117415
Let \(f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right),-1\lt x\lt 1\) and \(g(x)=\sqrt{3+4 x-4 x^2}\), then the domain \((f+g)\) (x)
1 \(\left[\frac{1}{2}, 1\right)\)
2 \(\left[\frac{-1}{2}, \frac{1}{2}\right)\)
3 \(\left[-\frac{1}{2}, 1\right)\)
4 \(\left[-\frac{1}{2},-1\right]\)
Explanation:
C According to the question- \(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})=\sqrt{3+4 \mathrm{x}-4 \mathrm{x}^2}+\frac{1}{2}-\tan \left(\frac{\pi \mathrm{x}}{2}\right)\) Therefore, for real values of \((f+g)\) - \(3+4 x-4 x^2 \geq 0\) \(4 x^2-4 x-3 \leq 0\) \(4 x^2-6 x+2 x-3 \leq 0\) \(2 x(2 x-3)+1(2 x-3) \leq 0\) \((2 x+1)(2 x-3) \leq 0\) \(x \leq \frac{-1}{2} \text { and } x \leq 3\) Therefore, \(\mathrm{x} \in\left[\frac{-1}{2}, 3\right]\) Also consider \(\tan \left(\frac{\pi \cdot x}{2}\right)\) since tanx is discontinuous for all \(x \in\left(\frac{(2 \mathrm{n}+1) \pi}{2}\right)\) Hence, for \(\tan \left(\frac{\pi \cdot x}{2}\right)\) the domain is \(R-\{1,3\), \(5 \ldots \ldots .(2 \mathrm{n}+1)\}\) From (i) \& (ii), we get - \(\mathrm{x} \in\left[-\frac{1}{2}, 1\right) \cup(1,3)\)Then, \(\quad \mathrm{x} \in\left[-\frac{1}{2}, 1\right)\)
TS EAMCET-19.07.2022
Sets, Relation and Function
117416
Let \(a>1\) be a constant. If \(f: A \rightarrow A\) and (x, \()\) ef satisfy \(\mathbf{a}^{\mathbf{x}}+\mathbf{a}^{\mathbf{y}}=\mathbf{a}\), then \(A=\)
1 \((0, a]\)
2 \([0, a]\)
3 \((-\infty, 1)\)
4 \((-\infty, a+1)\)
Explanation:
C Given- \(a^x+a^y=a\) \(a^{x-1}+a^{y-1}=1\) Let, we take \(x \geq 1\) and \(y \geq 1\) then these value not satisfy \(\mathrm{eq}^{\mathrm{n}} \cdot \mathrm{a}^{\mathrm{x}-1}+\mathrm{a}^{\mathrm{y}-1}=1\) Hence, \(\mathrm{x}-1\lt 0\) and \(\mathrm{y}-1\lt 0\) satisfy these equation - \(\Rightarrow \mathrm{x}\lt 1 \text { and } \mathrm{y}\lt 1\) \(\Rightarrow \mathrm{x} \in(-\infty, 1) \text { and } \mathrm{y} \in(-\infty, 1)\) \(\Rightarrow \mathrm{A}=(-\infty, 1) \)
TS EAMCET-05.08.2021
Sets, Relation and Function
117417
Let \([x]\) denote the greatest integer not more than \(x\). If \(A\) and \(B\) are the domains of the functions \(f(x)=\frac{x-[x]}{\sqrt{|x|-x}}\) and \(g(x)=\frac{x-[x]}{\sqrt{|x|+x}}\) respectively, then
1 \(\mathrm{A} \cup \mathrm{B}=\mathrm{R}\)
2 \(\mathrm{A} \cap \mathrm{B}=\phi\)
3 \(\mathrm{A}-\mathrm{B}=(-\infty, 0)\)
4 \(\mathrm{B}-\mathrm{A}=(0, \infty)\)
Explanation:
