117428
The domain of the function \(f(x)=\frac{1}{\log _{10}(1-x)}\) is
1 \((-\infty, 1]-\{0\}\)
2 \((-\infty, 1)-\{0\}\)
3 \((-\infty, 2)\)
4 \((0, \infty)\)
Explanation:
B Given that, \(f(x)=\frac{1}{\log _{10}(1-x)}\) For domain, \(1-\mathrm{x}>0\) \(\mathrm{x}\lt 1\) And, \(\log _{10}(1-\mathrm{x}) \neq 0\) From equation (i) and (ii), we get- Domain of \(f(x)=(-\infty, 1)-\{0\}\).
J and K CET-2010
Sets, Relation and Function
117429
For any real number \(y\) the greatest integer not exceeding \(y\) is denoted by [y]. If \(f: F \rightarrow R\) is denoted by \(f(x)=[2 x]-2[x]\) for \(x \in R\), then the range of \(f\) is:
D Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) and \(\mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) We take two cases to find range. Case - (i) : When \(x\) is integer then \([x]=x\) Let, \(\mathrm{x}=\mathrm{p}\) is integer Then, \(\mathrm{f}(\mathrm{x}) =[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[2 \mathrm{p}]-2[\mathrm{p}]\) \(=2 \mathrm{p}-2 \mathrm{p}=0\) \(\therefore \mathrm{f}(\mathrm{x})=0\), when \(\mathrm{x} \in \mathrm{I}\) Case - (ii) : When \(x\) is not integer Let, \(\mathrm{x}=\mathrm{p}+\mathrm{q}\), where \(\mathrm{p}\) is integer and \(\mathrm{q}\) is decimal part. And, \(0 \leq \mathrm{q}\lt 1\) Then, \([\mathrm{x}]=[\mathrm{p}+\mathrm{q}]=\mathrm{p}+[\mathrm{q}]\) Now, \(\mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[2 .(\mathrm{p}+\mathrm{q})]-2[\mathrm{p}+\mathrm{q}]\) \(=[2 \mathrm{p}+2 \mathrm{q}]-2(\mathrm{p}+[\mathrm{q}])\) \(=2 \mathrm{p}+[2 \mathrm{q}]-2 \mathrm{p}-2[\mathrm{q}]\) \(=[2 \mathrm{q}]-2[\mathrm{q}]\) \(=0-2 \times 0\) \(=0\), if \(0 \leq \mathrm{q} \leq 0.5\) or \(0 \leq 2 \mathrm{q}\lt 1\) \(=1-2 \times 0=1\), if \(0.5 \leq \mathrm{q}\lt 1\) or \(1 \leq 2 \mathrm{q}\lt 2\) \(\therefore \mathrm{f}(\mathrm{x})=0\) or 1 if \(\mathrm{x}\) is not integer From (i) and (ii) : for \(x \in R\), Range \(=\{0,1\}\). \(\therefore\) Range of \(\mathrm{f}=\{0,1\}\)
J and K CET-2006
Sets, Relation and Function
117432
The period of the function \(f(x)=\sin ^4 x+\cos ^4 x\) is
1 \(\pi\)
2 \(\frac{\pi}{2}\)
3 \(2 \pi\)
4 None of these
Explanation:
B Given that, \(\mathrm{f}(\mathrm{x})=\sin ^4 \mathrm{x}+\cos ^4 \mathrm{x}\) \(\mathrm{f}(\mathrm{x})=\left(\sin ^2 \mathrm{x}+\cos ^2 \mathrm{x}\right)^2-2 \sin ^2 \mathrm{x} \cos ^2 \mathrm{x}\) \(\mathrm{f}(\mathrm{x})=1-\frac{1}{2}(\sin 2 \mathrm{x})^2\) \(\mathrm{f}(\mathrm{x})=1-\frac{1}{2}\left(\frac{1-\cos 4 \mathrm{x}}{2}\right)\) \(\mathrm{f}(\mathrm{x})=\frac{3}{4}-\frac{1}{4} \cos 4 \mathrm{x}\) Ans: b Exp: B Given that, Therefore, here \(\cos 4 \mathrm{x}\) is a periodic function with period \(\frac{2 \pi}{4}=\frac{\pi}{2}\) Hence, the period of \(\mathrm{f}(\mathrm{x})\) is \(\frac{\pi}{2}\)
