117288
The domain of the real valued function \(f(x)=\sqrt{\frac{x-2}{3-x}}\) is
1 \([2,3]\)
2 \((2,3]\)
3 \([2,3)\)
4 \((2,3)\)
Explanation:
C Given, \(f(x)=\sqrt{\frac{x-2}{3-x}}\) Then, \(f(x)\) is defined if \(\frac{x-2}{3-x} \geq 0\) and \(x \neq 3\) \(\frac{x-2}{3-x} \geq 0\) if \(x-2 \geq 0\) and \(3-x>0\) or \(x-2 \leq 0\) and \(3-\mathrm{x}\lt 0\) \(\therefore \mathrm{x} \geq 2\) and \(3>\mathrm{x}\) or \(\mathrm{x} \leq 2\) and \(3\lt \mathrm{x}\) \(\therefore \mathrm{x} \geq 2\) and \(\mathrm{x}\lt 3\) or \(\mathrm{x} \leq 2\) and \(\mathrm{x}>3\) \(\therefore 2 \leq \mathrm{x}\lt 3\) it is impossible. So, \(\mathrm{x} \in[2,3)\)
MHT-CET 20
Sets, Relation and Function
117290
The domain of the function \(f(x)=\sqrt{x-\sqrt{1-x^2}}\) is
D For \(\mathrm{f}(\mathrm{x})\) to be defined, we must have \(\mathrm{x}-\sqrt{1-\mathrm{x}^2} \geq 0\) or \(\mathrm{x} \geq \sqrt{1-\mathrm{x}^2}>0\) Then, \(x^2 \geq 1-x^2\) or \(x^2 \geq \frac{1}{2}\). Also, \(1-\mathrm{x}^2 \geq 0\) or \(\mathrm{x}^2 \leq 1\). Now, \(x^2 \geq \frac{1}{2} \Rightarrow\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right) \geq 0\) \(\Rightarrow \mathrm{x} \leq-\frac{1}{\sqrt{2}}\) or \(\mathrm{x} \geq \frac{1}{\sqrt{2}}\) Also, \(\mathrm{x}^2 \leq 1 \Rightarrow(\mathrm{x}-1)(\mathrm{x}+1) \leq 0\) \(\Rightarrow-1 \leq \mathrm{x} \leq 1\) Thus, \(x>0, x^2 \geq \frac{1}{2}\) and \(x^2 \leq 1\) \(\Rightarrow \mathrm{x} \in\left[\frac{1}{\sqrt{2}}, 1\right]\)
BITSAT-2020
Sets, Relation and Function
117292
If \(e^x+e^{f(x)}=e\), then the domain of \(f(x)\) is
1 \((-\infty, 1)\)
2 \((-\infty, 0)\)
3 \((1, \infty)\)
4 None
Explanation:
A We have, \(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\mathrm{e}\) \(e^{f(x)}=e-e^x=e\left(1-e^{x-1}\right)\) \(f(x)=1+\log _e\left(1-e^{x-1}\right)\) Clearly, for \(\mathrm{f}(\mathrm{x})\) to be real, we must have - \(1-\mathrm{e}^{\mathrm{x}-1}>0\) \(\Rightarrow \mathrm{e}^{\mathrm{x}}\lt \mathrm{e} \Rightarrow \mathrm{x}\lt 1 \Rightarrow \mathrm{x} \in(-\infty, 1)\)
117288
The domain of the real valued function \(f(x)=\sqrt{\frac{x-2}{3-x}}\) is
1 \([2,3]\)
2 \((2,3]\)
3 \([2,3)\)
4 \((2,3)\)
Explanation:
C Given, \(f(x)=\sqrt{\frac{x-2}{3-x}}\) Then, \(f(x)\) is defined if \(\frac{x-2}{3-x} \geq 0\) and \(x \neq 3\) \(\frac{x-2}{3-x} \geq 0\) if \(x-2 \geq 0\) and \(3-x>0\) or \(x-2 \leq 0\) and \(3-\mathrm{x}\lt 0\) \(\therefore \mathrm{x} \geq 2\) and \(3>\mathrm{x}\) or \(\mathrm{x} \leq 2\) and \(3\lt \mathrm{x}\) \(\therefore \mathrm{x} \geq 2\) and \(\mathrm{x}\lt 3\) or \(\mathrm{x} \leq 2\) and \(\mathrm{x}>3\) \(\therefore 2 \leq \mathrm{x}\lt 3\) it is impossible. So, \(\mathrm{x} \in[2,3)\)
MHT-CET 20
Sets, Relation and Function