B Given that, function \(f(x)=\frac{x-[x]}{\sqrt{|x|-x}}\) will define, if \(|x|>x \Rightarrow x\lt 0 \Rightarrow x \in(-\infty, 0)\). And, the function, \(g(x)=\frac{x-[x]}{\sqrt{|x|+x}}\) will define, if \(|\mathrm{x}|+\mathrm{x}>0 \Rightarrow \mathrm{x}>0 \Rightarrow \mathrm{x} \in(0, \infty)\) \(\therefore \quad \mathrm{A}=\{\mathrm{x} \mid \mathrm{x} \in \mathrm{R}\) and \(\mathrm{x}\lt 0\}\) And, \(\quad \mathrm{B}=\{\mathrm{x} \mid \mathrm{x} \in \mathrm{R}\) and \(\mathrm{x}>0\}\) \(\therefore \quad \mathrm{A} \cap \mathrm{B}=\phi\)
TS EAMCET-11.09.2020
Sets, Relation and Function
117418
Domain of \(\cos ^{-1}\left[\log _5\left(x^2+7 x+15\right)\right]\) is
1 The set of all real numbers
2 \([-\infty,-5] \cup[-2, \infty)\)
3 \(\mathrm{R}-\{-5,-2\}\), where \(\mathrm{R}\) is the set of real numbers
117415
Let \(f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right),-1\lt x\lt 1\) and \(g(x)=\sqrt{3+4 x-4 x^2}\), then the domain \((f+g)\) (x)
1 \(\left[\frac{1}{2}, 1\right)\)
2 \(\left[\frac{-1}{2}, \frac{1}{2}\right)\)
3 \(\left[-\frac{1}{2}, 1\right)\)
4 \(\left[-\frac{1}{2},-1\right]\)
Explanation:
C According to the question- \(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})=\sqrt{3+4 \mathrm{x}-4 \mathrm{x}^2}+\frac{1}{2}-\tan \left(\frac{\pi \mathrm{x}}{2}\right)\) Therefore, for real values of \((f+g)\) - \(3+4 x-4 x^2 \geq 0\) \(4 x^2-4 x-3 \leq 0\) \(4 x^2-6 x+2 x-3 \leq 0\) \(2 x(2 x-3)+1(2 x-3) \leq 0\) \((2 x+1)(2 x-3) \leq 0\) \(x \leq \frac{-1}{2} \text { and } x \leq 3\) Therefore, \(\mathrm{x} \in\left[\frac{-1}{2}, 3\right]\) Also consider \(\tan \left(\frac{\pi \cdot x}{2}\right)\) since tanx is discontinuous for all \(x \in\left(\frac{(2 \mathrm{n}+1) \pi}{2}\right)\) Hence, for \(\tan \left(\frac{\pi \cdot x}{2}\right)\) the domain is \(R-\{1,3\), \(5 \ldots \ldots .(2 \mathrm{n}+1)\}\) From (i) \& (ii), we get - \(\mathrm{x} \in\left[-\frac{1}{2}, 1\right) \cup(1,3)\)Then, \(\quad \mathrm{x} \in\left[-\frac{1}{2}, 1\right)\)
TS EAMCET-19.07.2022
Sets, Relation and Function
117416
Let \(a>1\) be a constant. If \(f: A \rightarrow A\) and (x, \()\) ef satisfy \(\mathbf{a}^{\mathbf{x}}+\mathbf{a}^{\mathbf{y}}=\mathbf{a}\), then \(A=\)
1 \((0, a]\)
2 \([0, a]\)
3 \((-\infty, 1)\)
4 \((-\infty, a+1)\)
Explanation:
C Given- \(a^x+a^y=a\) \(a^{x-1}+a^{y-1}=1\) Let, we take \(x \geq 1\) and \(y \geq 1\) then these value not satisfy \(\mathrm{eq}^{\mathrm{n}} \cdot \mathrm{a}^{\mathrm{x}-1}+\mathrm{a}^{\mathrm{y}-1}=1\) Hence, \(\mathrm{x}-1\lt 0\) and \(\mathrm{y}-1\lt 0\) satisfy these equation - \(\Rightarrow \mathrm{x}\lt 1 \text { and } \mathrm{y}\lt 1\) \(\Rightarrow \mathrm{x} \in(-\infty, 1) \text { and } \mathrm{y} \in(-\infty, 1)\) \(\Rightarrow \mathrm{A}=(-\infty, 1) \)
