117275
The domain of the function, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)} \text { is }\)
1 \([0,1]\)
2 \([1,4]\)
3 \([4,5]\)
4 \((-\infty, \infty)\)
Explanation:
B We have, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) For \(\mathrm{f}(\mathrm{x})\) to be defined \(\log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0 \text { and } \frac{5 x-x^2}{4}>0\) \(\Rightarrow \quad \frac{5 x-x^2}{4} \geq 1 \text { and } \frac{5 x-x^2}{4}>0\) \(\Rightarrow \quad \frac{5 x-x^2}{4} \geq 1\) \(\Rightarrow \quad 5 \mathrm{x}-\mathrm{x}^2 \geq 4 \Rightarrow \mathrm{x}^2-5 \mathrm{x}+4 \leq 0\) \(\Rightarrow \quad(\mathrm{x}-4)(\mathrm{x}-1) \leq 0\) So, domain of \(f(x)\) is \([1,4]\)
Shift-I
Sets, Relation and Function
117276
The domain of \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) is
1 \(0 \leq x \leq 1\)
2 \(0 \leq x \leq 4\)
3 \(1 \leq x \leq 4\)
4 \(4 \leq x \leq 6\)
Explanation:
C Given, \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) We know that, Domain of \(\sin ^{-1} \mathrm{x} \Rightarrow-1 \leq \mathrm{x} \leq 1\) \(\Rightarrow-1 \leq\left(\log _2\left(\frac{\mathrm{x}}{2}\right)\right) \leq 1\) \(\Rightarrow 2^{-1} \leq \frac{x}{2} \leq 2^1 \Rightarrow \frac{1}{2} \leq \frac{x}{2} \leq 2\) \(\Rightarrow 1 \leq \mathrm{x} \leq 4\) So, the domain of \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) is \(1 \leq x \leq 4\).
Karnataka CET 2011
Sets, Relation and Function
117277
If \(f(x)=\frac{1-x}{1+x}\) the domain of \(f^{-1}(x)\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{-1\}\)
3 \((-\infty,-1)\)
4 \((-1, \infty)\)
Explanation:
B Let, \(\mathrm{f}(\mathrm{x})=\mathrm{y}\). Then, \(\quad \frac{1-\mathrm{x}}{1+\mathrm{x}}=\mathrm{y}\) \(\mathrm{x}=\frac{1-\mathrm{y}}{1+\mathrm{y}} \Rightarrow \mathrm{f}^{-1}(\mathrm{y})=\frac{1-\mathrm{y}}{1+\mathrm{y}}\) Thus, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{1-\mathrm{x}}{1+\mathrm{x}}\) Clearly, \(\mathrm{f}^{-1}(\mathrm{x})\) is defined for \(1+\mathrm{x} \neq 0\) Hence, domain of \(\mathrm{f}^{-1}(\mathrm{x})\) is \(\mathrm{R}-\{-1\}\)
BITSAT-2011
Sets, Relation and Function
117278
The range of the function \(f(x)=\sqrt{9-x^2}\) is
1 \([0,3]\)
2 \((0,3]\)
3 \((0,3)\)
4 \([0,3)\)
Explanation:
A Given the function \(f(x)=\sqrt{9-x^2}\) We know that \(9-x^2 \geq 0\) \(\Rightarrow \quad x^2 \leq 9\) \(\therefore \quad \mathrm{x} \leq 3\) For maximum \(f(\mathrm{x}), \mathrm{x}=0=f(\mathrm{x})=\sqrt{9-0}=3\) For minimum \(f(\mathrm{x}), \mathrm{x}=3=f(\mathrm{x})=\sqrt{9-9}=0\) Hence, the range of \(f(x)=[0,3]\)
Karnataka CET-2017
Sets, Relation and Function
117279
The domain of the function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) is
1 \([2,3]\)
2 \([2,3)\)
3 \([1,2]\)
4 \([1,2)\)
Explanation:
:Given, \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) Then, \(\quad f(x)\) is defined for- \(\Rightarrow-1 \leq x-3 \leq 1 \text { and } 9-x^2>0\) \(\Rightarrow-1+3 \leq x \leq 1+3 \text { and } 9>x^2\) \(\Rightarrow 2 \leq x \leq 4 \text { and }-3\lt x\lt 3\) \(\Rightarrow 2 \leq x\lt 3\)So, domain of \(f(x)=[2,3)\)
