117400
The domain of definition of the function \(f(x)=\frac{1}{\sqrt{|[|x|-1]|-5}}\) is
1 \((-\infty,-7] \cup[7, \infty)\)
2 \((-\infty, \infty)\)
3 \((7, \infty)\)
4 \((-7, \infty)\)
Explanation:
A Given that, \(f(x)=\frac{1}{\sqrt{|[|x|-1]|-5 \mid}}\) We know that the value in square roots is always greater than 0 . \(|[|\mathrm{x}|-1]|-5>0\) Let, \([|\mathrm{x}|-1]=\mathrm{t}\) \(|t|-5>0\) \(|t|>5\) \(t>5 \text { or } t\lt -5\) Putting the value of \(t\), we get - \({[|x|-1]>5 \text { or }[|x|-1]\lt -5}\) \({[|x|]-1>5 \text { or }[|x|]-1\lt -5}\) \({[|x|]>6 \text { or }[|x|]\lt -4}\) Since, \(|\mathrm{x}| \geq 0\) hence, we neglected \([|\mathrm{x}|]\lt -4\). Therefore, \([|x|]>6\) \(|x| \geq 7\) Hence, \(\mathrm{x} \geq 7\) or \(\mathrm{x} \leq-7\) \(\mathrm{x} \in[7, \infty)\) or \(\mathrm{x} \in(-\infty,-7]\) Therefore, \(x \in(-\infty,-7] \cup[7, \infty)\)
Manipal UGET-2016
Sets, Relation and Function
117401
The range of the function \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) is
1 \([0,3]\)
2 \([0, \sqrt{3}]\)
3 \((-\infty, \infty)\)
4 None of these
Explanation:
B Given, that, \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) Define the \(f(x)\) - \(\frac{\pi^2}{9}-x^2 \geq 0\) \(-x^2 \geq-\frac{\pi^2}{9}\) \(x^2 \leq \frac{\pi^2}{9}\) \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}\) \(x \in\left[\frac{-\pi}{3}, \frac{\pi}{3}\right]\) Since, \(\tan \mathrm{x}\) is an increasing function in \(\left[0, \frac{\pi}{2}\right]\) \(0 \leq \mathrm{x}^2 \leq \frac{\pi^2}{9}\) \(0 \geq-\mathrm{x}^2 \geq-\frac{\pi^2}{9}\) \(\frac{\pi^2}{9} \geq \frac{\pi^2}{9}-\mathrm{x}^2 \geq 0\) \(\frac{\pi}{3} \geq \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq 0\) \(\tan \frac{\pi}{3} \geq \tan \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq \tan 0\) \(\sqrt{3} \geq \tan \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq 0\)Hence, the range of \(\mathrm{f}(\mathrm{x})[0, \sqrt{3}]\).
Rajasthan PET-2012
Sets, Relation and Function
117402
The domain of the function \(f(x)=\frac{1}{\sqrt{|x|-x}}\) is
1 \((-\infty, 0)\)
2 \((-\infty, \infty)-\{0\}\)
3 \((0, \infty)\)
4 \((-\infty, 1)-\{0\}\)
Explanation:
A Given that, \(f(x)=\frac{1}{\sqrt{|x|-x}}\) The function \(f(x)\) is defined, Case - I: \(|\mathrm{x}|-\mathrm{x}>0\) \(\mathrm{x}-\mathrm{x}>0\) \(0>0\) It is not possible Case II: \(\quad-\mathrm{x}-\mathrm{x}>0\) \(-2 x >0\) \(x \in(-\infty, 0)\) Hence, domain of \(f(x)\) - \(x \in(-\infty, 0)\)
AIEEE-2011
Sets, Relation and Function
117403
The domain of the function \(f(x)=\) \(\sqrt{x-\sqrt{1-x^2}}\) is
117400
The domain of definition of the function \(f(x)=\frac{1}{\sqrt{|[|x|-1]|-5}}\) is
1 \((-\infty,-7] \cup[7, \infty)\)
2 \((-\infty, \infty)\)
3 \((7, \infty)\)
4 \((-7, \infty)\)
Explanation:
A Given that, \(f(x)=\frac{1}{\sqrt{|[|x|-1]|-5 \mid}}\) We know that the value in square roots is always greater than 0 . \(|[|\mathrm{x}|-1]|-5>0\) Let, \([|\mathrm{x}|-1]=\mathrm{t}\) \(|t|-5>0\) \(|t|>5\) \(t>5 \text { or } t\lt -5\) Putting the value of \(t\), we get - \({[|x|-1]>5 \text { or }[|x|-1]\lt -5}\) \({[|x|]-1>5 \text { or }[|x|]-1\lt -5}\) \({[|x|]>6 \text { or }[|x|]\lt -4}\) Since, \(|\mathrm{x}| \geq 0\) hence, we neglected \([|\mathrm{x}|]\lt -4\). Therefore, \([|x|]>6\) \(|x| \geq 7\) Hence, \(\mathrm{x} \geq 7\) or \(\mathrm{x} \leq-7\) \(\mathrm{x} \in[7, \infty)\) or \(\mathrm{x} \in(-\infty,-7]\) Therefore, \(x \in(-\infty,-7] \cup[7, \infty)\)
Manipal UGET-2016
Sets, Relation and Function
117401
The range of the function \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) is
1 \([0,3]\)
2 \([0, \sqrt{3}]\)
3 \((-\infty, \infty)\)
4 None of these
Explanation:
B Given, that, \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) Define the \(f(x)\) - \(\frac{\pi^2}{9}-x^2 \geq 0\) \(-x^2 \geq-\frac{\pi^2}{9}\) \(x^2 \leq \frac{\pi^2}{9}\) \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}\) \(x \in\left[\frac{-\pi}{3}, \frac{\pi}{3}\right]\) Since, \(\tan \mathrm{x}\) is an increasing function in \(\left[0, \frac{\pi}{2}\right]\) \(0 \leq \mathrm{x}^2 \leq \frac{\pi^2}{9}\) \(0 \geq-\mathrm{x}^2 \geq-\frac{\pi^2}{9}\) \(\frac{\pi^2}{9} \geq \frac{\pi^2}{9}-\mathrm{x}^2 \geq 0\) \(\frac{\pi}{3} \geq \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq 0\) \(\tan \frac{\pi}{3} \geq \tan \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq \tan 0\) \(\sqrt{3} \geq \tan \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq 0\)Hence, the range of \(\mathrm{f}(\mathrm{x})[0, \sqrt{3}]\).
