117393
Find the domain of the function \(f(x)=\frac{\left(x^2+1\right)}{\left(x^2-3 x+3\right)}\)
1 \(\mathrm{R}-\{1,2\}\)
2 \(\mathrm{R}-\{1,4\}\)
3 \(\mathrm{R}\)
4 \(\mathrm{R}-\{1\}\)
Explanation:
C Given, function is- \(f(x)=\frac{x^2+1}{x^2-3 x+3}\) Since, \(x^2-3 x+3>0\) Also, \(x^2+1>0, x \in R\) So, domain of \(f(x)=R\)
J and K CET-2014
Sets, Relation and Function
117394
Find the range of the function \(f:[0,1] \rightarrow R, f(x)=x^3-x^2+4 x+2 \sin ^{-1} x .\)
1 \([-(\pi+2), 0]\)
2 \([0,4+\pi]\)
3 \([2,3]\)
4 \((0,2+\pi)\)
Explanation:
B We have, \(f(x)=x^3-x^2+4 x+2 \sin ^{-1} x\) For, \(\mathrm{x}=0\), \(f(0)=0+2 \sin ^{-1}(0)=0\) For, \(\mathrm{x}=1\) \(f(1)=1-1+4+2 \sin ^{-1}\) \(=4+2 \times \frac{\pi}{2}=4+\pi\) Since, \(f(x)\) is an increasing function, So, its range \((R)=[0,4+\pi]\)
J and K CET-2014
Sets, Relation and Function
117395
Range of the function \(f(x)=\frac{x^2+x+2}{x^2+x+1}, x \in R\) is
1 \((1, \infty)\)
2 \(\left(1, \frac{11}{7}\right)\)
3 \(\left(1, \frac{7}{3}\right)\)
4 \(\left(1, \frac{7}{5}\right)\)
Explanation:
C Given, \(f(x)=\frac{x^2+x+2}{x^2+x+1}, x \in R\) Let, \(y=f(x)\), So, \(y=\frac{x^2+x+2}{x^2+x+1}\) \(y=1+\frac{1}{x^2+x+1} \quad[\because y>1] .\) \(\mathrm{yx}^2+\mathrm{yx}+\mathrm{y}=\mathrm{x}^2+\mathrm{x}+2\) \(x^2(y-1)+x(y-1)+(y-2)=6\) \(\because \mathrm{x} \in \mathrm{R} \Rightarrow \mathrm{D}>0\) \((y-1)^2-4(y-1)(y-2)>0\) \((\mathrm{y}-1)[(\mathrm{y}-1)-4(\mathrm{y}-2)]>0\) \((\mathrm{y}-1)(-3 \mathrm{y}+7)>0\) \((\mathrm{y}-1)(3 \mathrm{y}-7)\lt 0\) \(-\infty \frac{+}{1-\int \frac{7}{3}}+\infty\) \(\mathrm{y} \in\left(1, \frac{7}{3}\right)\)
Jamia Millia Islamia-2013
Sets, Relation and Function
117404
The domain of the function \(f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}\) is (where \([x]\) denotes the greatest integer less than or equal to \(x\) )
1 \((-\infty,-2) \cup(5, \infty)\)
2 \((-\infty,-3] \cup[6, \infty)\)
3 \((-\infty,-2) \cup[6, \infty)\)
4 \((-\infty,-3] \cup(5, \infty)\)
Explanation:
C Given that, \(f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}\) Define the function \(\mathrm{f}(\mathrm{x})\) - \({[\mathrm{x}]^2-3[\mathrm{x}]-10>0}\) \([\mathrm{x}]-5)([\mathrm{x}]+2)>0\) \(\mathrm{x} \in(-\infty,-2) \cup(5, \infty)\) \((6, \infty)\) Domain of function \(\mathrm{f}(\mathrm{x})\), \(x \in(-\infty,-2) \cup[6, \infty)\)
JEE Main-11.04.2023
Sets, Relation and Function
117431
The domain of \(\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]\) is
1 \([1,9]\)
2 \([-1,9]\)
3 \([-9,1]\)
4 \([-9,-1]\)
Explanation:
A Let, \(f(x)=\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]\) We know that, the domain of \(\sin ^{-1} \mathrm{x}\) is \([-1,1]\). Then, \(-1 \leq \log _3\left(\frac{x}{3}\right) \leq 1\) \(3^{-1} \leq \frac{x}{3} \leq 3\) \(\frac{1}{3} \leq \frac{x}{3} \leq 3\) \(1 \leq x \leq 9\) \(x \in[1,9]\)Hence, domain of \(f(x)\) is \([1,9]\).
