117316
The range of \(\alpha\), for which the point \((\alpha, \alpha)\) lies inside the region bounded by the curves \(y=\sqrt{1-x^2}\) and \(x+y=1\) is
1 \(\frac{1}{2}\lt \alpha\lt \frac{1}{\sqrt{2}}\)
2 \(\frac{1}{2}\lt \alpha\lt \frac{1}{3}\)
3 \(\frac{1}{3}\lt \alpha\lt \frac{1}{\sqrt{3}}\)
4 \(\frac{1}{4}\lt \alpha\lt \frac{1}{2}\)
Explanation:
A Given, \(y-\sqrt{1-x^2}=0 \text { and } x+y-1=0\) And, also given point \((\alpha, \alpha)\) lies but much both lines - \(\alpha-\sqrt{1-\alpha^2}\lt 0\) \(\alpha\lt \sqrt{1-\alpha^2}\) Both side squaring - \(\alpha^2\lt 1-\alpha^2\) \(2 \alpha^2\lt 1\) \(\alpha\lt \frac{1}{\sqrt{2}}\) \(\alpha+\alpha-1>0\) \(2 \alpha>1\) \(\alpha>\frac{1}{2}\)So, \(\quad \frac{1}{2}\lt \alpha\lt \frac{1}{\sqrt{2}}\)
JCECE-2013
Sets, Relation and Function
117317
The domain of the function \(f(x)=\sqrt{3-x}+\cos ^{-1}\left(\frac{3-2 x}{5}\right)\), is
1 \([-1,3]\)
2 \((-1,3]\)
3 \([-1,3)\)
4 None of these
Explanation:
A Given, \(f(x)=\sqrt{3-x}+\cos ^{-1}\left(\frac{3-2 x}{5}\right)\) For, \(\quad 3-\mathrm{x} \geq 0\) \(x \leq 3\) \(-1 \leq \frac{3-2 x}{5} \leq 1\) \(-5 \leq 3-2 x \leq 5\) \(-8 \leq-2 \mathrm{x} \leq 2\) \(-1 \leq \mathrm{x} \leq 4\) From equation (i) and (ii) - \((-\infty, 3] \cap[-1,4]=[-1,3]\)
JCECE-2010
Sets, Relation and Function
117318
Domain of \(f(x)=y=\sqrt{\log _3\{\cos (\sin x)\}}\) is
117319
Range of the function \(f(x)=\frac{x}{1+x^2}\) is
1 \((-\infty, \infty)\)
2 \([-1,1]\)
3 \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
4 \([-\sqrt{2}, \sqrt{2}]\)
Explanation:
C Given, \(f(x)=\frac{x}{1+x^2}\) \(y=\frac{x}{1+x^2}\) \(x^2 y-x+y=0\) \(y x^2-x+y=0\) For \(x\) to be real \(1-4 y^2 \geq 0\) \((1-2 y)(1+2 y) \geq 0\) Now, for real \(\mathrm{x}\), we have - \(\left(\frac{1}{2}-y\right)\left(\frac{1}{2}+y\right) \geq 0\) \(\frac{-1}{2} \leq y \leq \frac{1}{2}\) \(y=f(x) \in\left[-\frac{1}{2}, \frac{1}{2}\right] .\)
117316
The range of \(\alpha\), for which the point \((\alpha, \alpha)\) lies inside the region bounded by the curves \(y=\sqrt{1-x^2}\) and \(x+y=1\) is
1 \(\frac{1}{2}\lt \alpha\lt \frac{1}{\sqrt{2}}\)
2 \(\frac{1}{2}\lt \alpha\lt \frac{1}{3}\)
3 \(\frac{1}{3}\lt \alpha\lt \frac{1}{\sqrt{3}}\)
4 \(\frac{1}{4}\lt \alpha\lt \frac{1}{2}\)
Explanation:
A Given, \(y-\sqrt{1-x^2}=0 \text { and } x+y-1=0\) And, also given point \((\alpha, \alpha)\) lies but much both lines - \(\alpha-\sqrt{1-\alpha^2}\lt 0\) \(\alpha\lt \sqrt{1-\alpha^2}\) Both side squaring - \(\alpha^2\lt 1-\alpha^2\) \(2 \alpha^2\lt 1\) \(\alpha\lt \frac{1}{\sqrt{2}}\) \(\alpha+\alpha-1>0\) \(2 \alpha>1\) \(\alpha>\frac{1}{2}\)So, \(\quad \frac{1}{2}\lt \alpha\lt \frac{1}{\sqrt{2}}\)
JCECE-2013
Sets, Relation and Function
117317
The domain of the function \(f(x)=\sqrt{3-x}+\cos ^{-1}\left(\frac{3-2 x}{5}\right)\), is
1 \([-1,3]\)
2 \((-1,3]\)
3 \([-1,3)\)
4 None of these
Explanation:
A Given, \(f(x)=\sqrt{3-x}+\cos ^{-1}\left(\frac{3-2 x}{5}\right)\) For, \(\quad 3-\mathrm{x} \geq 0\) \(x \leq 3\) \(-1 \leq \frac{3-2 x}{5} \leq 1\) \(-5 \leq 3-2 x \leq 5\) \(-8 \leq-2 \mathrm{x} \leq 2\) \(-1 \leq \mathrm{x} \leq 4\) From equation (i) and (ii) - \((-\infty, 3] \cap[-1,4]=[-1,3]\)
JCECE-2010
Sets, Relation and Function
117318
Domain of \(f(x)=y=\sqrt{\log _3\{\cos (\sin x)\}}\) is
117319
Range of the function \(f(x)=\frac{x}{1+x^2}\) is
1 \((-\infty, \infty)\)
2 \([-1,1]\)
3 \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
4 \([-\sqrt{2}, \sqrt{2}]\)
Explanation:
C Given, \(f(x)=\frac{x}{1+x^2}\) \(y=\frac{x}{1+x^2}\) \(x^2 y-x+y=0\) \(y x^2-x+y=0\) For \(x\) to be real \(1-4 y^2 \geq 0\) \((1-2 y)(1+2 y) \geq 0\) Now, for real \(\mathrm{x}\), we have - \(\left(\frac{1}{2}-y\right)\left(\frac{1}{2}+y\right) \geq 0\) \(\frac{-1}{2} \leq y \leq \frac{1}{2}\) \(y=f(x) \in\left[-\frac{1}{2}, \frac{1}{2}\right] .\)
117316
The range of \(\alpha\), for which the point \((\alpha, \alpha)\) lies inside the region bounded by the curves \(y=\sqrt{1-x^2}\) and \(x+y=1\) is
1 \(\frac{1}{2}\lt \alpha\lt \frac{1}{\sqrt{2}}\)
2 \(\frac{1}{2}\lt \alpha\lt \frac{1}{3}\)
3 \(\frac{1}{3}\lt \alpha\lt \frac{1}{\sqrt{3}}\)
4 \(\frac{1}{4}\lt \alpha\lt \frac{1}{2}\)
Explanation:
A Given, \(y-\sqrt{1-x^2}=0 \text { and } x+y-1=0\) And, also given point \((\alpha, \alpha)\) lies but much both lines - \(\alpha-\sqrt{1-\alpha^2}\lt 0\) \(\alpha\lt \sqrt{1-\alpha^2}\) Both side squaring - \(\alpha^2\lt 1-\alpha^2\) \(2 \alpha^2\lt 1\) \(\alpha\lt \frac{1}{\sqrt{2}}\) \(\alpha+\alpha-1>0\) \(2 \alpha>1\) \(\alpha>\frac{1}{2}\)So, \(\quad \frac{1}{2}\lt \alpha\lt \frac{1}{\sqrt{2}}\)
JCECE-2013
Sets, Relation and Function
117317
The domain of the function \(f(x)=\sqrt{3-x}+\cos ^{-1}\left(\frac{3-2 x}{5}\right)\), is
1 \([-1,3]\)
2 \((-1,3]\)
3 \([-1,3)\)
4 None of these