AIEEE-2002
Sets, Relation and Function
117433
Domain of definition of the function \(f(x)=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right)\), is
1 \((1,2)\)
2 \((-1,0) \cup(1,2)\)
3 \((1,2) \cup(2, \infty)\)
4 \((-1,0) \cup(1,2) \cup(2, \infty)\)
Explanation:
D Given that, \(f(x)=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right)\) Define the function \(\mathrm{f}(\mathrm{x})\) is, \(4-x^2 \neq 0\) \(x^2 \neq 4\) \(x \neq \pm 2\) And, \(\mathrm{x}^3-\mathrm{x}>0\) \(\mathrm{x}\left(\mathrm{x}^2-1\right)>0\) \(\mathrm{x}(\mathrm{x}+1)(\mathrm{x}-1)>0\) \(\mathrm{x} \in(-1,0) \cup(1, \infty)\) From equation (i) and (ii) we get \(\mathrm{x} \in(-1,0) \cup(1,2) \cup(2, \infty)\)Hence, the domain of \(\mathrm{f}(\mathrm{x})\) is \((-1,0) \cup(1,2) \cup(2, \infty)\)
117428
The domain of the function \(f(x)=\frac{1}{\log _{10}(1-x)}\) is
1 \((-\infty, 1]-\{0\}\)
2 \((-\infty, 1)-\{0\}\)
3 \((-\infty, 2)\)
4 \((0, \infty)\)
Explanation:
B Given that, \(f(x)=\frac{1}{\log _{10}(1-x)}\) For domain, \(1-\mathrm{x}>0\) \(\mathrm{x}\lt 1\) And, \(\log _{10}(1-\mathrm{x}) \neq 0\) From equation (i) and (ii), we get- Domain of \(f(x)=(-\infty, 1)-\{0\}\).
J and K CET-2010
Sets, Relation and Function
117429
For any real number \(y\) the greatest integer not exceeding \(y\) is denoted by [y]. If \(f: F \rightarrow R\) is denoted by \(f(x)=[2 x]-2[x]\) for \(x \in R\), then the range of \(f\) is:
D Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) and \(\mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) We take two cases to find range. Case - (i) : When \(x\) is integer then \([x]=x\) Let, \(\mathrm{x}=\mathrm{p}\) is integer Then, \(\mathrm{f}(\mathrm{x}) =[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[2 \mathrm{p}]-2[\mathrm{p}]\) \(=2 \mathrm{p}-2 \mathrm{p}=0\) \(\therefore \mathrm{f}(\mathrm{x})=0\), when \(\mathrm{x} \in \mathrm{I}\) Case - (ii) : When \(x\) is not integer Let, \(\mathrm{x}=\mathrm{p}+\mathrm{q}\), where \(\mathrm{p}\) is integer and \(\mathrm{q}\) is decimal part. And, \(0 \leq \mathrm{q}\lt 1\) Then, \([\mathrm{x}]=[\mathrm{p}+\mathrm{q}]=\mathrm{p}+[\mathrm{q}]\) Now, \(\mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[2 .(\mathrm{p}+\mathrm{q})]-2[\mathrm{p}+\mathrm{q}]\) \(=[2 \mathrm{p}+2 \mathrm{q}]-2(\mathrm{p}+[\mathrm{q}])\) \(=2 \mathrm{p}+[2 \mathrm{q}]-2 \mathrm{p}-2[\mathrm{q}]\) \(=[2 \mathrm{q}]-2[\mathrm{q}]\) \(=0-2 \times 0\) \(=0\), if \(0 \leq \mathrm{q} \leq 0.5\) or \(0 \leq 2 \mathrm{q}\lt 1\) \(=1-2 \times 0=1\), if \(0.5 \leq \mathrm{q}\lt 1\) or \(1 \leq 2 \mathrm{q}\lt 2\) \(\therefore \mathrm{f}(\mathrm{x})=0\) or 1 if \(\mathrm{x}\) is not integer From (i) and (ii) : for \(x \in R\), Range \(=\{0,1\}\). \(\therefore\) Range of \(\mathrm{f}=\{0,1\}\)