117393
Find the domain of the function \(f(x)=\frac{\left(x^2+1\right)}{\left(x^2-3 x+3\right)}\)
1 \(\mathrm{R}-\{1,2\}\)
2 \(\mathrm{R}-\{1,4\}\)
3 \(\mathrm{R}\)
4 \(\mathrm{R}-\{1\}\)
Explanation:
C Given, function is- \(f(x)=\frac{x^2+1}{x^2-3 x+3}\) Since, \(x^2-3 x+3>0\) Also, \(x^2+1>0, x \in R\) So, domain of \(f(x)=R\)
J and K CET-2014
Sets, Relation and Function
117394
Find the range of the function \(f:[0,1] \rightarrow R, f(x)=x^3-x^2+4 x+2 \sin ^{-1} x .\)
1 \([-(\pi+2), 0]\)
2 \([0,4+\pi]\)
3 \([2,3]\)
4 \((0,2+\pi)\)
Explanation:
B We have, \(f(x)=x^3-x^2+4 x+2 \sin ^{-1} x\) For, \(\mathrm{x}=0\), \(f(0)=0+2 \sin ^{-1}(0)=0\) For, \(\mathrm{x}=1\) \(f(1)=1-1+4+2 \sin ^{-1}\) \(=4+2 \times \frac{\pi}{2}=4+\pi\) Since, \(f(x)\) is an increasing function, So, its range \((R)=[0,4+\pi]\)
J and K CET-2014
Sets, Relation and Function
117395
Range of the function \(f(x)=\frac{x^2+x+2}{x^2+x+1}, x \in R\) is
1 \((1, \infty)\)
2 \(\left(1, \frac{11}{7}\right)\)
3 \(\left(1, \frac{7}{3}\right)\)
4 \(\left(1, \frac{7}{5}\right)\)
Explanation:
C Given, \(f(x)=\frac{x^2+x+2}{x^2+x+1}, x \in R\) Let, \(y=f(x)\), So, \(y=\frac{x^2+x+2}{x^2+x+1}\) \(y=1+\frac{1}{x^2+x+1} \quad[\because y>1] .\) \(\mathrm{yx}^2+\mathrm{yx}+\mathrm{y}=\mathrm{x}^2+\mathrm{x}+2\) \(x^2(y-1)+x(y-1)+(y-2)=6\) \(\because \mathrm{x} \in \mathrm{R} \Rightarrow \mathrm{D}>0\) \((y-1)^2-4(y-1)(y-2)>0\) \((\mathrm{y}-1)[(\mathrm{y}-1)-4(\mathrm{y}-2)]>0\) \((\mathrm{y}-1)(-3 \mathrm{y}+7)>0\) \((\mathrm{y}-1)(3 \mathrm{y}-7)\lt 0\) \(-\infty \frac{+}{1-\int \frac{7}{3}}+\infty\) \(\mathrm{y} \in\left(1, \frac{7}{3}\right)\)
Jamia Millia Islamia-2013
Sets, Relation and Function
117404
The domain of the function \(f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}\) is (where \([x]\) denotes the greatest integer less than or equal to \(x\) )
1 \((-\infty,-2) \cup(5, \infty)\)
2 \((-\infty,-3] \cup[6, \infty)\)
3 \((-\infty,-2) \cup[6, \infty)\)
4 \((-\infty,-3] \cup(5, \infty)\)
Explanation:
C Given that, \(f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}\) Define the function \(\mathrm{f}(\mathrm{x})\) - \({[\mathrm{x}]^2-3[\mathrm{x}]-10>0}\) \([\mathrm{x}]-5)([\mathrm{x}]+2)>0\) \(\mathrm{x} \in(-\infty,-2) \cup(5, \infty)\) \((6, \infty)\) Domain of function \(\mathrm{f}(\mathrm{x})\), \(x \in(-\infty,-2) \cup[6, \infty)\)
JEE Main-11.04.2023
Sets, Relation and Function
117431
The domain of \(\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]\) is
1 \([1,9]\)
2 \([-1,9]\)
3 \([-9,1]\)
4 \([-9,-1]\)
Explanation:
A Let, \(f(x)=\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]\) We know that, the domain of \(\sin ^{-1} \mathrm{x}\) is \([-1,1]\). Then, \(-1 \leq \log _3\left(\frac{x}{3}\right) \leq 1\) \(3^{-1} \leq \frac{x}{3} \leq 3\) \(\frac{1}{3} \leq \frac{x}{3} \leq 3\) \(1 \leq x \leq 9\) \(x \in[1,9]\)Hence, domain of \(f(x)\) is \([1,9]\).