117290
The domain of the function \(f(x)=\sqrt{x-\sqrt{1-x^2}}\) is
D For \(\mathrm{f}(\mathrm{x})\) to be defined, we must have \(\mathrm{x}-\sqrt{1-\mathrm{x}^2} \geq 0\) or \(\mathrm{x} \geq \sqrt{1-\mathrm{x}^2}>0\) Then, \(x^2 \geq 1-x^2\) or \(x^2 \geq \frac{1}{2}\). Also, \(1-\mathrm{x}^2 \geq 0\) or \(\mathrm{x}^2 \leq 1\). Now, \(x^2 \geq \frac{1}{2} \Rightarrow\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right) \geq 0\) \(\Rightarrow \mathrm{x} \leq-\frac{1}{\sqrt{2}}\) or \(\mathrm{x} \geq \frac{1}{\sqrt{2}}\) Also, \(\mathrm{x}^2 \leq 1 \Rightarrow(\mathrm{x}-1)(\mathrm{x}+1) \leq 0\) \(\Rightarrow-1 \leq \mathrm{x} \leq 1\) Thus, \(x>0, x^2 \geq \frac{1}{2}\) and \(x^2 \leq 1\) \(\Rightarrow \mathrm{x} \in\left[\frac{1}{\sqrt{2}}, 1\right]\)
BITSAT-2020
Sets, Relation and Function
117292
If \(e^x+e^{f(x)}=e\), then the domain of \(f(x)\) is
1 \((-\infty, 1)\)
2 \((-\infty, 0)\)
3 \((1, \infty)\)
4 None
Explanation:
A We have, \(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\mathrm{e}\) \(e^{f(x)}=e-e^x=e\left(1-e^{x-1}\right)\) \(f(x)=1+\log _e\left(1-e^{x-1}\right)\) Clearly, for \(\mathrm{f}(\mathrm{x})\) to be real, we must have - \(1-\mathrm{e}^{\mathrm{x}-1}>0\) \(\Rightarrow \mathrm{e}^{\mathrm{x}}\lt \mathrm{e} \Rightarrow \mathrm{x}\lt 1 \Rightarrow \mathrm{x} \in(-\infty, 1)\)
117288
The domain of the real valued function \(f(x)=\sqrt{\frac{x-2}{3-x}}\) is
1 \([2,3]\)
2 \((2,3]\)
3 \([2,3)\)
4 \((2,3)\)
Explanation:
C Given, \(f(x)=\sqrt{\frac{x-2}{3-x}}\) Then, \(f(x)\) is defined if \(\frac{x-2}{3-x} \geq 0\) and \(x \neq 3\) \(\frac{x-2}{3-x} \geq 0\) if \(x-2 \geq 0\) and \(3-x>0\) or \(x-2 \leq 0\) and \(3-\mathrm{x}\lt 0\) \(\therefore \mathrm{x} \geq 2\) and \(3>\mathrm{x}\) or \(\mathrm{x} \leq 2\) and \(3\lt \mathrm{x}\) \(\therefore \mathrm{x} \geq 2\) and \(\mathrm{x}\lt 3\) or \(\mathrm{x} \leq 2\) and \(\mathrm{x}>3\) \(\therefore 2 \leq \mathrm{x}\lt 3\) it is impossible. So, \(\mathrm{x} \in[2,3)\)
MHT-CET 20
Sets, Relation and Function
117290
The domain of the function \(f(x)=\sqrt{x-\sqrt{1-x^2}}\) is
D For \(\mathrm{f}(\mathrm{x})\) to be defined, we must have \(\mathrm{x}-\sqrt{1-\mathrm{x}^2} \geq 0\) or \(\mathrm{x} \geq \sqrt{1-\mathrm{x}^2}>0\) Then, \(x^2 \geq 1-x^2\) or \(x^2 \geq \frac{1}{2}\). Also, \(1-\mathrm{x}^2 \geq 0\) or \(\mathrm{x}^2 \leq 1\). Now, \(x^2 \geq \frac{1}{2} \Rightarrow\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right) \geq 0\) \(\Rightarrow \mathrm{x} \leq-\frac{1}{\sqrt{2}}\) or \(\mathrm{x} \geq \frac{1}{\sqrt{2}}\) Also, \(\mathrm{x}^2 \leq 1 \Rightarrow(\mathrm{x}-1)(\mathrm{x}+1) \leq 0\) \(\Rightarrow-1 \leq \mathrm{x} \leq 1\) Thus, \(x>0, x^2 \geq \frac{1}{2}\) and \(x^2 \leq 1\) \(\Rightarrow \mathrm{x} \in\left[\frac{1}{\sqrt{2}}, 1\right]\)
