117283
Domain of the real values function \(f(x)=\frac{x+2}{9-x^2}\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{3\}\)
3 \(\mathrm{R}-\{-3,3\}\)
4 \(-3 \leq \mathrm{x} \leq 3\)
Explanation:
C Given, \(f(x)=\frac{x+2}{9-x^2}\) is not defined if \(9-x^2=0 \Rightarrow x= \pm 3\) So, required domain \(=\mathrm{R}-\{-3,3\}\)
MHT-CET 20
Sets, Relation and Function
117287
If \(f: \mathbf{R}-\{2\} \rightarrow \mathbf{R}\) is function defined by \(f(x)=\frac{x^2-4}{x-2}\), then its range is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{2\}\)
3 \(\mathrm{R}-\{4\}\)
4 \(\mathrm{R}-\{-2,2\}\)
Explanation:
C We have, \(f(x)=\frac{x^2-4}{x-2} \Rightarrow y=\frac{(x-2)(x+2)}{x-2}=x+2\) \(\therefore \mathrm{x}=\mathrm{y}-2\) and domain of \(\mathrm{f}(\mathrm{x})\) is \(\mathrm{R}-\{2\}\) When \(x=2\), we get \(y=2+2=4\) So, range of \(f(x)\) is \(R-\{4\}\)
MHT-CET 20
Sets, Relation and Function
117289
The range of function \(f(x)=\sin x+\operatorname{cosec} x\) is
1 \([-1,1]\)
2 \((-1,1)\)
3 \(\mathrm{R}-[-2,2]\)
4 \(\mathrm{R}-(-2,2)\)
Explanation:
D Given, \(f(x)=\sin x+\operatorname{cosec} x=\sin x+\frac{1}{\sin x}\) We know that, the sum of the number and their reciprocal is either less than or equal to -2 or greater than or equal to 2 . i.e. \(x+\frac{1}{x} \leq-2\) or \(x+\frac{1}{x} \geq 2\). This range of \(f(x)\) is \(\mathrm{R}-(-2,2)\)
MHT-CET 20
Sets, Relation and Function
117302
The domain of the definition of the function \(y=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) is
1 \(x \geq-2\)
2 \(-3\lt x \leq-2\)
3 \(-2 \leq x\lt 0\)
4 \(-2 \leq x\lt 1\)
Explanation:
C Given,\(y=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) For this function domain to be defined \((1-x)>0,(1-x)>1\) and \((x+2)>0\) \(x\lt 1, x\lt 0\) and \(x \geq-2\) Common interval \(=-2 \leq \mathrm{x}\lt 0=\) Domain of \(\mathrm{y}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117283
Domain of the real values function \(f(x)=\frac{x+2}{9-x^2}\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{3\}\)
3 \(\mathrm{R}-\{-3,3\}\)
4 \(-3 \leq \mathrm{x} \leq 3\)
Explanation:
C Given, \(f(x)=\frac{x+2}{9-x^2}\) is not defined if \(9-x^2=0 \Rightarrow x= \pm 3\) So, required domain \(=\mathrm{R}-\{-3,3\}\)
MHT-CET 20
Sets, Relation and Function
117287
If \(f: \mathbf{R}-\{2\} \rightarrow \mathbf{R}\) is function defined by \(f(x)=\frac{x^2-4}{x-2}\), then its range is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{2\}\)
3 \(\mathrm{R}-\{4\}\)
4 \(\mathrm{R}-\{-2,2\}\)
Explanation:
C We have, \(f(x)=\frac{x^2-4}{x-2} \Rightarrow y=\frac{(x-2)(x+2)}{x-2}=x+2\) \(\therefore \mathrm{x}=\mathrm{y}-2\) and domain of \(\mathrm{f}(\mathrm{x})\) is \(\mathrm{R}-\{2\}\) When \(x=2\), we get \(y=2+2=4\) So, range of \(f(x)\) is \(R-\{4\}\)
MHT-CET 20
Sets, Relation and Function
117289
The range of function \(f(x)=\sin x+\operatorname{cosec} x\) is
1 \([-1,1]\)
2 \((-1,1)\)
3 \(\mathrm{R}-[-2,2]\)
4 \(\mathrm{R}-(-2,2)\)
Explanation:
D Given, \(f(x)=\sin x+\operatorname{cosec} x=\sin x+\frac{1}{\sin x}\) We know that, the sum of the number and their reciprocal is either less than or equal to -2 or greater than or equal to 2 . i.e. \(x+\frac{1}{x} \leq-2\) or \(x+\frac{1}{x} \geq 2\). This range of \(f(x)\) is \(\mathrm{R}-(-2,2)\)
MHT-CET 20
Sets, Relation and Function
117302
The domain of the definition of the function \(y=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) is
1 \(x \geq-2\)
2 \(-3\lt x \leq-2\)
3 \(-2 \leq x\lt 0\)
4 \(-2 \leq x\lt 1\)
Explanation:
C Given,\(y=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) For this function domain to be defined \((1-x)>0,(1-x)>1\) and \((x+2)>0\) \(x\lt 1, x\lt 0\) and \(x \geq-2\) Common interval \(=-2 \leq \mathrm{x}\lt 0=\) Domain of \(\mathrm{y}\)