117275
The domain of the function, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)} \text { is }\)
1 \([0,1]\)
2 \([1,4]\)
3 \([4,5]\)
4 \((-\infty, \infty)\)
Explanation:
B We have, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) For \(\mathrm{f}(\mathrm{x})\) to be defined \(\log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0 \text { and } \frac{5 x-x^2}{4}>0\) \(\Rightarrow \quad \frac{5 x-x^2}{4} \geq 1 \text { and } \frac{5 x-x^2}{4}>0\) \(\Rightarrow \quad \frac{5 x-x^2}{4} \geq 1\) \(\Rightarrow \quad 5 \mathrm{x}-\mathrm{x}^2 \geq 4 \Rightarrow \mathrm{x}^2-5 \mathrm{x}+4 \leq 0\) \(\Rightarrow \quad(\mathrm{x}-4)(\mathrm{x}-1) \leq 0\) So, domain of \(f(x)\) is \([1,4]\)
Shift-I
Sets, Relation and Function
117276
The domain of \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) is
1 \(0 \leq x \leq 1\)
2 \(0 \leq x \leq 4\)
3 \(1 \leq x \leq 4\)
4 \(4 \leq x \leq 6\)
Explanation:
C Given, \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) We know that, Domain of \(\sin ^{-1} \mathrm{x} \Rightarrow-1 \leq \mathrm{x} \leq 1\) \(\Rightarrow-1 \leq\left(\log _2\left(\frac{\mathrm{x}}{2}\right)\right) \leq 1\) \(\Rightarrow 2^{-1} \leq \frac{x}{2} \leq 2^1 \Rightarrow \frac{1}{2} \leq \frac{x}{2} \leq 2\) \(\Rightarrow 1 \leq \mathrm{x} \leq 4\) So, the domain of \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) is \(1 \leq x \leq 4\).
Karnataka CET 2011
Sets, Relation and Function
117277
If \(f(x)=\frac{1-x}{1+x}\) the domain of \(f^{-1}(x)\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{-1\}\)
3 \((-\infty,-1)\)
4 \((-1, \infty)\)
Explanation:
B Let, \(\mathrm{f}(\mathrm{x})=\mathrm{y}\). Then, \(\quad \frac{1-\mathrm{x}}{1+\mathrm{x}}=\mathrm{y}\) \(\mathrm{x}=\frac{1-\mathrm{y}}{1+\mathrm{y}} \Rightarrow \mathrm{f}^{-1}(\mathrm{y})=\frac{1-\mathrm{y}}{1+\mathrm{y}}\) Thus, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{1-\mathrm{x}}{1+\mathrm{x}}\) Clearly, \(\mathrm{f}^{-1}(\mathrm{x})\) is defined for \(1+\mathrm{x} \neq 0\) Hence, domain of \(\mathrm{f}^{-1}(\mathrm{x})\) is \(\mathrm{R}-\{-1\}\)
BITSAT-2011
Sets, Relation and Function
117278
The range of the function \(f(x)=\sqrt{9-x^2}\) is
1 \([0,3]\)
2 \((0,3]\)
3 \((0,3)\)
4 \([0,3)\)
Explanation:
A Given the function \(f(x)=\sqrt{9-x^2}\) We know that \(9-x^2 \geq 0\) \(\Rightarrow \quad x^2 \leq 9\) \(\therefore \quad \mathrm{x} \leq 3\) For maximum \(f(\mathrm{x}), \mathrm{x}=0=f(\mathrm{x})=\sqrt{9-0}=3\) For minimum \(f(\mathrm{x}), \mathrm{x}=3=f(\mathrm{x})=\sqrt{9-9}=0\) Hence, the range of \(f(x)=[0,3]\)
Karnataka CET-2017
Sets, Relation and Function
117279
The domain of the function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) is
1 \([2,3]\)
2 \([2,3)\)
3 \([1,2]\)
4 \([1,2)\)
Explanation:
:Given, \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) Then, \(\quad f(x)\) is defined for- \(\Rightarrow-1 \leq x-3 \leq 1 \text { and } 9-x^2>0\) \(\Rightarrow-1+3 \leq x \leq 1+3 \text { and } 9>x^2\) \(\Rightarrow 2 \leq x \leq 4 \text { and }-3\lt x\lt 3\) \(\Rightarrow 2 \leq x\lt 3\)So, domain of \(f(x)=[2,3)\)
