NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117137
If \(f: R \rightarrow R\) is given by \(f(x)=x^3+3\), then \(f^{-1}(x)\) is equal to
1 \(x^{1 / 3}-3\)
2 \(\mathrm{x}^{1 / 3}+3\)
3 \((\mathrm{x}-3)\)
4 \((x-3)^{1 / 3}\)
Explanation:
Exp: (D) : Given, Function is \(f: R \rightarrow R\) is given by \(f(x)=x^3+3\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) \(y =x^3+3\) \(y -3=x^3\) \(x =\sqrt[3]{y-3}\) \(f^{-1}(y) =(y-3)^{1 / 3}\) \(\therefore f^{-1}(x) =(x-3)^{1 / 3}\)
[SRMJEEE-2009]
Sets, Relation and Function
117138
\(R\) is a relation from \(\{11,12,13\}\) to \(\{8,10,12\}\) defined by \(y=x-3\). Then \(R^{-1}\) is
1 \(\{(8,11),(10,13)\}\)
2 \(\{11,18),(13,10)\}\)
3 \(\{10,13),(8,11)\}\)
4 None of the above
Explanation:
Exp: (A) : Since \(\mathrm{R}\) is a relation from \(\{11,12,13\}\) to \(\{8\), \(10,12\}\) Defined by \(y=x-3\) if \(\mathrm{x}=11 \rightarrow \mathrm{y}=11-3=8\) if \(x=12 \rightarrow y=12-3=9\) if \(x=13 \rightarrow y=13-3=10\) Then from the relation \(\mathrm{R}\) the ordered pairs obtain are \(\therefore \mathrm{R}=\{(11,8),(13,10)\}\) Neglect the pairs \((12,9)\) as it does not belong to the given set . \(\Rightarrow \mathrm{R}^{-1}=\{(8,11),(10,13)\}\)
Jamia Millia Islamia-2008]
Sets, Relation and Function
117146
Inverse of function \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) is
Exp: (B) : Given, \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) Let, \(\mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) \(y \times 10^x+y \times 10^{-x}=10^x-10^{-x}\) \(y \times 10^x-10^x=-10^{-x}-y \times 10^{-x}\) \(10^x(y-1)=-10^{-x}(1+y)\) \(10^x(1-y)=10^{-x}(1+y)\) \(\frac{10^x}{10^{-x}}=\frac{1+y}{1-y}\) \(10^{2 x}=\frac{1+y}{1-y}\) Taking both side \(\log _{10}\), we get - \(\log _{10} 10^{2 \mathrm{x}}=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(2 \mathrm{x} \times \log _{10} 10=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(2 \mathrm{x}=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(\mathrm{x}=\frac{1}{2} \log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) So, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{1}{2} \log \frac{1+\mathrm{x}}{1-\mathrm{x}}\) Hence, inverse of function \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) is \(\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right) .\)
[UPSEE-2010
Sets, Relation and Function
117147
Let \(S\) be a finite set containing \(n\) elements. Then the total number of binary operations on \(S\) is
1 \(\mathrm{n}^{\mathrm{n}}\)
2 \(2^{\mathrm{n}^2}\)
3 \(\mathrm{n}^{\mathrm{n}^2}\)
4 \(\mathrm{n}^2\)
Explanation:
Exp: (C) : Given that, \(\mathrm{S}\) be a finite set containing \(\mathrm{n}\) element. Since, commutative binary operation- \(\mathrm{S} \times \mathrm{S} \rightarrow \mathrm{S}\) Therefore, \(|S|^{|S| \times|S|}=|S|^{S^2}\) Then, the number of binary operations on \(\mathrm{S}\) is \(=\mathrm{n}^{\mathrm{n}^2}\)
117137
If \(f: R \rightarrow R\) is given by \(f(x)=x^3+3\), then \(f^{-1}(x)\) is equal to
1 \(x^{1 / 3}-3\)
2 \(\mathrm{x}^{1 / 3}+3\)
3 \((\mathrm{x}-3)\)
4 \((x-3)^{1 / 3}\)
Explanation:
Exp: (D) : Given, Function is \(f: R \rightarrow R\) is given by \(f(x)=x^3+3\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) \(y =x^3+3\) \(y -3=x^3\) \(x =\sqrt[3]{y-3}\) \(f^{-1}(y) =(y-3)^{1 / 3}\) \(\therefore f^{-1}(x) =(x-3)^{1 / 3}\)
[SRMJEEE-2009]
Sets, Relation and Function
117138
\(R\) is a relation from \(\{11,12,13\}\) to \(\{8,10,12\}\) defined by \(y=x-3\). Then \(R^{-1}\) is
1 \(\{(8,11),(10,13)\}\)
2 \(\{11,18),(13,10)\}\)
3 \(\{10,13),(8,11)\}\)
4 None of the above
Explanation:
Exp: (A) : Since \(\mathrm{R}\) is a relation from \(\{11,12,13\}\) to \(\{8\), \(10,12\}\) Defined by \(y=x-3\) if \(\mathrm{x}=11 \rightarrow \mathrm{y}=11-3=8\) if \(x=12 \rightarrow y=12-3=9\) if \(x=13 \rightarrow y=13-3=10\) Then from the relation \(\mathrm{R}\) the ordered pairs obtain are \(\therefore \mathrm{R}=\{(11,8),(13,10)\}\) Neglect the pairs \((12,9)\) as it does not belong to the given set . \(\Rightarrow \mathrm{R}^{-1}=\{(8,11),(10,13)\}\)