117393
Find the domain of the function \(f(x)=\frac{\left(x^2+1\right)}{\left(x^2-3 x+3\right)}\)
1 \(\mathrm{R}-\{1,2\}\)
2 \(\mathrm{R}-\{1,4\}\)
3 \(\mathrm{R}\)
4 \(\mathrm{R}-\{1\}\)
Explanation:
C Given, function is- \(f(x)=\frac{x^2+1}{x^2-3 x+3}\) Since, \(x^2-3 x+3>0\) Also, \(x^2+1>0, x \in R\) So, domain of \(f(x)=R\)
J and K CET-2014
Sets, Relation and Function
117394
Find the range of the function \(f:[0,1] \rightarrow R, f(x)=x^3-x^2+4 x+2 \sin ^{-1} x .\)
1 \([-(\pi+2), 0]\)
2 \([0,4+\pi]\)
3 \([2,3]\)
4 \((0,2+\pi)\)
Explanation:
B We have, \(f(x)=x^3-x^2+4 x+2 \sin ^{-1} x\) For, \(\mathrm{x}=0\), \(f(0)=0+2 \sin ^{-1}(0)=0\) For, \(\mathrm{x}=1\) \(f(1)=1-1+4+2 \sin ^{-1}\) \(=4+2 \times \frac{\pi}{2}=4+\pi\) Since, \(f(x)\) is an increasing function, So, its range \((R)=[0,4+\pi]\)
J and K CET-2014
Sets, Relation and Function
117395
Range of the function \(f(x)=\frac{x^2+x+2}{x^2+x+1}, x \in R\) is
1 \((1, \infty)\)
2 \(\left(1, \frac{11}{7}\right)\)
3 \(\left(1, \frac{7}{3}\right)\)
4 \(\left(1, \frac{7}{5}\right)\)
Explanation:
C Given, \(f(x)=\frac{x^2+x+2}{x^2+x+1}, x \in R\) Let, \(y=f(x)\), So, \(y=\frac{x^2+x+2}{x^2+x+1}\) \(y=1+\frac{1}{x^2+x+1} \quad[\because y>1] .\) \(\mathrm{yx}^2+\mathrm{yx}+\mathrm{y}=\mathrm{x}^2+\mathrm{x}+2\) \(x^2(y-1)+x(y-1)+(y-2)=6\) \(\because \mathrm{x} \in \mathrm{R} \Rightarrow \mathrm{D}>0\) \((y-1)^2-4(y-1)(y-2)>0\) \((\mathrm{y}-1)[(\mathrm{y}-1)-4(\mathrm{y}-2)]>0\) \((\mathrm{y}-1)(-3 \mathrm{y}+7)>0\) \((\mathrm{y}-1)(3 \mathrm{y}-7)\lt 0\) \(-\infty \frac{+}{1-\int \frac{7}{3}}+\infty\) \(\mathrm{y} \in\left(1, \frac{7}{3}\right)\)
Jamia Millia Islamia-2013
Sets, Relation and Function
117404
The domain of the function \(f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}\) is (where \([x]\) denotes the greatest integer less than or equal to \(x\) )
1 \((-\infty,-2) \cup(5, \infty)\)
2 \((-\infty,-3] \cup[6, \infty)\)
3 \((-\infty,-2) \cup[6, \infty)\)
4 \((-\infty,-3] \cup(5, \infty)\)
Explanation:
C Given that, \(f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}\) Define the function \(\mathrm{f}(\mathrm{x})\) - \({[\mathrm{x}]^2-3[\mathrm{x}]-10>0}\) \([\mathrm{x}]-5)([\mathrm{x}]+2)>0\) \(\mathrm{x} \in(-\infty,-2) \cup(5, \infty)\) \((6, \infty)\) Domain of function \(\mathrm{f}(\mathrm{x})\), \(x \in(-\infty,-2) \cup[6, \infty)\)
JEE Main-11.04.2023
Sets, Relation and Function
117431
The domain of \(\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]\) is
1 \([1,9]\)
2 \([-1,9]\)
3 \([-9,1]\)
4 \([-9,-1]\)
Explanation:
A Let, \(f(x)=\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]\) We know that, the domain of \(\sin ^{-1} \mathrm{x}\) is \([-1,1]\). Then, \(-1 \leq \log _3\left(\frac{x}{3}\right) \leq 1\) \(3^{-1} \leq \frac{x}{3} \leq 3\) \(\frac{1}{3} \leq \frac{x}{3} \leq 3\) \(1 \leq x \leq 9\) \(x \in[1,9]\)Hence, domain of \(f(x)\) is \([1,9]\).