TS EAMCET-05.08.2021
Sets, Relation and Function
117417
Let \([x]\) denote the greatest integer not more than \(x\). If \(A\) and \(B\) are the domains of the functions \(f(x)=\frac{x-[x]}{\sqrt{|x|-x}}\) and \(g(x)=\frac{x-[x]}{\sqrt{|x|+x}}\) respectively, then
1 \(\mathrm{A} \cup \mathrm{B}=\mathrm{R}\)
2 \(\mathrm{A} \cap \mathrm{B}=\phi\)
3 \(\mathrm{A}-\mathrm{B}=(-\infty, 0)\)
4 \(\mathrm{B}-\mathrm{A}=(0, \infty)\)
Explanation:
B Given that, function \(f(x)=\frac{x-[x]}{\sqrt{|x|-x}}\) will define, if \(|x|>x \Rightarrow x\lt 0 \Rightarrow x \in(-\infty, 0)\). And, the function, \(g(x)=\frac{x-[x]}{\sqrt{|x|+x}}\) will define, if \(|\mathrm{x}|+\mathrm{x}>0 \Rightarrow \mathrm{x}>0 \Rightarrow \mathrm{x} \in(0, \infty)\) \(\therefore \quad \mathrm{A}=\{\mathrm{x} \mid \mathrm{x} \in \mathrm{R}\) and \(\mathrm{x}\lt 0\}\) And, \(\quad \mathrm{B}=\{\mathrm{x} \mid \mathrm{x} \in \mathrm{R}\) and \(\mathrm{x}>0\}\) \(\therefore \quad \mathrm{A} \cap \mathrm{B}=\phi\)
TS EAMCET-11.09.2020
Sets, Relation and Function
117418
Domain of \(\cos ^{-1}\left[\log _5\left(x^2+7 x+15\right)\right]\) is
1 The set of all real numbers
2 \([-\infty,-5] \cup[-2, \infty)\)
3 \(\mathrm{R}-\{-5,-2\}\), where \(\mathrm{R}\) is the set of real numbers
117415
Let \(f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right),-1\lt x\lt 1\) and \(g(x)=\sqrt{3+4 x-4 x^2}\), then the domain \((f+g)\) (x)
1 \(\left[\frac{1}{2}, 1\right)\)
2 \(\left[\frac{-1}{2}, \frac{1}{2}\right)\)
3 \(\left[-\frac{1}{2}, 1\right)\)
4 \(\left[-\frac{1}{2},-1\right]\)
Explanation:
C According to the question- \(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})=\sqrt{3+4 \mathrm{x}-4 \mathrm{x}^2}+\frac{1}{2}-\tan \left(\frac{\pi \mathrm{x}}{2}\right)\) Therefore, for real values of \((f+g)\) - \(3+4 x-4 x^2 \geq 0\) \(4 x^2-4 x-3 \leq 0\) \(4 x^2-6 x+2 x-3 \leq 0\) \(2 x(2 x-3)+1(2 x-3) \leq 0\) \((2 x+1)(2 x-3) \leq 0\) \(x \leq \frac{-1}{2} \text { and } x \leq 3\) Therefore, \(\mathrm{x} \in\left[\frac{-1}{2}, 3\right]\) Also consider \(\tan \left(\frac{\pi \cdot x}{2}\right)\) since tanx is discontinuous for all \(x \in\left(\frac{(2 \mathrm{n}+1) \pi}{2}\right)\) Hence, for \(\tan \left(\frac{\pi \cdot x}{2}\right)\) the domain is \(R-\{1,3\), \(5 \ldots \ldots .(2 \mathrm{n}+1)\}\) From (i) \& (ii), we get - \(\mathrm{x} \in\left[-\frac{1}{2}, 1\right) \cup(1,3)\)Then, \(\quad \mathrm{x} \in\left[-\frac{1}{2}, 1\right)\)
TS EAMCET-19.07.2022
Sets, Relation and Function
117416