117275
The domain of the function, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)} \text { is }\)
1 \([0,1]\)
2 \([1,4]\)
3 \([4,5]\)
4 \((-\infty, \infty)\)
Explanation:
B We have, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) For \(\mathrm{f}(\mathrm{x})\) to be defined \(\log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0 \text { and } \frac{5 x-x^2}{4}>0\) \(\Rightarrow \quad \frac{5 x-x^2}{4} \geq 1 \text { and } \frac{5 x-x^2}{4}>0\) \(\Rightarrow \quad \frac{5 x-x^2}{4} \geq 1\) \(\Rightarrow \quad 5 \mathrm{x}-\mathrm{x}^2 \geq 4 \Rightarrow \mathrm{x}^2-5 \mathrm{x}+4 \leq 0\) \(\Rightarrow \quad(\mathrm{x}-4)(\mathrm{x}-1) \leq 0\) So, domain of \(f(x)\) is \([1,4]\)
Shift-I
Sets, Relation and Function
117276
The domain of \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) is
1 \(0 \leq x \leq 1\)
2 \(0 \leq x \leq 4\)
3 \(1 \leq x \leq 4\)
4 \(4 \leq x \leq 6\)
Explanation:
C Given, \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) We know that, Domain of \(\sin ^{-1} \mathrm{x} \Rightarrow-1 \leq \mathrm{x} \leq 1\) \(\Rightarrow-1 \leq\left(\log _2\left(\frac{\mathrm{x}}{2}\right)\right) \leq 1\) \(\Rightarrow 2^{-1} \leq \frac{x}{2} \leq 2^1 \Rightarrow \frac{1}{2} \leq \frac{x}{2} \leq 2\) \(\Rightarrow 1 \leq \mathrm{x} \leq 4\) So, the domain of \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) is \(1 \leq x \leq 4\).
Karnataka CET 2011
Sets, Relation and Function
117277
If \(f(x)=\frac{1-x}{1+x}\) the domain of \(f^{-1}(x)\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{-1\}\)
3 \((-\infty,-1)\)
4 \((-1, \infty)\)
Explanation:
B Let, \(\mathrm{f}(\mathrm{x})=\mathrm{y}\). Then, \(\quad \frac{1-\mathrm{x}}{1+\mathrm{x}}=\mathrm{y}\) \(\mathrm{x}=\frac{1-\mathrm{y}}{1+\mathrm{y}} \Rightarrow \mathrm{f}^{-1}(\mathrm{y})=\frac{1-\mathrm{y}}{1+\mathrm{y}}\) Thus, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{1-\mathrm{x}}{1+\mathrm{x}}\) Clearly, \(\mathrm{f}^{-1}(\mathrm{x})\) is defined for \(1+\mathrm{x} \neq 0\) Hence, domain of \(\mathrm{f}^{-1}(\mathrm{x})\) is \(\mathrm{R}-\{-1\}\)
BITSAT-2011
Sets, Relation and Function
117278
The range of the function \(f(x)=\sqrt{9-x^2}\) is
1 \([0,3]\)
2 \((0,3]\)
3 \((0,3)\)
4 \([0,3)\)
Explanation:
A Given the function \(f(x)=\sqrt{9-x^2}\) We know that \(9-x^2 \geq 0\) \(\Rightarrow \quad x^2 \leq 9\) \(\therefore \quad \mathrm{x} \leq 3\) For maximum \(f(\mathrm{x}), \mathrm{x}=0=f(\mathrm{x})=\sqrt{9-0}=3\) For minimum \(f(\mathrm{x}), \mathrm{x}=3=f(\mathrm{x})=\sqrt{9-9}=0\) Hence, the range of \(f(x)=[0,3]\)
Karnataka CET-2017
Sets, Relation and Function
117279
The domain of the function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) is
1 \([2,3]\)
2 \([2,3)\)
3 \([1,2]\)
4 \([1,2)\)
Explanation:
:Given, \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) Then, \(\quad f(x)\) is defined for- \(\Rightarrow-1 \leq x-3 \leq 1 \text { and } 9-x^2>0\) \(\Rightarrow-1+3 \leq x \leq 1+3 \text { and } 9>x^2\) \(\Rightarrow 2 \leq x \leq 4 \text { and }-3\lt x\lt 3\) \(\Rightarrow 2 \leq x\lt 3\)So, domain of \(f(x)=[2,3)\)