117393
Find the domain of the function \(f(x)=\frac{\left(x^2+1\right)}{\left(x^2-3 x+3\right)}\)
1 \(\mathrm{R}-\{1,2\}\)
2 \(\mathrm{R}-\{1,4\}\)
3 \(\mathrm{R}\)
4 \(\mathrm{R}-\{1\}\)
Explanation:
C Given, function is- \(f(x)=\frac{x^2+1}{x^2-3 x+3}\) Since, \(x^2-3 x+3>0\) Also, \(x^2+1>0, x \in R\) So, domain of \(f(x)=R\)
J and K CET-2014
Sets, Relation and Function
117394
Find the range of the function \(f:[0,1] \rightarrow R, f(x)=x^3-x^2+4 x+2 \sin ^{-1} x .\)
1 \([-(\pi+2), 0]\)
2 \([0,4+\pi]\)
3 \([2,3]\)
4 \((0,2+\pi)\)
Explanation:
B We have, \(f(x)=x^3-x^2+4 x+2 \sin ^{-1} x\) For, \(\mathrm{x}=0\), \(f(0)=0+2 \sin ^{-1}(0)=0\) For, \(\mathrm{x}=1\) \(f(1)=1-1+4+2 \sin ^{-1}\) \(=4+2 \times \frac{\pi}{2}=4+\pi\) Since, \(f(x)\) is an increasing function, So, its range \((R)=[0,4+\pi]\)
J and K CET-2014
Sets, Relation and Function
117395
Range of the function \(f(x)=\frac{x^2+x+2}{x^2+x+1}, x \in R\) is
1 \((1, \infty)\)
2 \(\left(1, \frac{11}{7}\right)\)
3 \(\left(1, \frac{7}{3}\right)\)
4 \(\left(1, \frac{7}{5}\right)\)
Explanation:
C Given, \(f(x)=\frac{x^2+x+2}{x^2+x+1}, x \in R\) Let, \(y=f(x)\), So, \(y=\frac{x^2+x+2}{x^2+x+1}\) \(y=1+\frac{1}{x^2+x+1} \quad[\because y>1] .\) \(\mathrm{yx}^2+\mathrm{yx}+\mathrm{y}=\mathrm{x}^2+\mathrm{x}+2\) \(x^2(y-1)+x(y-1)+(y-2)=6\) \(\because \mathrm{x} \in \mathrm{R} \Rightarrow \mathrm{D}>0\) \((y-1)^2-4(y-1)(y-2)>0\) \((\mathrm{y}-1)[(\mathrm{y}-1)-4(\mathrm{y}-2)]>0\) \((\mathrm{y}-1)(-3 \mathrm{y}+7)>0\) \((\mathrm{y}-1)(3 \mathrm{y}-7)\lt 0\) \(-\infty \frac{+}{1-\int \frac{7}{3}}+\infty\) \(\mathrm{y} \in\left(1, \frac{7}{3}\right)\)
Jamia Millia Islamia-2013
Sets, Relation and Function
117404
The domain of the function \(f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}\) is (where \([x]\) denotes the greatest integer less than or equal to \(x\) )
1 \((-\infty,-2) \cup(5, \infty)\)
2 \((-\infty,-3] \cup[6, \infty)\)
3 \((-\infty,-2) \cup[6, \infty)\)
4 \((-\infty,-3] \cup(5, \infty)\)
Explanation:
C Given that, \(f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}\) Define the function \(\mathrm{f}(\mathrm{x})\) - \({[\mathrm{x}]^2-3[\mathrm{x}]-10>0}\) \([\mathrm{x}]-5)([\mathrm{x}]+2)>0\) \(\mathrm{x} \in(-\infty,-2) \cup(5, \infty)\) \((6, \infty)\) Domain of function \(\mathrm{f}(\mathrm{x})\), \(x \in(-\infty,-2) \cup[6, \infty)\)
JEE Main-11.04.2023
Sets, Relation and Function
117431
The domain of \(\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]\) is
1 \([1,9]\)
2 \([-1,9]\)
3 \([-9,1]\)
4 \([-9,-1]\)
Explanation:
A Let, \(f(x)=\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]\) We know that, the domain of \(\sin ^{-1} \mathrm{x}\) is \([-1,1]\). Then, \(-1 \leq \log _3\left(\frac{x}{3}\right) \leq 1\) \(3^{-1} \leq \frac{x}{3} \leq 3\) \(\frac{1}{3} \leq \frac{x}{3} \leq 3\) \(1 \leq x \leq 9\) \(x \in[1,9]\)Hence, domain of \(f(x)\) is \([1,9]\).