Let \(a>1\) be a constant. If \(f: A \rightarrow A\) and (x, \()\) ef satisfy \(\mathbf{a}^{\mathbf{x}}+\mathbf{a}^{\mathbf{y}}=\mathbf{a}\), then \(A=\)
1 \((0, a]\)
2 \([0, a]\)
3 \((-\infty, 1)\)
4 \((-\infty, a+1)\)
Explanation:
C Given- \(a^x+a^y=a\) \(a^{x-1}+a^{y-1}=1\) Let, we take \(x \geq 1\) and \(y \geq 1\) then these value not satisfy \(\mathrm{eq}^{\mathrm{n}} \cdot \mathrm{a}^{\mathrm{x}-1}+\mathrm{a}^{\mathrm{y}-1}=1\) Hence, \(\mathrm{x}-1\lt 0\) and \(\mathrm{y}-1\lt 0\) satisfy these equation - \(\Rightarrow \mathrm{x}\lt 1 \text { and } \mathrm{y}\lt 1\) \(\Rightarrow \mathrm{x} \in(-\infty, 1) \text { and } \mathrm{y} \in(-\infty, 1)\) \(\Rightarrow \mathrm{A}=(-\infty, 1) \)
TS EAMCET-05.08.2021
Sets, Relation and Function
117417
Let \([x]\) denote the greatest integer not more than \(x\). If \(A\) and \(B\) are the domains of the functions \(f(x)=\frac{x-[x]}{\sqrt{|x|-x}}\) and \(g(x)=\frac{x-[x]}{\sqrt{|x|+x}}\) respectively, then
1 \(\mathrm{A} \cup \mathrm{B}=\mathrm{R}\)
2 \(\mathrm{A} \cap \mathrm{B}=\phi\)
3 \(\mathrm{A}-\mathrm{B}=(-\infty, 0)\)
4 \(\mathrm{B}-\mathrm{A}=(0, \infty)\)
Explanation:
B Given that, function \(f(x)=\frac{x-[x]}{\sqrt{|x|-x}}\) will define, if \(|x|>x \Rightarrow x\lt 0 \Rightarrow x \in(-\infty, 0)\). And, the function, \(g(x)=\frac{x-[x]}{\sqrt{|x|+x}}\) will define, if \(|\mathrm{x}|+\mathrm{x}>0 \Rightarrow \mathrm{x}>0 \Rightarrow \mathrm{x} \in(0, \infty)\) \(\therefore \quad \mathrm{A}=\{\mathrm{x} \mid \mathrm{x} \in \mathrm{R}\) and \(\mathrm{x}\lt 0\}\) And, \(\quad \mathrm{B}=\{\mathrm{x} \mid \mathrm{x} \in \mathrm{R}\) and \(\mathrm{x}>0\}\) \(\therefore \quad \mathrm{A} \cap \mathrm{B}=\phi\)
TS EAMCET-11.09.2020
Sets, Relation and Function
117418
Domain of \(\cos ^{-1}\left[\log _5\left(x^2+7 x+15\right)\right]\) is
1 The set of all real numbers
2 \([-\infty,-5] \cup[-2, \infty)\)
3 \(\mathrm{R}-\{-5,-2\}\), where \(\mathrm{R}\) is the set of real numbers
117415
Let \(f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right),-1\lt x\lt 1\) and \(g(x)=\sqrt{3+4 x-4 x^2}\), then the domain \((f+g)\) (x)
1 \(\left[\frac{1}{2}, 1\right)\)
2 \(\left[\frac{-1}{2}, \frac{1}{2}\right)\)
3 \(\left[-\frac{1}{2}, 1\right)\)
4 \(\left[-\frac{1}{2},-1\right]\)
Explanation:
C According to the question- \(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})=\sqrt{3+4 \mathrm{x}-4 \mathrm{x}^2}+\frac{1}{2}-\tan \left(\frac{\pi \mathrm{x}}{2}\right)\) Therefore, for real values of \((f+g)\) - \(3+4 x-4 x^2 \geq 0\) \(4 x^2-4 x-3 \leq 0\) \(4 x^2-6 x+2 x-3 \leq 0\) \(2 x(2 x-3)+1(2 x-3) \leq 0\) \((2 x+1)(2 x-3) \leq 0\) \(x \leq \frac{-1}{2} \text { and } x \leq 3\) Therefore, \(\mathrm{x} \in\left[\frac{-1}{2}, 3\right]\) Also consider \(\tan \left(\frac{\pi \cdot x}{2}\right)\) since tanx is discontinuous for all \(x \in\left(\frac{(2 \mathrm{n}+1) \pi}{2}\right)\) Hence, for \(\tan \left(\frac{\pi \cdot x}{2}\right)\) the domain is \(R-\{1,3\), \(5 \ldots \ldots .(2 \mathrm{n}+1)\}\) From (i) \& (ii), we get - \(\mathrm{x} \in\left[-\frac{1}{2}, 1\right) \cup(1,3)\)Then, \(\quad \mathrm{x} \in\left[-\frac{1}{2}, 1\right)\)
TS EAMCET-19.07.2022
Sets, Relation and Function
117416
Let \(a>1\) be a constant. If \(f: A \rightarrow A\) and (x, \()\) ef satisfy \(\mathbf{a}^{\mathbf{x}}+\mathbf{a}^{\mathbf{y}}=\mathbf{a}\), then \(A=\)
1 \((0, a]\)
2 \([0, a]\)
3 \((-\infty, 1)\)
4 \((-\infty, a+1)\)
Explanation:
C Given- \(a^x+a^y=a\) \(a^{x-1}+a^{y-1}=1\) Let, we take \(x \geq 1\) and \(y \geq 1\) then these value not satisfy \(\mathrm{eq}^{\mathrm{n}} \cdot \mathrm{a}^{\mathrm{x}-1}+\mathrm{a}^{\mathrm{y}-1}=1\) Hence, \(\mathrm{x}-1\lt 0\) and \(\mathrm{y}-1\lt 0\) satisfy these equation - \(\Rightarrow \mathrm{x}\lt 1 \text { and } \mathrm{y}\lt 1\) \(\Rightarrow \mathrm{x} \in(-\infty, 1) \text { and } \mathrm{y} \in(-\infty, 1)\) \(\Rightarrow \mathrm{A}=(-\infty, 1) \)
TS EAMCET-05.08.2021
Sets, Relation and Function
117417
Let \([x]\) denote the greatest integer not more than \(x\). If \(A\) and \(B\) are the domains of the functions \(f(x)=\frac{x-[x]}{\sqrt{|x|-x}}\) and \(g(x)=\frac{x-[x]}{\sqrt{|x|+x}}\) respectively, then
1 \(\mathrm{A} \cup \mathrm{B}=\mathrm{R}\)
2 \(\mathrm{A} \cap \mathrm{B}=\phi\)
3 \(\mathrm{A}-\mathrm{B}=(-\infty, 0)\)
4 \(\mathrm{B}-\mathrm{A}=(0, \infty)\)
Explanation:
B Given that, function \(f(x)=\frac{x-[x]}{\sqrt{|x|-x}}\) will define, if \(|x|>x \Rightarrow x\lt 0 \Rightarrow x \in(-\infty, 0)\). And, the function, \(g(x)=\frac{x-[x]}{\sqrt{|x|+x}}\) will define, if \(|\mathrm{x}|+\mathrm{x}>0 \Rightarrow \mathrm{x}>0 \Rightarrow \mathrm{x} \in(0, \infty)\) \(\therefore \quad \mathrm{A}=\{\mathrm{x} \mid \mathrm{x} \in \mathrm{R}\) and \(\mathrm{x}\lt 0\}\) And, \(\quad \mathrm{B}=\{\mathrm{x} \mid \mathrm{x} \in \mathrm{R}\) and \(\mathrm{x}>0\}\) \(\therefore \quad \mathrm{A} \cap \mathrm{B}=\phi\)
TS EAMCET-11.09.2020
Sets, Relation and Function
117418
Domain of \(\cos ^{-1}\left[\log _5\left(x^2+7 x+15\right)\right]\) is
1 The set of all real numbers
2 \([-\infty,-5] \cup[-2, \infty)\)
3 \(\mathrm{R}-\{-5,-2\}\), where \(\mathrm{R}\) is the set of real numbers