Explanation:
A Given, \(f(x)=\sqrt{3-x}+\cos ^{-1}\left(\frac{3-2 x}{5}\right)\) For, \(\quad 3-\mathrm{x} \geq 0\) \(x \leq 3\) \(-1 \leq \frac{3-2 x}{5} \leq 1\) \(-5 \leq 3-2 x \leq 5\) \(-8 \leq-2 \mathrm{x} \leq 2\) \(-1 \leq \mathrm{x} \leq 4\) From equation (i) and (ii) - \((-\infty, 3] \cap[-1,4]=[-1,3]\)
JCECE-2010
Sets, Relation and Function
117318
Domain of \(f(x)=y=\sqrt{\log _3\{\cos (\sin x)\}}\) is
117319
Range of the function \(f(x)=\frac{x}{1+x^2}\) is
1 \((-\infty, \infty)\)
2 \([-1,1]\)
3 \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
4 \([-\sqrt{2}, \sqrt{2}]\)
Explanation:
C Given, \(f(x)=\frac{x}{1+x^2}\) \(y=\frac{x}{1+x^2}\) \(x^2 y-x+y=0\) \(y x^2-x+y=0\) For \(x\) to be real \(1-4 y^2 \geq 0\) \((1-2 y)(1+2 y) \geq 0\) Now, for real \(\mathrm{x}\), we have - \(\left(\frac{1}{2}-y\right)\left(\frac{1}{2}+y\right) \geq 0\) \(\frac{-1}{2} \leq y \leq \frac{1}{2}\) \(y=f(x) \in\left[-\frac{1}{2}, \frac{1}{2}\right] .\)
117316
The range of \(\alpha\), for which the point \((\alpha, \alpha)\) lies inside the region bounded by the curves \(y=\sqrt{1-x^2}\) and \(x+y=1\) is
1 \(\frac{1}{2}\lt \alpha\lt \frac{1}{\sqrt{2}}\)
2 \(\frac{1}{2}\lt \alpha\lt \frac{1}{3}\)
3 \(\frac{1}{3}\lt \alpha\lt \frac{1}{\sqrt{3}}\)
4 \(\frac{1}{4}\lt \alpha\lt \frac{1}{2}\)
Explanation:
A Given, \(y-\sqrt{1-x^2}=0 \text { and } x+y-1=0\) And, also given point \((\alpha, \alpha)\) lies but much both lines - \(\alpha-\sqrt{1-\alpha^2}\lt 0\) \(\alpha\lt \sqrt{1-\alpha^2}\) Both side squaring - \(\alpha^2\lt 1-\alpha^2\) \(2 \alpha^2\lt 1\) \(\alpha\lt \frac{1}{\sqrt{2}}\) \(\alpha+\alpha-1>0\) \(2 \alpha>1\) \(\alpha>\frac{1}{2}\)So, \(\quad \frac{1}{2}\lt \alpha\lt \frac{1}{\sqrt{2}}\)
JCECE-2013
Sets, Relation and Function
117317
The domain of the function \(f(x)=\sqrt{3-x}+\cos ^{-1}\left(\frac{3-2 x}{5}\right)\), is
1 \([-1,3]\)
2 \((-1,3]\)
3 \([-1,3)\)
4 None of these
Explanation:
A Given, \(f(x)=\sqrt{3-x}+\cos ^{-1}\left(\frac{3-2 x}{5}\right)\) For, \(\quad 3-\mathrm{x} \geq 0\) \(x \leq 3\) \(-1 \leq \frac{3-2 x}{5} \leq 1\) \(-5 \leq 3-2 x \leq 5\) \(-8 \leq-2 \mathrm{x} \leq 2\) \(-1 \leq \mathrm{x} \leq 4\) From equation (i) and (ii) - \((-\infty, 3] \cap[-1,4]=[-1,3]\)
JCECE-2010
Sets, Relation and Function
117318
Domain of \(f(x)=y=\sqrt{\log _3\{\cos (\sin x)\}}\) is
117319
Range of the function \(f(x)=\frac{x}{1+x^2}\) is