Jamia Millia Islamia-2008]
Sets, Relation and Function
117146
Inverse of function \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) is
Exp: (B) : Given, \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) Let, \(\mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) \(y \times 10^x+y \times 10^{-x}=10^x-10^{-x}\) \(y \times 10^x-10^x=-10^{-x}-y \times 10^{-x}\) \(10^x(y-1)=-10^{-x}(1+y)\) \(10^x(1-y)=10^{-x}(1+y)\) \(\frac{10^x}{10^{-x}}=\frac{1+y}{1-y}\) \(10^{2 x}=\frac{1+y}{1-y}\) Taking both side \(\log _{10}\), we get - \(\log _{10} 10^{2 \mathrm{x}}=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(2 \mathrm{x} \times \log _{10} 10=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(2 \mathrm{x}=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(\mathrm{x}=\frac{1}{2} \log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) So, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{1}{2} \log \frac{1+\mathrm{x}}{1-\mathrm{x}}\) Hence, inverse of function \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) is \(\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right) .\)
[UPSEE-2010
Sets, Relation and Function
117147
Let \(S\) be a finite set containing \(n\) elements. Then the total number of binary operations on \(S\) is
1 \(\mathrm{n}^{\mathrm{n}}\)
2 \(2^{\mathrm{n}^2}\)
3 \(\mathrm{n}^{\mathrm{n}^2}\)
4 \(\mathrm{n}^2\)
Explanation:
Exp: (C) : Given that, \(\mathrm{S}\) be a finite set containing \(\mathrm{n}\) element. Since, commutative binary operation- \(\mathrm{S} \times \mathrm{S} \rightarrow \mathrm{S}\) Therefore, \(|S|^{|S| \times|S|}=|S|^{S^2}\) Then, the number of binary operations on \(\mathrm{S}\) is \(=\mathrm{n}^{\mathrm{n}^2}\)
117137
If \(f: R \rightarrow R\) is given by \(f(x)=x^3+3\), then \(f^{-1}(x)\) is equal to
1 \(x^{1 / 3}-3\)
2 \(\mathrm{x}^{1 / 3}+3\)
3 \((\mathrm{x}-3)\)
4 \((x-3)^{1 / 3}\)
Explanation:
Exp: (D) : Given, Function is \(f: R \rightarrow R\) is given by \(f(x)=x^3+3\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) \(y =x^3+3\) \(y -3=x^3\) \(x =\sqrt[3]{y-3}\) \(f^{-1}(y) =(y-3)^{1 / 3}\) \(\therefore f^{-1}(x) =(x-3)^{1 / 3}\)
[SRMJEEE-2009]
Sets, Relation and Function
117138
\(R\) is a relation from \(\{11,12,13\}\) to \(\{8,10,12\}\) defined by \(y=x-3\). Then \(R^{-1}\) is
1 \(\{(8,11),(10,13)\}\)
2 \(\{11,18),(13,10)\}\)
3 \(\{10,13),(8,11)\}\)
4 None of the above
Explanation:
Exp: (A) : Since \(\mathrm{R}\) is a relation from \(\{11,12,13\}\) to \(\{8\), \(10,12\}\) Defined by \(y=x-3\) if \(\mathrm{x}=11 \rightarrow \mathrm{y}=11-3=8\) if \(x=12 \rightarrow y=12-3=9\) if \(x=13 \rightarrow y=13-3=10\) Then from the relation \(\mathrm{R}\) the ordered pairs obtain are \(\therefore \mathrm{R}=\{(11,8),(13,10)\}\) Neglect the pairs \((12,9)\) as it does not belong to the given set . \(\Rightarrow \mathrm{R}^{-1}=\{(8,11),(10,13)\}\)
Jamia Millia Islamia-2008]
Sets, Relation and Function
117146
Inverse of function \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) is
Exp: (B) : Given, \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) Let, \(\mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) \(y \times 10^x+y \times 10^{-x}=10^x-10^{-x}\) \(y \times 10^x-10^x=-10^{-x}-y \times 10^{-x}\) \(10^x(y-1)=-10^{-x}(1+y)\) \(10^x(1-y)=10^{-x}(1+y)\) \(\frac{10^x}{10^{-x}}=\frac{1+y}{1-y}\) \(10^{2 x}=\frac{1+y}{1-y}\) Taking both side \(\log _{10}\), we get - \(\log _{10} 10^{2 \mathrm{x}}=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(2 \mathrm{x} \times \log _{10} 10=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(2 \mathrm{x}=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(\mathrm{x}=\frac{1}{2} \log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) So, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{1}{2} \log \frac{1+\mathrm{x}}{1-\mathrm{x}}\) Hence, inverse of function \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) is \(\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right) .\)