J and K CET-2006
Sets, Relation and Function
117432
The period of the function \(f(x)=\sin ^4 x+\cos ^4 x\) is
1 \(\pi\)
2 \(\frac{\pi}{2}\)
3 \(2 \pi\)
4 None of these
Explanation:
B Given that, \(\mathrm{f}(\mathrm{x})=\sin ^4 \mathrm{x}+\cos ^4 \mathrm{x}\) \(\mathrm{f}(\mathrm{x})=\left(\sin ^2 \mathrm{x}+\cos ^2 \mathrm{x}\right)^2-2 \sin ^2 \mathrm{x} \cos ^2 \mathrm{x}\) \(\mathrm{f}(\mathrm{x})=1-\frac{1}{2}(\sin 2 \mathrm{x})^2\) \(\mathrm{f}(\mathrm{x})=1-\frac{1}{2}\left(\frac{1-\cos 4 \mathrm{x}}{2}\right)\) \(\mathrm{f}(\mathrm{x})=\frac{3}{4}-\frac{1}{4} \cos 4 \mathrm{x}\) Ans: b Exp: B Given that, Therefore, here \(\cos 4 \mathrm{x}\) is a periodic function with period \(\frac{2 \pi}{4}=\frac{\pi}{2}\) Hence, the period of \(\mathrm{f}(\mathrm{x})\) is \(\frac{\pi}{2}\)
AIEEE-2002
Sets, Relation and Function
117433
Domain of definition of the function \(f(x)=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right)\), is
1 \((1,2)\)
2 \((-1,0) \cup(1,2)\)
3 \((1,2) \cup(2, \infty)\)
4 \((-1,0) \cup(1,2) \cup(2, \infty)\)
Explanation:
D Given that, \(f(x)=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right)\) Define the function \(\mathrm{f}(\mathrm{x})\) is, \(4-x^2 \neq 0\) \(x^2 \neq 4\) \(x \neq \pm 2\) And, \(\mathrm{x}^3-\mathrm{x}>0\) \(\mathrm{x}\left(\mathrm{x}^2-1\right)>0\) \(\mathrm{x}(\mathrm{x}+1)(\mathrm{x}-1)>0\) \(\mathrm{x} \in(-1,0) \cup(1, \infty)\) From equation (i) and (ii) we get \(\mathrm{x} \in(-1,0) \cup(1,2) \cup(2, \infty)\)Hence, the domain of \(\mathrm{f}(\mathrm{x})\) is \((-1,0) \cup(1,2) \cup(2, \infty)\)
117428
The domain of the function \(f(x)=\frac{1}{\log _{10}(1-x)}\) is
1 \((-\infty, 1]-\{0\}\)
2 \((-\infty, 1)-\{0\}\)
3 \((-\infty, 2)\)
4 \((0, \infty)\)
Explanation:
B Given that, \(f(x)=\frac{1}{\log _{10}(1-x)}\) For domain, \(1-\mathrm{x}>0\) \(\mathrm{x}\lt 1\) And, \(\log _{10}(1-\mathrm{x}) \neq 0\) From equation (i) and (ii), we get- Domain of \(f(x)=(-\infty, 1)-\{0\}\).
J and K CET-2010
Sets, Relation and Function
117429
For any real number \(y\) the greatest integer not exceeding \(y\) is denoted by [y]. If \(f: F \rightarrow R\) is denoted by \(f(x)=[2 x]-2[x]\) for \(x \in R\), then the range of \(f\) is:
D Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) and \(\mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) We take two cases to find range. Case - (i) : When \(x\) is integer then \([x]=x\) Let, \(\mathrm{x}=\mathrm{p}\) is integer Then, \(\mathrm{f}(\mathrm{x}) =[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[2 \mathrm{p}]-2[\mathrm{p}]\) \(=2 \mathrm{p}-2 \mathrm{p}=0\) \(\therefore \mathrm{f}(\mathrm{x})=0\), when \(\mathrm{x} \in \mathrm{I}\) Case - (ii) : When \(x\) is not integer Let, \(\mathrm{x}=\mathrm{p}+\mathrm{q}\), where \(\mathrm{p}\) is integer and \(\mathrm{q}\) is decimal part. And, \(0 \leq \mathrm{q}\lt 1\) Then, \([\mathrm{x}]=[\mathrm{p}+\mathrm{q}]=\mathrm{p}+[\mathrm{q}]\) Now, \(\mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[2 .