Rajasthan PET-2012
Sets, Relation and Function
117402
The domain of the function \(f(x)=\frac{1}{\sqrt{|x|-x}}\) is
1 \((-\infty, 0)\)
2 \((-\infty, \infty)-\{0\}\)
3 \((0, \infty)\)
4 \((-\infty, 1)-\{0\}\)
Explanation:
A Given that, \(f(x)=\frac{1}{\sqrt{|x|-x}}\) The function \(f(x)\) is defined, Case - I: \(|\mathrm{x}|-\mathrm{x}>0\) \(\mathrm{x}-\mathrm{x}>0\) \(0>0\) It is not possible Case II: \(\quad-\mathrm{x}-\mathrm{x}>0\) \(-2 x >0\) \(x \in(-\infty, 0)\) Hence, domain of \(f(x)\) - \(x \in(-\infty, 0)\)
AIEEE-2011
Sets, Relation and Function
117403
The domain of the function \(f(x)=\) \(\sqrt{x-\sqrt{1-x^2}}\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117400
The domain of definition of the function \(f(x)=\frac{1}{\sqrt{|[|x|-1]|-5}}\) is
1 \((-\infty,-7] \cup[7, \infty)\)
2 \((-\infty, \infty)\)
3 \((7, \infty)\)
4 \((-7, \infty)\)
Explanation:
A Given that, \(f(x)=\frac{1}{\sqrt{|[|x|-1]|-5 \mid}}\) We know that the value in square roots is always greater than 0 . \(|[|\mathrm{x}|-1]|-5>0\) Let, \([|\mathrm{x}|-1]=\mathrm{t}\) \(|t|-5>0\) \(|t|>5\) \(t>5 \text { or } t\lt -5\) Putting the value of \(t\), we get - \({[|x|-1]>5 \text { or }[|x|-1]\lt -5}\) \({[|x|]-1>5 \text { or }[|x|]-1\lt -5}\) \({[|x|]>6 \text { or }[|x|]\lt -4}\) Since, \(|\mathrm{x}| \geq 0\) hence, we neglected \([|\mathrm{x}|]\lt -4\). Therefore, \([|x|]>6\) \(|x| \geq 7\) Hence, \(\mathrm{x} \geq 7\) or \(\mathrm{x} \leq-7\) \(\mathrm{x} \in[7, \infty)\) or \(\mathrm{x} \in(-\infty,-7]\) Therefore, \(x \in(-\infty,-7] \cup[7, \infty)\)
Manipal UGET-2016
Sets, Relation and Function
117401
The range of the function \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) is
1 \([0,3]\)
2 \([0, \sqrt{3}]\)
3 \((-\infty, \infty)\)
4 None of these
Explanation:
B Given, that, \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) Define the \(f(x)\) - \(\frac{\pi^2}{9}-x^2 \geq 0\) \(-x^2 \geq-\frac{\pi^2}{9}\) \(x^2 \leq \frac{\pi^2}{9}\) \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}\) \(x \in\left[\frac{-\pi}{3}, \frac{\pi}{3}\right]\) Since, \(\tan \mathrm{x}\) is an increasing function in \(\left[0, \frac{\pi}{2}\right]\) \(0 \leq \mathrm{x}^2 \leq \frac{\pi^2}{9}\) \(0 \geq-\mathrm{x}^2 \geq-\frac{\pi^2}{9}\) \(\frac{\pi^2}{9} \geq \frac{\pi^2}{9}-\mathrm{x}^2 \geq 0\) \(\frac{\pi}{3} \geq \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq 0\) \(\tan \frac{\pi}{3} \geq \tan \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq \tan 0\) \(\sqrt{3} \geq \tan \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq 0\)Hence, the range of \(\mathrm{f}(\mathrm{x})[0, \sqrt{3}]\).