117275
The domain of the function, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)} \text { is }\)
1 \([0,1]\)
2 \([1,4]\)
3 \([4,5]\)
4 \((-\infty, \infty)\)
Explanation:
B We have, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) For \(\mathrm{f}(\mathrm{x})\) to be defined \(\log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0 \text { and } \frac{5 x-x^2}{4}>0\) \(\Rightarrow \quad \frac{5 x-x^2}{4} \geq 1 \text { and } \frac{5 x-x^2}{4}>0\) \(\Rightarrow \quad \frac{5 x-x^2}{4} \geq 1\) \(\Rightarrow \quad 5 \mathrm{x}-\mathrm{x}^2 \geq 4 \Rightarrow \mathrm{x}^2-5 \mathrm{x}+4 \leq 0\) \(\Rightarrow \quad(\mathrm{x}-4)(\mathrm{x}-1) \leq 0\) So, domain of \(f(x)\) is \([1,4]\)
Shift-I
Sets, Relation and Function
117276
The domain of \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) is
1 \(0 \leq x \leq 1\)
2 \(0 \leq x \leq 4\)
3 \(1 \leq x \leq 4\)
4 \(4 \leq x \leq 6\)
Explanation:
C Given, \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) We know that, Domain of \(\sin ^{-1} \mathrm{x} \Rightarrow-1 \leq \mathrm{x} \leq 1\) \(\Rightarrow-1 \leq\left(\log _2\left(\frac{\mathrm{x}}{2}\right)\right) \leq 1\) \(\Rightarrow 2^{-1} \leq \frac{x}{2} \leq 2^1 \Rightarrow \frac{1}{2} \leq \frac{x}{2} \leq 2\) \(\Rightarrow 1 \leq \mathrm{x} \leq 4\) So, the domain of \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) is \(1 \leq x \leq 4\).
Karnataka CET 2011
Sets, Relation and Function
117277
If \(f(x)=\frac{1-x}{1+x}\) the domain of \(f^{-1}(x)\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{-1\}\)
3 \((-\infty,-1)\)
4 \((-1, \infty)\)
Explanation:
B Let, \(\mathrm{f}(\mathrm{x})=\mathrm{y}\). Then, \(\quad \frac{1-\mathrm{x}}{1+\mathrm{x}}=\mathrm{y}\) \(\mathrm{x}=\frac{1-\mathrm{y}}{1+\mathrm{y}} \Rightarrow \mathrm{f}^{-1}(\mathrm{y})=\frac{1-\mathrm{y}}{1+\mathrm{y}}\) Thus, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{1-\mathrm{x}}{1+\mathrm{x}}\) Clearly, \(\mathrm{f}^{-1}(\mathrm{x})\) is defined for \(1+\mathrm{x} \neq 0\) Hence, domain of \(\mathrm{f}^{-1}(\mathrm{x})\) is \(\mathrm{R}-\{-1\}\)
BITSAT-2011
Sets, Relation and Function
117278
The range of the function \(f(x)=\sqrt{9-x^2}\) is
1 \([0,3]\)
2 \((0,3]\)
3 \((0,3)\)
4 \([0,3)\)
Explanation:
A Given the function \(f(x)=\sqrt{9-x^2}\) We know that \(9-x^2 \geq 0\) \(\Rightarrow \quad x^2 \leq 9\) \(\therefore \quad \mathrm{x} \leq 3\) For maximum \(f(\mathrm{x}), \mathrm{x}=0=f(\mathrm{x})=\sqrt{9-0}=3\) For minimum \(f(\mathrm{x}), \mathrm{x}=3=f(\mathrm{x})=\sqrt{9-9}=0\) Hence, the range of \(f(x)=[0,3]\)
Karnataka CET-2017
Sets, Relation and Function
117279
The domain of the function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) is
1 \([2,3]\)
2 \([2,3)\)
3 \([1,2]\)
4 \([1,2)\)
Explanation:
:Given, \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) Then, \(\quad f(x)\) is defined for- \(\Rightarrow-1 \leq x-3 \leq 1 \text { and } 9-x^2>0\) \(\Rightarrow-1+3 \leq x \leq 1+3 \text { and } 9>x^2\) \(\Rightarrow 2 \leq x \leq 4 \text { and }-3\lt x\lt 3\) \(\Rightarrow 2 \leq x\lt 3\)So, domain of \(f(x)=[2,3)\)