(\mathrm{p}+\mathrm{q})]-2[\mathrm{p}+\mathrm{q}]\) \(=[2 \mathrm{p}+2 \mathrm{q}]-2(\mathrm{p}+[\mathrm{q}])\) \(=2 \mathrm{p}+[2 \mathrm{q}]-2 \mathrm{p}-2[\mathrm{q}]\) \(=[2 \mathrm{q}]-2[\mathrm{q}]\) \(=0-2 \times 0\) \(=0\), if \(0 \leq \mathrm{q} \leq 0.5\) or \(0 \leq 2 \mathrm{q}\lt 1\) \(=1-2 \times 0=1\), if \(0.5 \leq \mathrm{q}\lt 1\) or \(1 \leq 2 \mathrm{q}\lt 2\) \(\therefore \mathrm{f}(\mathrm{x})=0\) or 1 if \(\mathrm{x}\) is not integer From (i) and (ii) : for \(x \in R\), Range \(=\{0,1\}\). \(\therefore\) Range of \(\mathrm{f}=\{0,1\}\)
J and K CET-2006
Sets, Relation and Function
117432
The period of the function \(f(x)=\sin ^4 x+\cos ^4 x\) is
1 \(\pi\)
2 \(\frac{\pi}{2}\)
3 \(2 \pi\)
4 None of these
Explanation:
B Given that, \(\mathrm{f}(\mathrm{x})=\sin ^4 \mathrm{x}+\cos ^4 \mathrm{x}\) \(\mathrm{f}(\mathrm{x})=\left(\sin ^2 \mathrm{x}+\cos ^2 \mathrm{x}\right)^2-2 \sin ^2 \mathrm{x} \cos ^2 \mathrm{x}\) \(\mathrm{f}(\mathrm{x})=1-\frac{1}{2}(\sin 2 \mathrm{x})^2\) \(\mathrm{f}(\mathrm{x})=1-\frac{1}{2}\left(\frac{1-\cos 4 \mathrm{x}}{2}\right)\) \(\mathrm{f}(\mathrm{x})=\frac{3}{4}-\frac{1}{4} \cos 4 \mathrm{x}\) Ans: b Exp: B Given that, Therefore, here \(\cos 4 \mathrm{x}\) is a periodic function with period \(\frac{2 \pi}{4}=\frac{\pi}{2}\) Hence, the period of \(\mathrm{f}(\mathrm{x})\) is \(\frac{\pi}{2}\)
AIEEE-2002
Sets, Relation and Function
117433
Domain of definition of the function \(f(x)=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right)\), is
1 \((1,2)\)
2 \((-1,0) \cup(1,2)\)
3 \((1,2) \cup(2, \infty)\)
4 \((-1,0) \cup(1,2) \cup(2, \infty)\)
Explanation:
D Given that, \(f(x)=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right)\) Define the function \(\mathrm{f}(\mathrm{x})\) is, \(4-x^2 \neq 0\) \(x^2 \neq 4\) \(x \neq \pm 2\) And, \(\mathrm{x}^3-\mathrm{x}>0\) \(\mathrm{x}\left(\mathrm{x}^2-1\right)>0\) \(\mathrm{x}(\mathrm{x}+1)(\mathrm{x}-1)>0\) \(\mathrm{x} \in(-1,0) \cup(1, \infty)\) From equation (i) and (ii) we get \(\mathrm{x} \in(-1,0) \cup(1,2) \cup(2, \infty)\)Hence, the domain of \(\mathrm{f}(\mathrm{x})\) is \((-1,0) \cup(1,2) \cup(2, \infty)\)
117428
The domain of the function \(f(x)=\frac{1}{\log _{10}(1-x)}\) is
1 \((-\infty, 1]-\{0\}\)
2 \((-\infty, 1)-\{0\}\)
3 \((-\infty, 2)\)
4 \((0, \infty)\)
Explanation:
B Given that, \(f(x)=\frac{1}{\log _{10}(1-x)}\) For domain, \(1-\mathrm{x}>0\) \(\mathrm{x}\lt 1\) And, \(\log _{10}(1-\mathrm{x}) \neq 0\) From equation (i) and (ii), we get- Domain of \(f(x)=(-\infty, 1)-\{0\}\).