117393
Find the domain of the function \(f(x)=\frac{\left(x^2+1\right)}{\left(x^2-3 x+3\right)}\)
1 \(\mathrm{R}-\{1,2\}\)
2 \(\mathrm{R}-\{1,4\}\)
3 \(\mathrm{R}\)
4 \(\mathrm{R}-\{1\}\)
Explanation:
C Given, function is- \(f(x)=\frac{x^2+1}{x^2-3 x+3}\) Since, \(x^2-3 x+3>0\) Also, \(x^2+1>0, x \in R\) So, domain of \(f(x)=R\)
J and K CET-2014
Sets, Relation and Function
117394
Find the range of the function \(f:[0,1] \rightarrow R, f(x)=x^3-x^2+4 x+2 \sin ^{-1} x .\)
1 \([-(\pi+2), 0]\)
2 \([0,4+\pi]\)
3 \([2,3]\)
4 \((0,2+\pi)\)
Explanation:
B We have, \(f(x)=x^3-x^2+4 x+2 \sin ^{-1} x\) For, \(\mathrm{x}=0\), \(f(0)=0+2 \sin ^{-1}(0)=0\) For, \(\mathrm{x}=1\) \(f(1)=1-1+4+2 \sin ^{-1}\) \(=4+2 \times \frac{\pi}{2}=4+\pi\) Since, \(f(x)\) is an increasing function, So, its range \((R)=[0,4+\pi]\)
J and K CET-2014
Sets, Relation and Function
117395
Range of the function \(f(x)=\frac{x^2+x+2}{x^2+x+1}, x \in R\) is
1 \((1, \infty)\)
2 \(\left(1, \frac{11}{7}\right)\)
3 \(\left(1, \frac{7}{3}\right)\)
4 \(\left(1, \frac{7}{5}\right)\)
Explanation:
C Given, \(f(x)=\frac{x^2+x+2}{x^2+x+1}, x \in R\) Let, \(y=f(x)\), So, \(y=\frac{x^2+x+2}{x^2+x+1}\) \(y=1+\frac{1}{x^2+x+1} \quad[\because y>1] .\) \(\mathrm{yx}^2+\mathrm{yx}+\mathrm{y}=\mathrm{x}^2+\mathrm{x}+2\) \(x^2(y-1)+x(y-1)+(y-2)=6\) \(\because \mathrm{x} \in \mathrm{R} \Rightarrow \mathrm{D}>0\) \((y-1)^2-4(y-1)(y-2)>0\) \((\mathrm{y}-1)[(\mathrm{y}-1)-4(\mathrm{y}-2)]>0\) \((\mathrm{y}-1)(-3 \mathrm{y}+7)>0\) \((\mathrm{y}-1)(3 \mathrm{y}-7)\lt 0\) \(-\infty \frac{+}{1-\int \frac{7}{3}}+\infty\) \(\mathrm{y} \in\left(1, \frac{7}{3}\right)\)
Jamia Millia Islamia-2013
Sets, Relation and Function
117404
The domain of the function \(f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}\) is (where \([x]\) denotes the greatest integer less than or equal to \(x\) )
1 \((-\infty,-2) \cup(5, \infty)\)
2 \((-\infty,-3] \cup[6, \infty)\)
3 \((-\infty,-2) \cup[6, \infty)\)
4 \((-\infty,-3] \cup(5, \infty)\)
Explanation:
C Given that, \(f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}\) Define the function \(\mathrm{f}(\mathrm{x})\) - \({[\mathrm{x}]^2-3[\mathrm{x}]-10>0}\) \([\mathrm{x}]-5)([\mathrm{x}]+2)>0\) \(\mathrm{x} \in(-\infty,-2) \cup(5, \infty)\) \((6, \infty)\) Domain of function \(\mathrm{f}(\mathrm{x})\), \(x \in(-\infty,-2) \cup[6, \infty)\)
JEE Main-11.04.2023
Sets, Relation and Function
117431
The domain of \(\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]\) is
1 \([1,9]\)
2 \([-1,9]\)
3 \([-9,1]\)
4 \([-9,-1]\)
Explanation:
A Let, \(f(x)=\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]\) We know that, the domain of \(\sin ^{-1} \mathrm{x}\) is \([-1,1]\). Then, \(-1 \leq \log _3\left(\frac{x}{3}\right) \leq 1\) \(3^{-1} \leq \frac{x}{3} \leq 3\) \(\frac{1}{3} \leq \frac{x}{3} \leq 3\) \(1 \leq x \leq 9\) \(x \in[1,9]\)Hence, domain of \(f(x)\) is \([1,9]\).