117275
The domain of the function, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)} \text { is }\)
1 \([0,1]\)
2 \([1,4]\)
3 \([4,5]\)
4 \((-\infty, \infty)\)
Explanation:
B We have, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) For \(\mathrm{f}(\mathrm{x})\) to be defined \(\log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0 \text { and } \frac{5 x-x^2}{4}>0\) \(\Rightarrow \quad \frac{5 x-x^2}{4} \geq 1 \text { and } \frac{5 x-x^2}{4}>0\) \(\Rightarrow \quad \frac{5 x-x^2}{4} \geq 1\) \(\Rightarrow \quad 5 \mathrm{x}-\mathrm{x}^2 \geq 4 \Rightarrow \mathrm{x}^2-5 \mathrm{x}+4 \leq 0\) \(\Rightarrow \quad(\mathrm{x}-4)(\mathrm{x}-1) \leq 0\) So, domain of \(f(x)\) is \([1,4]\)
Shift-I
Sets, Relation and Function
117276
The domain of \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) is
1 \(0 \leq x \leq 1\)
2 \(0 \leq x \leq 4\)
3 \(1 \leq x \leq 4\)
4 \(4 \leq x \leq 6\)
Explanation:
C Given, \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) We know that, Domain of \(\sin ^{-1} \mathrm{x} \Rightarrow-1 \leq \mathrm{x} \leq 1\) \(\Rightarrow-1 \leq\left(\log _2\left(\frac{\mathrm{x}}{2}\right)\right) \leq 1\) \(\Rightarrow 2^{-1} \leq \frac{x}{2} \leq 2^1 \Rightarrow \frac{1}{2} \leq \frac{x}{2} \leq 2\) \(\Rightarrow 1 \leq \mathrm{x} \leq 4\) So, the domain of \(f(x)=\sin ^{-1}\left(\log _2\left(\frac{x}{2}\right)\right)\) is \(1 \leq x \leq 4\).
Karnataka CET 2011
Sets, Relation and Function
117277
If \(f(x)=\frac{1-x}{1+x}\) the domain of \(f^{-1}(x)\) is
1 \(\mathrm{R}\)
2 \(\mathrm{R}-\{-1\}\)
3 \((-\infty,-1)\)
4 \((-1, \infty)\)
Explanation:
B Let, \(\mathrm{f}(\mathrm{x})=\mathrm{y}\). Then, \(\quad \frac{1-\mathrm{x}}{1+\mathrm{x}}=\mathrm{y}\) \(\mathrm{x}=\frac{1-\mathrm{y}}{1+\mathrm{y}} \Rightarrow \mathrm{f}^{-1}(\mathrm{y})=\frac{1-\mathrm{y}}{1+\mathrm{y}}\) Thus, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{1-\mathrm{x}}{1+\mathrm{x}}\) Clearly, \(\mathrm{f}^{-1}(\mathrm{x})\) is defined for \(1+\mathrm{x} \neq 0\) Hence, domain of \(\mathrm{f}^{-1}(\mathrm{x})\) is \(\mathrm{R}-\{-1\}\)
BITSAT-2011
Sets, Relation and Function
117278
The range of the function \(f(x)=\sqrt{9-x^2}\) is
1 \([0,3]\)
2 \((0,3]\)
3 \((0,3)\)
4 \([0,3)\)
Explanation:
A Given the function \(f(x)=\sqrt{9-x^2}\) We know that \(9-x^2 \geq 0\) \(\Rightarrow \quad x^2 \leq 9\) \(\therefore \quad \mathrm{x} \leq 3\) For maximum \(f(\mathrm{x}), \mathrm{x}=0=f(\mathrm{x})=\sqrt{9-0}=3\) For minimum \(f(\mathrm{x}), \mathrm{x}=3=f(\mathrm{x})=\sqrt{9-9}=0\) Hence, the range of \(f(x)=[0,3]\)
Karnataka CET-2017
Sets, Relation and Function
117279
The domain of the function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) is
1 \([2,3]\)
2 \([2,3)\)
3 \([1,2]\)
4 \([1,2)\)
Explanation:
:Given, \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}\) Then, \(\quad f(x)\) is defined for- \(\Rightarrow-1 \leq x-3 \leq 1 \text { and } 9-x^2>0\) \(\Rightarrow-1+3 \leq x \leq 1+3 \text { and } 9>x^2\) \(\Rightarrow 2 \leq x \leq 4 \text { and }-3\lt x\lt 3\) \(\Rightarrow 2 \leq x\lt 3\)So, domain of \(f(x)=[2,3)\)
