117309
The domain of definition of function \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) is
1 \(\mathrm{R}\)
2 \((-4,4)\)
3 \(\mathrm{R}^{+}\)
4 \((-4,0) \cup(0, \infty)\)
Explanation:
D Given, \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) is It is also written as - \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) \(f(x)=\frac{1+\frac{2}{\sqrt{x+4}}}{2-\sqrt{x+4}}+\sqrt{x+4}+4 \sqrt{x+4}\) \(f(x)\) is defined for \(x+4>0\) and \(x \neq 0\) domain of \(f(x)\) is \((-4,0) \cup(0, \infty)\).
UPSEE-2010
Sets, Relation and Function
117311
The domain of the function \(f(x)=\frac{1}{\sqrt{x+|x|}} \text { is }\)
1 \(\mathrm{R}\) (reals)
2 \(\mathrm{R}^{+}\)(+vereals)
3 \(\mathrm{R}^{-}\)(-vereals)
4 \(\mathrm{N}\) (natural numbers)
Explanation:
:Given, \(f(x)=\frac{1}{\sqrt{x+|x|}}\) For the function to be defined, the denominator should not be equal to 0 and negative \(x+|x|>0\) \(|x|>-x\) It is true for all \(\mathrm{R}^{+}\). \(\left(\mathrm{R}^{+}\right.\)means set of all positive real numbers) Domain of the given function \(=\mathrm{R}^{+}\)(Real number)
JCECE-2019
Sets, Relation and Function
117313
The values of a for which \(\left(a^2-1\right) x^2+2(a-1) x+\) 2 is positive for any \(x\), are
1 \(a \geq 1\)
2 \(\mathrm{a} \leq 1\)
3 \(a>-3\)
4 \(\mathrm{a}\lt -3\) or \(\mathrm{a}>1\)
Explanation:
D Given, \(\left(a^2-1\right) x^2+2(a-1) x+2\) \(a x^2+b x+c>0\) For all a, b, c, \(\mathrm{x}\lt 0\) And, \(\quad b^2\lt 4 \mathrm{ac}\) \(\left(a^2-1\right) x^2+2(a-1) x+2\) is positive for all \(\mathrm{x}\) \(a^2-1>0 \text { and } 4(a-1)^2-8\left(a^2-1\right)\lt 0\) \(a^2-1>0 \text { and }-4(a-1)(a+3)\lt 0\) \(a^2-1>0 \text { and }(a-1)(a+3)\lt 0\) \(a^2>1 \text { and } a\lt -3 \text { of } a>1\)So, \(\quad \mathrm{a}\lt -3\) or \(\mathrm{a}>1\)
JCECE-2014
Sets, Relation and Function
117314
Domain of function \(f(x)=\sin ^{-1} 5 x\) is
1 \(\left(-\frac{1}{5}, \frac{1}{5}\right)\)
2 \(\left[-\frac{1}{5}, \frac{1}{5}\right]\)
3 \(\mathrm{R}\)
4 \(\left(0, \frac{1}{5}\right)\)
Explanation:
B Given, \(f(x)=\sin ^{-1} 5 x\) \(-1 \leq 5 x \leq 1\) \(\frac{-1}{5} \leq x \leq \frac{1}{5}\) Hence, domain of function \(f(x)=\sin ^{-1} 5 x\) \(\text { is }\left[-\frac{1}{5}, \frac{1}{5}\right] \text {. }\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117309
The domain of definition of function \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) is
1 \(\mathrm{R}\)
2 \((-4,4)\)
3 \(\mathrm{R}^{+}\)
4 \((-4,0) \cup(0, \infty)\)
Explanation:
D Given, \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) is It is also written as - \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) \(f(x)=\frac{1+\frac{2}{\sqrt{x+4}}}{2-\sqrt{x+4}}+\sqrt{x+4}+4 \sqrt{x+4}\) \(f(x)\) is defined for \(x+4>0\) and \(x \neq 0\) domain of \(f(x)\) is \((-4,0) \cup(0, \infty)\).