J and K CET-2010
Sets, Relation and Function
117429
For any real number \(y\) the greatest integer not exceeding \(y\) is denoted by [y]. If \(f: F \rightarrow R\) is denoted by \(f(x)=[2 x]-2[x]\) for \(x \in R\), then the range of \(f\) is:
D Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) and \(\mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) We take two cases to find range. Case - (i) : When \(x\) is integer then \([x]=x\) Let, \(\mathrm{x}=\mathrm{p}\) is integer Then, \(\mathrm{f}(\mathrm{x}) =[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[2 \mathrm{p}]-2[\mathrm{p}]\) \(=2 \mathrm{p}-2 \mathrm{p}=0\) \(\therefore \mathrm{f}(\mathrm{x})=0\), when \(\mathrm{x} \in \mathrm{I}\) Case - (ii) : When \(x\) is not integer Let, \(\mathrm{x}=\mathrm{p}+\mathrm{q}\), where \(\mathrm{p}\) is integer and \(\mathrm{q}\) is decimal part. And, \(0 \leq \mathrm{q}\lt 1\) Then, \([\mathrm{x}]=[\mathrm{p}+\mathrm{q}]=\mathrm{p}+[\mathrm{q}]\) Now, \(\mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[2 .(\mathrm{p}+\mathrm{q})]-2[\mathrm{p}+\mathrm{q}]\) \(=[2 \mathrm{p}+2 \mathrm{q}]-2(\mathrm{p}+[\mathrm{q}])\) \(=2 \mathrm{p}+[2 \mathrm{q}]-2 \mathrm{p}-2[\mathrm{q}]\) \(=[2 \mathrm{q}]-2[\mathrm{q}]\) \(=0-2 \times 0\) \(=0\), if \(0 \leq \mathrm{q} \leq 0.5\) or \(0 \leq 2 \mathrm{q}\lt 1\) \(=1-2 \times 0=1\), if \(0.5 \leq \mathrm{q}\lt 1\) or \(1 \leq 2 \mathrm{q}\lt 2\) \(\therefore \mathrm{f}(\mathrm{x})=0\) or 1 if \(\mathrm{x}\) is not integer From (i) and (ii) : for \(x \in R\), Range \(=\{0,1\}\). \(\therefore\) Range of \(\mathrm{f}=\{0,1\}\)
J and K CET-2006
Sets, Relation and Function
117432
The period of the function \(f(x)=\sin ^4 x+\cos ^4 x\) is
1 \(\pi\)
2 \(\frac{\pi}{2}\)
3 \(2 \pi\)
4 None of these
Explanation:
B Given that, \(\mathrm{f}(\mathrm{x})=\sin ^4 \mathrm{x}+\cos ^4 \mathrm{x}\) \(\mathrm{f}(\mathrm{x})=\left(\sin ^2 \mathrm{x}+\cos ^2 \mathrm{x}\right)^2-2 \sin ^2 \mathrm{x} \cos ^2 \mathrm{x}\) \(\mathrm{f}(\mathrm{x})=1-\frac{1}{2}(\sin 2 \mathrm{x})^2\) \(\mathrm{f}(\mathrm{x})=1-\frac{1}{2}\left(\frac{1-\cos 4 \mathrm{x}}{2}\right)\) \(\mathrm{f}(\mathrm{x})=\frac{3}{4}-\frac{1}{4} \cos 4 \mathrm{x}\) Ans: b Exp: B Given that, Therefore, here \(\cos 4 \mathrm{x}\) is a periodic function with period \(\frac{2 \pi}{4}=\frac{\pi}{2}\) Hence, the period of \(\mathrm{f}(\mathrm{x})\) is \(\frac{\pi}{2}\)
AIEEE-2002
Sets, Relation and Function
117433
Domain of definition of the function \(f(x)=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right)\), is
1 \((1,2)\)
2 \((-1,0) \cup(1,2)\)
3 \((1,2) \cup(2, \infty)\)
4 \((-1,0) \cup(1,2) \cup(2, \infty)\)
Explanation:
D Given that, \(f(x)=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right)\) Define the function \(\mathrm{f}(\mathrm{x})\) is, \(4-x^2 \neq 0\) \(x^2 \neq 4\) \(x \neq \pm 2\) And, \(\mathrm{x}^3-\mathrm{x}>0\) \(\mathrm{x}\left(\mathrm{x}^2-1\right)>0\) \(\mathrm{x}(\mathrm{x}+1)(\mathrm{x}-1)>0\) \(\mathrm{x} \in(-1,0) \cup(1, \infty)\) From equation (i) and (ii) we get \(\mathrm{x} \in(-1,0) \cup(1,2) \cup(2, \infty)\)Hence, the domain of \(\mathrm{f}(\mathrm{x})\) is \((-1,0) \cup(1,2) \cup(2, \infty)\)