117283
Domain of the real values function \(f(x)=\frac{x+2}{9-x^2}\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{3\}\)
3 \(\mathrm{R}-\{-3,3\}\)
4 \(-3 \leq \mathrm{x} \leq 3\)
Explanation:
C Given, \(f(x)=\frac{x+2}{9-x^2}\) is not defined if \(9-x^2=0 \Rightarrow x= \pm 3\) So, required domain \(=\mathrm{R}-\{-3,3\}\)
MHT-CET 20
Sets, Relation and Function
117287
If \(f: \mathbf{R}-\{2\} \rightarrow \mathbf{R}\) is function defined by \(f(x)=\frac{x^2-4}{x-2}\), then its range is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{2\}\)
3 \(\mathrm{R}-\{4\}\)
4 \(\mathrm{R}-\{-2,2\}\)
Explanation:
C We have, \(f(x)=\frac{x^2-4}{x-2} \Rightarrow y=\frac{(x-2)(x+2)}{x-2}=x+2\) \(\therefore \mathrm{x}=\mathrm{y}-2\) and domain of \(\mathrm{f}(\mathrm{x})\) is \(\mathrm{R}-\{2\}\) When \(x=2\), we get \(y=2+2=4\) So, range of \(f(x)\) is \(R-\{4\}\)
MHT-CET 20
Sets, Relation and Function
117289
The range of function \(f(x)=\sin x+\operatorname{cosec} x\) is
1 \([-1,1]\)
2 \((-1,1)\)
3 \(\mathrm{R}-[-2,2]\)
4 \(\mathrm{R}-(-2,2)\)
Explanation:
D Given, \(f(x)=\sin x+\operatorname{cosec} x=\sin x+\frac{1}{\sin x}\) We know that, the sum of the number and their reciprocal is either less than or equal to -2 or greater than or equal to 2 . i.e. \(x+\frac{1}{x} \leq-2\) or \(x+\frac{1}{x} \geq 2\). This range of \(f(x)\) is \(\mathrm{R}-(-2,2)\)
MHT-CET 20
Sets, Relation and Function
117302
The domain of the definition of the function \(y=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) is
1 \(x \geq-2\)
2 \(-3\lt x \leq-2\)
3 \(-2 \leq x\lt 0\)
4 \(-2 \leq x\lt 1\)
Explanation:
C Given,\(y=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) For this function domain to be defined \((1-x)>0,(1-x)>1\) and \((x+2)>0\) \(x\lt 1, x\lt 0\) and \(x \geq-2\) Common interval \(=-2 \leq \mathrm{x}\lt 0=\) Domain of \(\mathrm{y}\)
117283
Domain of the real values function \(f(x)=\frac{x+2}{9-x^2}\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{3\}\)
3 \(\mathrm{R}-\{-3,3\}\)
4 \(-3 \leq \mathrm{x} \leq 3\)
Explanation:
C Given, \(f(x)=\frac{x+2}{9-x^2}\) is not defined if \(9-x^2=0 \Rightarrow x= \pm 3\) So, required domain \(=\mathrm{R}-\{-3,3\}\)
MHT-CET 20
Sets, Relation and Function
117287
If \(f: \mathbf{R}-\{2\} \rightarrow \mathbf{R}\) is function defined by \(f(x)=\frac{x^2-4}{x-2}\), then its range is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{2\}\)
3 \(\mathrm{R}-\{4\}\)
4 \(\mathrm{R}-\{-2,2\}\)
Explanation:
C We have, \(f(x)=\frac{x^2-4}{x-2} \Rightarrow y=\frac{(x-2)(x+2)}{x-2}=x+2\) \(\therefore \mathrm{x}=\mathrm{y}-2\) and domain of \(\mathrm{f}(\mathrm{x})\) is \(\mathrm{R}-\{2\}\) When \(x=2\), we get \(y=2+2=4\) So, range of \(f(x)\) is \(R-\{4\}\)
MHT-CET 20
Sets, Relation and Function
117289
The range of function \(f(x)=\sin x+\operatorname{cosec} x\) is
1 \([-1,1]\)
2 \((-1,1)\)
3 \(\mathrm{R}-[-2,2]\)
4 \(\mathrm{R}-(-2,2)\)
Explanation:
D Given, \(f(x)=\sin x+\operatorname{cosec} x=\sin x+\frac{1}{\sin x}\) We know that, the sum of the number and their reciprocal is either less than or equal to -2 or greater than or equal to 2 . i.e. \(x+\frac{1}{x} \leq-2\) or \(x+\frac{1}{x} \geq 2\). This range of \(f(x)\) is \(\mathrm{R}-(-2,2)\)
MHT-CET 20
Sets, Relation and Function
117302
The domain of the definition of the function \(y=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) is
1 \(x \geq-2\)
2 \(-3\lt x \leq-2\)
3 \(-2 \leq x\lt 0\)
4 \(-2 \leq x\lt 1\)
Explanation:
C Given,\(y=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) For this function domain to be defined \((1-x)>0,(1-x)>1\) and \((x+2)>0\) \(x\lt 1, x\lt 0\) and \(x \geq-2\) Common interval \(=-2 \leq \mathrm{x}\lt 0=\) Domain of \(\mathrm{y}\)