UPSEE-2010
Sets, Relation and Function
117311
The domain of the function \(f(x)=\frac{1}{\sqrt{x+|x|}} \text { is }\)
1 \(\mathrm{R}\) (reals)
2 \(\mathrm{R}^{+}\)(+vereals)
3 \(\mathrm{R}^{-}\)(-vereals)
4 \(\mathrm{N}\) (natural numbers)
Explanation:
:Given, \(f(x)=\frac{1}{\sqrt{x+|x|}}\) For the function to be defined, the denominator should not be equal to 0 and negative \(x+|x|>0\) \(|x|>-x\) It is true for all \(\mathrm{R}^{+}\). \(\left(\mathrm{R}^{+}\right.\)means set of all positive real numbers) Domain of the given function \(=\mathrm{R}^{+}\)(Real number)
JCECE-2019
Sets, Relation and Function
117313
The values of a for which \(\left(a^2-1\right) x^2+2(a-1) x+\) 2 is positive for any \(x\), are
1 \(a \geq 1\)
2 \(\mathrm{a} \leq 1\)
3 \(a>-3\)
4 \(\mathrm{a}\lt -3\) or \(\mathrm{a}>1\)
Explanation:
D Given, \(\left(a^2-1\right) x^2+2(a-1) x+2\) \(a x^2+b x+c>0\) For all a, b, c, \(\mathrm{x}\lt 0\) And, \(\quad b^2\lt 4 \mathrm{ac}\) \(\left(a^2-1\right) x^2+2(a-1) x+2\) is positive for all \(\mathrm{x}\) \(a^2-1>0 \text { and } 4(a-1)^2-8\left(a^2-1\right)\lt 0\) \(a^2-1>0 \text { and }-4(a-1)(a+3)\lt 0\) \(a^2-1>0 \text { and }(a-1)(a+3)\lt 0\) \(a^2>1 \text { and } a\lt -3 \text { of } a>1\)So, \(\quad \mathrm{a}\lt -3\) or \(\mathrm{a}>1\)
JCECE-2014
Sets, Relation and Function
117314
Domain of function \(f(x)=\sin ^{-1} 5 x\) is
1 \(\left(-\frac{1}{5}, \frac{1}{5}\right)\)
2 \(\left[-\frac{1}{5}, \frac{1}{5}\right]\)
3 \(\mathrm{R}\)
4 \(\left(0, \frac{1}{5}\right)\)
Explanation:
B Given, \(f(x)=\sin ^{-1} 5 x\) \(-1 \leq 5 x \leq 1\) \(\frac{-1}{5} \leq x \leq \frac{1}{5}\) Hence, domain of function \(f(x)=\sin ^{-1} 5 x\) \(\text { is }\left[-\frac{1}{5}, \frac{1}{5}\right] \text {. }\)
117309
The domain of definition of function \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) is
1 \(\mathrm{R}\)
2 \((-4,4)\)
3 \(\mathrm{R}^{+}\)
4 \((-4,0) \cup(0, \infty)\)
Explanation:
D Given, \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) is It is also written as - \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) \(f(x)=\frac{1+\frac{2}{\sqrt{x+4}}}{2-\sqrt{x+4}}+\sqrt{x+4}+4 \sqrt{x+4}\) \(f(x)\) is defined for \(x+4>0\) and \(x \neq 0\) domain of \(f(x)\) is \((-4,0) \cup(0, \infty)\).
UPSEE-2010
Sets, Relation and Function
117311
The domain of the function \(f(x)=\frac{1}{\sqrt{x+|x|}} \text { is }\)
1 \(\mathrm{R}\) (reals)
2 \(\mathrm{R}^{+}\)(+vereals)
3 \(\mathrm{R}^{-}\)(-vereals)
4 \(\mathrm{N}\) (natural numbers)
Explanation:
:Given, \(f(x)=\frac{1}{\sqrt{x+|x|}}\) For the function to be defined, the denominator should not be equal to 0 and negative \(x+|x|>0\) \(|x|>-x\) It is true for all \(\mathrm{R}^{+}\). \(\left(\mathrm{R}^{+}\right.\)means set of all positive real numbers) Domain of the given function \(=\mathrm{R}^{+}\)(Real number)
JCECE-2019
Sets, Relation and Function
117313
The values of a for which \(\left(a^2-1\right) x^2+2(a-1) x+\) 2 is positive for any \(x\), are
1 \(a \geq 1\)
2 \(\mathrm{a} \leq 1\)
3 \(a>-3\)
4 \(\mathrm{a}\lt -3\) or \(\mathrm{a}>1\)
Explanation:
D Given, \(\left(a^2-1\right) x^2+2(a-1) x+2\) \(a x^2+b x+c>0\) For all a, b, c, \(\mathrm{x}\lt 0\) And, \(\quad b^2\lt 4 \mathrm{ac}\) \(\left(a^2-1\right) x^2+2(a-1) x+2\) is positive for all \(\mathrm{x}\) \(a^2-1>0 \text { and } 4(a-1)^2-8\left(a^2-1\right)\lt 0\) \(a^2-1>0 \text { and }-4(a-1)(a+3)\lt 0\) \(a^2-1>0 \text { and }(a-1)(a+3)\lt 0\) \(a^2>1 \text { and } a\lt -3 \text { of } a>1\)So, \(\quad \mathrm{a}\lt -3\) or \(\mathrm{a}>1\)
JCECE-2014
Sets, Relation and Function
117314
Domain of function \(f(x)=\sin ^{-1} 5 x\) is
1 \(\left(-\frac{1}{5}, \frac{1}{5}\right)\)
2 \(\left[-\frac{1}{5}, \frac{1}{5}\right]\)
3 \(\mathrm{R}\)
4 \(\left(0, \frac{1}{5}\right)\)
Explanation:
B Given, \(f(x)=\sin ^{-1} 5 x\) \(-1 \leq 5 x \leq 1\) \(\frac{-1}{5} \leq x \leq \frac{1}{5}\) Hence, domain of function \(f(x)=\sin ^{-1} 5 x\) \(\text { is }\left[-\frac{1}{5}, \frac{1}{5}\right] \text {. }\)
117309
The domain of definition of function \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) is
1 \(\mathrm{R}\)
2 \((-4,4)\)
3 \(\mathrm{R}^{+}\)
4 \((-4,0) \cup(0, \infty)\)
Explanation:
D Given, \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) is It is also written as - \(f(x)=\frac{1+2(x+4)^{-0.5}}{2-(x+4)^{0.5}}+(x+4)^{0.5}+4(x+4)^{0.5}\) \(f(x)=\frac{1+\frac{2}{\sqrt{x+4}}}{2-\sqrt{x+4}}+\sqrt{x+4}+4 \sqrt{x+4}\) \(f(x)\) is defined for \(x+4>0\) and \(x \neq 0\) domain of \(f(x)\) is \((-4,0) \cup(0, \infty)\).
UPSEE-2010
Sets, Relation and Function
117311
The domain of the function \(f(x)=\frac{1}{\sqrt{x+|x|}} \text { is }\)
1 \(\mathrm{R}\) (reals)
2 \(\mathrm{R}^{+}\)(+vereals)
3 \(\mathrm{R}^{-}\)(-vereals)
4 \(\mathrm{N}\) (natural numbers)
Explanation:
:Given, \(f(x)=\frac{1}{\sqrt{x+|x|}}\) For the function to be defined, the denominator should not be equal to 0 and negative \(x+|x|>0\) \(|x|>-x\) It is true for all \(\mathrm{R}^{+}\). \(\left(\mathrm{R}^{+}\right.\)means set of all positive real numbers) Domain of the given function \(=\mathrm{R}^{+}\)(Real number)
JCECE-2019
Sets, Relation and Function
117313
The values of a for which \(\left(a^2-1\right) x^2+2(a-1) x+\) 2 is positive for any \(x\), are
1 \(a \geq 1\)
2 \(\mathrm{a} \leq 1\)
3 \(a>-3\)
4 \(\mathrm{a}\lt -3\) or \(\mathrm{a}>1\)
Explanation:
D Given, \(\left(a^2-1\right) x^2+2(a-1) x+2\) \(a x^2+b x+c>0\) For all a, b, c, \(\mathrm{x}\lt 0\) And, \(\quad b^2\lt 4 \mathrm{ac}\) \(\left(a^2-1\right) x^2+2(a-1) x+2\) is positive for all \(\mathrm{x}\) \(a^2-1>0 \text { and } 4(a-1)^2-8\left(a^2-1\right)\lt 0\) \(a^2-1>0 \text { and }-4(a-1)(a+3)\lt 0\) \(a^2-1>0 \text { and }(a-1)(a+3)\lt 0\) \(a^2>1 \text { and } a\lt -3 \text { of } a>1\)So, \(\quad \mathrm{a}\lt -3\) or \(\mathrm{a}>1\)
JCECE-2014
Sets, Relation and Function
117314
Domain of function \(f(x)=\sin ^{-1} 5 x\) is
1 \(\left(-\frac{1}{5}, \frac{1}{5}\right)\)
2 \(\left[-\frac{1}{5}, \frac{1}{5}\right]\)
3 \(\mathrm{R}\)
4 \(\left(0, \frac{1}{5}\right)\)
Explanation:
B Given, \(f(x)=\sin ^{-1} 5 x\) \(-1 \leq 5 x \leq 1\) \(\frac{-1}{5} \leq x \leq \frac{1}{5}\) Hence, domain of function \(f(x)=\sin ^{-1} 5 x\) \(\text { is }\left[-\frac{1}{5}, \frac{1}{5}\right] \text {. }\)