Rajasthan PET-2012
Sets, Relation and Function
117402
The domain of the function \(f(x)=\frac{1}{\sqrt{|x|-x}}\) is
1 \((-\infty, 0)\)
2 \((-\infty, \infty)-\{0\}\)
3 \((0, \infty)\)
4 \((-\infty, 1)-\{0\}\)
Explanation:
A Given that, \(f(x)=\frac{1}{\sqrt{|x|-x}}\) The function \(f(x)\) is defined, Case - I: \(|\mathrm{x}|-\mathrm{x}>0\) \(\mathrm{x}-\mathrm{x}>0\) \(0>0\) It is not possible Case II: \(\quad-\mathrm{x}-\mathrm{x}>0\) \(-2 x >0\) \(x \in(-\infty, 0)\) Hence, domain of \(f(x)\) - \(x \in(-\infty, 0)\)
AIEEE-2011
Sets, Relation and Function
117403
The domain of the function \(f(x)=\) \(\sqrt{x-\sqrt{1-x^2}}\) is
117400
The domain of definition of the function \(f(x)=\frac{1}{\sqrt{|[|x|-1]|-5}}\) is
1 \((-\infty,-7] \cup[7, \infty)\)
2 \((-\infty, \infty)\)
3 \((7, \infty)\)
4 \((-7, \infty)\)
Explanation:
A Given that, \(f(x)=\frac{1}{\sqrt{|[|x|-1]|-5 \mid}}\) We know that the value in square roots is always greater than 0 . \(|[|\mathrm{x}|-1]|-5>0\) Let, \([|\mathrm{x}|-1]=\mathrm{t}\) \(|t|-5>0\) \(|t|>5\) \(t>5 \text { or } t\lt -5\) Putting the value of \(t\), we get - \({[|x|-1]>5 \text { or }[|x|-1]\lt -5}\) \({[|x|]-1>5 \text { or }[|x|]-1\lt -5}\) \({[|x|]>6 \text { or }[|x|]\lt -4}\) Since, \(|\mathrm{x}| \geq 0\) hence, we neglected \([|\mathrm{x}|]\lt -4\). Therefore, \([|x|]>6\) \(|x| \geq 7\) Hence, \(\mathrm{x} \geq 7\) or \(\mathrm{x} \leq-7\) \(\mathrm{x} \in[7, \infty)\) or \(\mathrm{x} \in(-\infty,-7]\) Therefore, \(x \in(-\infty,-7] \cup[7, \infty)\)
Manipal UGET-2016
Sets, Relation and Function
117401
The range of the function \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) is
1 \([0,3]\)
2 \([0, \sqrt{3}]\)
3 \((-\infty, \infty)\)
4 None of these
Explanation:
B Given, that, \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) Define the \(f(x)\) - \(\frac{\pi^2}{9}-x^2 \geq 0\) \(-x^2 \geq-\frac{\pi^2}{9}\) \(x^2 \leq \frac{\pi^2}{9}\) \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}\) \(x \in\left[\frac{-\pi}{3}, \frac{\pi}{3}\right]\) Since, \(\tan \mathrm{x}\) is an increasing function in \(\left[0, \frac{\pi}{2}\right]\) \(0 \leq \mathrm{x}^2 \leq \frac{\pi^2}{9}\) \(0 \geq-\mathrm{x}^2 \geq-\frac{\pi^2}{9}\) \(\frac{\pi^2}{9} \geq \frac{\pi^2}{9}-\mathrm{x}^2 \geq 0\) \(\frac{\pi}{3} \geq \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq 0\) \(\tan \frac{\pi}{3} \geq \tan \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq \tan 0\) \(\sqrt{3} \geq \tan \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \geq 0\)Hence, the range of \(\mathrm{f}(\mathrm{x})[0, \sqrt{3}]\).
Rajasthan PET-2012
Sets, Relation and Function
117402
The domain of the function \(f(x)=\frac{1}{\sqrt{|x|-x}}\) is
1 \((-\infty, 0)\)
2 \((-\infty, \infty)-\{0\}\)
3 \((0, \infty)\)
4 \((-\infty, 1)-\{0\}\)
Explanation:
A Given that, \(f(x)=\frac{1}{\sqrt{|x|-x}}\) The function \(f(x)\) is defined, Case - I: \(|\mathrm{x}|-\mathrm{x}>0\) \(\mathrm{x}-\mathrm{x}>0\) \(0>0\) It is not possible Case II: \(\quad-\mathrm{x}-\mathrm{x}>0\) \(-2 x >0\) \(x \in(-\infty, 0)\) Hence, domain of \(f(x)\) - \(x \in(-\infty, 0)\)
AIEEE-2011
Sets, Relation and Function
117403
The domain of the function \(f(x)=\) \(\sqrt{x-\sqrt{1-x^2}}\) is