BITSAT-2020
Sets, Relation and Function
117292
If \(e^x+e^{f(x)}=e\), then the domain of \(f(x)\) is
1 \((-\infty, 1)\)
2 \((-\infty, 0)\)
3 \((1, \infty)\)
4 None
Explanation:
A We have, \(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\mathrm{e}\) \(e^{f(x)}=e-e^x=e\left(1-e^{x-1}\right)\) \(f(x)=1+\log _e\left(1-e^{x-1}\right)\) Clearly, for \(\mathrm{f}(\mathrm{x})\) to be real, we must have - \(1-\mathrm{e}^{\mathrm{x}-1}>0\) \(\Rightarrow \mathrm{e}^{\mathrm{x}}\lt \mathrm{e} \Rightarrow \mathrm{x}\lt 1 \Rightarrow \mathrm{x} \in(-\infty, 1)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117288
The domain of the real valued function \(f(x)=\sqrt{\frac{x-2}{3-x}}\) is
1 \([2,3]\)
2 \((2,3]\)
3 \([2,3)\)
4 \((2,3)\)
Explanation:
C Given, \(f(x)=\sqrt{\frac{x-2}{3-x}}\) Then, \(f(x)\) is defined if \(\frac{x-2}{3-x} \geq 0\) and \(x \neq 3\) \(\frac{x-2}{3-x} \geq 0\) if \(x-2 \geq 0\) and \(3-x>0\) or \(x-2 \leq 0\) and \(3-\mathrm{x}\lt 0\) \(\therefore \mathrm{x} \geq 2\) and \(3>\mathrm{x}\) or \(\mathrm{x} \leq 2\) and \(3\lt \mathrm{x}\) \(\therefore \mathrm{x} \geq 2\) and \(\mathrm{x}\lt 3\) or \(\mathrm{x} \leq 2\) and \(\mathrm{x}>3\) \(\therefore 2 \leq \mathrm{x}\lt 3\) it is impossible. So, \(\mathrm{x} \in[2,3)\)
MHT-CET 20
Sets, Relation and Function
117290
The domain of the function \(f(x)=\sqrt{x-\sqrt{1-x^2}}\) is
D For \(\mathrm{f}(\mathrm{x})\) to be defined, we must have \(\mathrm{x}-\sqrt{1-\mathrm{x}^2} \geq 0\) or \(\mathrm{x} \geq \sqrt{1-\mathrm{x}^2}>0\) Then, \(x^2 \geq 1-x^2\) or \(x^2 \geq \frac{1}{2}\). Also, \(1-\mathrm{x}^2 \geq 0\) or \(\mathrm{x}^2 \leq 1\). Now, \(x^2 \geq \frac{1}{2} \Rightarrow\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right) \geq 0\) \(\Rightarrow \mathrm{x} \leq-\frac{1}{\sqrt{2}}\) or \(\mathrm{x} \geq \frac{1}{\sqrt{2}}\) Also, \(\mathrm{x}^2 \leq 1 \Rightarrow(\mathrm{x}-1)(\mathrm{x}+1) \leq 0\) \(\Rightarrow-1 \leq \mathrm{x} \leq 1\) Thus, \(x>0, x^2 \geq \frac{1}{2}\) and \(x^2 \leq 1\) \(\Rightarrow \mathrm{x} \in\left[\frac{1}{\sqrt{2}}, 1\right]\)
BITSAT-2020
Sets, Relation and Function
117292
If \(e^x+e^{f(x)}=e\), then the domain of \(f(x)\) is
1 \((-\infty, 1)\)
2 \((-\infty, 0)\)
3 \((1, \infty)\)
4 None
Explanation:
A We have, \(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\mathrm{e}\) \(e^{f(x)}=e-e^x=e\left(1-e^{x-1}\right)\) \(f(x)=1+\log _e\left(1-e^{x-1}\right)\) Clearly, for \(\mathrm{f}(\mathrm{x})\) to be real, we must have - \(1-\mathrm{e}^{\mathrm{x}-1}>0\) \(\Rightarrow \mathrm{e}^{\mathrm{x}}\lt \mathrm{e} \Rightarrow \mathrm{x}\lt 1 \Rightarrow \mathrm{x} \in(-\infty, 1)\)