1 \((-\infty, \infty)\)
2 \([-1,1]\)
3 \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
4 \([-\sqrt{2}, \sqrt{2}]\)
Explanation:
C Given, \(f(x)=\frac{x}{1+x^2}\) \(y=\frac{x}{1+x^2}\) \(x^2 y-x+y=0\) \(y x^2-x+y=0\) For \(x\) to be real \(1-4 y^2 \geq 0\) \((1-2 y)(1+2 y) \geq 0\) Now, for real \(\mathrm{x}\), we have - \(\left(\frac{1}{2}-y\right)\left(\frac{1}{2}+y\right) \geq 0\) \(\frac{-1}{2} \leq y \leq \frac{1}{2}\) \(y=f(x) \in\left[-\frac{1}{2}, \frac{1}{2}\right] .\)
117316
The range of \(\alpha\), for which the point \((\alpha, \alpha)\) lies inside the region bounded by the curves \(y=\sqrt{1-x^2}\) and \(x+y=1\) is
1 \(\frac{1}{2}\lt \alpha\lt \frac{1}{\sqrt{2}}\)
2 \(\frac{1}{2}\lt \alpha\lt \frac{1}{3}\)
3 \(\frac{1}{3}\lt \alpha\lt \frac{1}{\sqrt{3}}\)
4 \(\frac{1}{4}\lt \alpha\lt \frac{1}{2}\)
Explanation:
A Given, \(y-\sqrt{1-x^2}=0 \text { and } x+y-1=0\) And, also given point \((\alpha, \alpha)\) lies but much both lines - \(\alpha-\sqrt{1-\alpha^2}\lt 0\) \(\alpha\lt \sqrt{1-\alpha^2}\) Both side squaring - \(\alpha^2\lt 1-\alpha^2\) \(2 \alpha^2\lt 1\) \(\alpha\lt \frac{1}{\sqrt{2}}\) \(\alpha+\alpha-1>0\) \(2 \alpha>1\) \(\alpha>\frac{1}{2}\)So, \(\quad \frac{1}{2}\lt \alpha\lt \frac{1}{\sqrt{2}}\)
JCECE-2013
Sets, Relation and Function
117317
The domain of the function \(f(x)=\sqrt{3-x}+\cos ^{-1}\left(\frac{3-2 x}{5}\right)\), is
1 \([-1,3]\)
2 \((-1,3]\)
3 \([-1,3)\)
4 None of these
Explanation:
A Given, \(f(x)=\sqrt{3-x}+\cos ^{-1}\left(\frac{3-2 x}{5}\right)\) For, \(\quad 3-\mathrm{x} \geq 0\) \(x \leq 3\) \(-1 \leq \frac{3-2 x}{5} \leq 1\) \(-5 \leq 3-2 x \leq 5\) \(-8 \leq-2 \mathrm{x} \leq 2\) \(-1 \leq \mathrm{x} \leq 4\) From equation (i) and (ii) - \((-\infty, 3] \cap[-1,4]=[-1,3]\)
JCECE-2010
Sets, Relation and Function
117318
Domain of \(f(x)=y=\sqrt{\log _3\{\cos (\sin x)\}}\) is
117319
Range of the function \(f(x)=\frac{x}{1+x^2}\) is
1 \((-\infty, \infty)\)
2 \([-1,1]\)
3 \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
4 \([-\sqrt{2}, \sqrt{2}]\)
Explanation:
C Given, \(f(x)=\frac{x}{1+x^2}\) \(y=\frac{x}{1+x^2}\) \(x^2 y-x+y=0\) \(y x^2-x+y=0\) For \(x\) to be real \(1-4 y^2 \geq 0\) \((1-2 y)(1+2 y) \geq 0\) Now, for real \(\mathrm{x}\), we have - \(\left(\frac{1}{2}-y\right)\left(\frac{1}{2}+y\right) \geq 0\) \(\frac{-1}{2} \leq y \leq \frac{1}{2}\) \(y=f(x) \in\left[-\frac{1}{2}, \frac{1}{2}\right] .\)