[UPSEE-2010
Sets, Relation and Function
117147
Let \(S\) be a finite set containing \(n\) elements. Then the total number of binary operations on \(S\) is
1 \(\mathrm{n}^{\mathrm{n}}\)
2 \(2^{\mathrm{n}^2}\)
3 \(\mathrm{n}^{\mathrm{n}^2}\)
4 \(\mathrm{n}^2\)
Explanation:
Exp: (C) : Given that, \(\mathrm{S}\) be a finite set containing \(\mathrm{n}\) element. Since, commutative binary operation- \(\mathrm{S} \times \mathrm{S} \rightarrow \mathrm{S}\) Therefore, \(|S|^{|S| \times|S|}=|S|^{S^2}\) Then, the number of binary operations on \(\mathrm{S}\) is \(=\mathrm{n}^{\mathrm{n}^2}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117137
If \(f: R \rightarrow R\) is given by \(f(x)=x^3+3\), then \(f^{-1}(x)\) is equal to
1 \(x^{1 / 3}-3\)
2 \(\mathrm{x}^{1 / 3}+3\)
3 \((\mathrm{x}-3)\)
4 \((x-3)^{1 / 3}\)
Explanation:
Exp: (D) : Given, Function is \(f: R \rightarrow R\) is given by \(f(x)=x^3+3\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) \(y =x^3+3\) \(y -3=x^3\) \(x =\sqrt[3]{y-3}\) \(f^{-1}(y) =(y-3)^{1 / 3}\) \(\therefore f^{-1}(x) =(x-3)^{1 / 3}\)
[SRMJEEE-2009]
Sets, Relation and Function
117138
\(R\) is a relation from \(\{11,12,13\}\) to \(\{8,10,12\}\) defined by \(y=x-3\). Then \(R^{-1}\) is
1 \(\{(8,11),(10,13)\}\)
2 \(\{11,18),(13,10)\}\)
3 \(\{10,13),(8,11)\}\)
4 None of the above
Explanation:
Exp: (A) : Since \(\mathrm{R}\) is a relation from \(\{11,12,13\}\) to \(\{8\), \(10,12\}\) Defined by \(y=x-3\) if \(\mathrm{x}=11 \rightarrow \mathrm{y}=11-3=8\) if \(x=12 \rightarrow y=12-3=9\) if \(x=13 \rightarrow y=13-3=10\) Then from the relation \(\mathrm{R}\) the ordered pairs obtain are \(\therefore \mathrm{R}=\{(11,8),(13,10)\}\) Neglect the pairs \((12,9)\) as it does not belong to the given set . \(\Rightarrow \mathrm{R}^{-1}=\{(8,11),(10,13)\}\)
Jamia Millia Islamia-2008]
Sets, Relation and Function
117146
Inverse of function \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) is
Exp: (B) : Given, \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) Let, \(\mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) \(y \times 10^x+y \times 10^{-x}=10^x-10^{-x}\) \(y \times 10^x-10^x=-10^{-x}-y \times 10^{-x}\) \(10^x(y-1)=-10^{-x}(1+y)\) \(10^x(1-y)=10^{-x}(1+y)\) \(\frac{10^x}{10^{-x}}=\frac{1+y}{1-y}\) \(10^{2 x}=\frac{1+y}{1-y}\) Taking both side \(\log _{10}\), we get - \(\log _{10} 10^{2 \mathrm{x}}=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(2 \mathrm{x} \times \log _{10} 10=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(2 \mathrm{x}=\log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) \(\mathrm{x}=\frac{1}{2} \log \frac{1+\mathrm{y}}{1-\mathrm{y}}\) So, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{1}{2} \log \frac{1+\mathrm{x}}{1-\mathrm{x}}\) Hence, inverse of function \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) is \(\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right) .\)
[UPSEE-2010
Sets, Relation and Function
117147
Let \(S\) be a finite set containing \(n\) elements. Then the total number of binary operations on \(S\) is
1 \(\mathrm{n}^{\mathrm{n}}\)
2 \(2^{\mathrm{n}^2}\)
3 \(\mathrm{n}^{\mathrm{n}^2}\)
4 \(\mathrm{n}^2\)
Explanation:
Exp: (C) : Given that, \(\mathrm{S}\) be a finite set containing \(\mathrm{n}\) element. Since, commutative binary operation- \(\mathrm{S} \times \mathrm{S} \rightarrow \mathrm{S}\) Therefore, \(|S|^{|S| \times|S|}=|S|^{S^2}\) Then, the number of binary operations on \(\mathrm{S}\) is \(=\mathrm{n}^{\mathrm{n}^2}\)