,0[\cup] 0,1[\)
3 \(] 0,1[\)
4 None of the above
Explanation:
B We have, \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) For this function to be define - \((1-x)>0, \neq 1 \text { and }(x+2) \geq 0\) \(\Rightarrow \quad \mathrm{x}\lt 1, \neq 0\) and \(\mathrm{x} \geq-2\) So, the domain of \(f(x)\) is - \(\mathrm{x} \in[-2,0[\cup] 0,1[\)
VITEEE-2013
Sets, Relation and Function
117296
The domain of the function \(f(x)=\frac{\sqrt{4-x^2}}{\sin ^{-1}(2-x)}\) is
1 \([0,2]\)
2 \([0,2)\)
3 \([1,2)\)
4 \([1,2]\)
Explanation:
C Case (I) - \(4-x^2 \geq 0\) \(x^2-4 \leq 0\) \((x-2)(x+2) \leq 0\) \(x \in[2,2]\) Case (II) - \(\sin ^{-1}(2-x) \neq 0 \text { and } 2-x \neq 0\) \(-1 \leq \sin ^{-1}(2-x) \leq 1 \text { and } x \neq 2\) \(-1 \leq(2-x) \leq 1\) \(-1-2 \leq-x \leq 1-2\) \(-3 \leq-x \leq-1\) \(3 \geq x \geq 1\) \(x \in[1,3]-\{2\}\) \({[1,2] \cup(2,3]}\)So, the domain of the given function is \([1,2)\).
VITEEE-2012
Sets, Relation and Function
117297
If \(D\) is the set of all real \(x\) such that \(1-e^{(1 / x)-1}\) is positive, then \(D\) is equal to
117294
The domain and range of the function \(f\) given by \(f(x)=2-|x-5|\) is
1 Domain \(=\mathrm{R}^{+}\), Range \(=(-\infty, 1]\)
2 Domain \(=\mathrm{R}\), Range \(=(-\infty, 2]\)
3 Domain \(=\mathrm{R}\), Range \(=(-\infty, 2)\)
4 Domain \(=\mathrm{R}^{+}\), Range \(=(-\infty, 2]\)
Explanation:
B Given, \(f(x)=2-|x-5|\) Domain of \(f(x)\) is defined for all real values of \(x\). Since, \(|x-5| \geq 0 \Rightarrow-|x-5| \leq 0\) \(\Rightarrow 2-|\mathrm{x}-5| \leq 2\) \(\Rightarrow \mathrm{f}(\mathrm{x}) \leq 2\) Hence, range of \(f(x)\) is \((-\infty, 2]\).
VITEEE-2016
Sets, Relation and Function
117295
The domain of the function \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2} \text { is }\)
1 \(]-3,-2.5[\cap]-2.5,-2[\)
2 \([-2,0[\cup] 0,1[\)
3 \(] 0,1[\)
4 None of the above
Explanation:
B We have, \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) For this function to be define - \((1-x)>0, \neq 1 \text { and }(x+2) \geq 0\) \(\Rightarrow \quad \mathrm{x}\lt 1, \neq 0\) and \(\mathrm{x} \geq-2\) So, the domain of \(f(x)\) is - \(\mathrm{x} \in[-2,0[\cup] 0,1[\)
VITEEE-2013
Sets, Relation and Function
117296
The domain of the function \(f(x)=\frac{\sqrt{4-x^2}}{\sin ^{-1}(2-x)}\) is
1 \([0,2]\)
2 \([0,2)\)
3 \([1,2)\)
4 \([1,2]\)
Explanation:
C Case (I) - \(4-x^2 \geq 0\) \(x^2-4 \leq 0\) \((x-2)(x+2) \leq 0\) \(x \in[2,2]\) Case (II) - \(\sin ^{-1}(2-x) \neq 0 \text { and } 2-x \neq 0\) \(-1 \leq \sin ^{-1}(2-x) \leq 1 \text { and } x \neq 2\) \(-1 \leq(2-x) \leq 1\) \(-1-2 \leq-x \leq 1-2\) \(-3 \leq-x \leq-1\) \(3 \geq x \geq 1\) \(x \in[1,3]-\{2\}\) \({[1,2] \cup(2,3]}\)So, the domain of the given function is \([1,2)\).
VITEEE-2012
Sets, Relation and Function
117297
If \(D\) is the set of all real \(x\) such that \(1-e^{(1 / x)-1}\) is positive, then \(D\) is equal to
117294
The domain and range of the function \(f\) given by \(f(x)=2-|x-5|\) is
1 Domain \(=\mathrm{R}^{+}\), Range \(=(-\infty, 1]\)
2 Domain \(=\mathrm{R}\), Range \(=(-\infty, 2]\)
3 Domain \(=\mathrm{R}\), Range \(=(-\infty, 2)\)
4 Domain \(=\mathrm{R}^{+}\), Range \(=(-\infty, 2]\)
Explanation:
B Given, \(f(x)=2-|x-5|\) Domain of \(f(x)\) is defined for all real values of \(x\). Since, \(|x-5| \geq 0 \Rightarrow-|x-5| \leq 0\) \(\Rightarrow 2-|\mathrm{x}-5| \leq 2\) \(\Rightarrow \mathrm{f}(\mathrm{x}) \leq 2\) Hence, range of \(f(x)\) is \((-\infty, 2]\).
VITEEE-2016
Sets, Relation and Function
117295
The domain of the function \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2} \text { is }\)
1 \(]-3,-2.5[\cap]-2.5,-2[\)
2 \([-2,0[\cup] 0,1[\)
3 \(] 0,1[\)
4 None of the above
Explanation:
B We have, \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) For this function to be define - \((1-x)>0, \neq 1 \text { and }(x+2) \geq 0\) \(\Rightarrow \quad \mathrm{x}\lt 1, \neq 0\) and \(\mathrm{x} \geq-2\) So, the domain of \(f(x)\) is - \(\mathrm{x} \in[-2,0[\cup] 0,1[\)
VITEEE-2013
Sets, Relation and Function
117296
The domain of the function \(f(x)=\frac{\sqrt{4-x^2}}{\sin ^{-1}(2-x)}\) is
1 \([0,2]\)
2 \([0,2)\)
3 \([1,2)\)
4 \([1,2]\)
Explanation:
C Case (I) - \(4-x^2 \geq 0\) \(x^2-4 \leq 0\) \((x-2)(x+2) \leq 0\) \(x \in[2,2]\) Case (II) - \(\sin ^{-1}(2-x) \neq 0 \text { and } 2-x \neq 0\) \(-1 \leq \sin ^{-1}(2-x) \leq 1 \text { and } x \neq 2\) \(-1 \leq(2-x) \leq 1\) \(-1-2 \leq-x \leq 1-2\) \(-3 \leq-x \leq-1\) \(3 \geq x \geq 1\) \(x \in[1,3]-\{2\}\) \({[1,2] \cup(2,3]}\)So, the domain of the given function is \([1,2)\).
VITEEE-2012
Sets, Relation and Function
117297
If \(D\) is the set of all real \(x\) such that \(1-e^{(1 / x)-1}\) is positive, then \(D\) is equal to
