117191
In \(Z\), the set of all integers, the inverse of -7 with respect to * defined by \(a * b=a+b+7\) for all \(a, b \in Z\) is
1 -14
2 7
3 14
4 -7
Explanation:
D Given, \([\mathrm{Z}, *]\) is a group Where \(*\) is defined by \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}+7\) \(\Rightarrow \mathrm{a}+\mathrm{e}+7=\mathrm{a} \Rightarrow \mathrm{e}+7=0\) \(\Rightarrow \mathrm{e}+7=0 \Rightarrow \mathrm{e}=-7\) \(\therefore-7\) is the identity element. So, identity element has its own inverse. \(\therefore(-7)^{-1}=-7\)
COMEDK 2011
Sets, Relation and Function
117193
If \(f:[1, \infty) \rightarrow[2, \infty)\) is given by \(f(x)=x+\frac{1}{x}\), then \(f^{-1}(x)\) is equal to
1 \(\frac{x+\sqrt{x^2-4}}{2}\)
2 \(\frac{x}{1+x^2}\)
3 \(\frac{x-\sqrt{x^2-4}}{2}\)
4 \(1+\sqrt{\mathrm{x}-4}\)
Explanation:
A Given, \(\mathrm{f}:[1, \infty) \rightarrow[2, \infty)\) is given by - \(f(x)=x+\frac{1}{x}\) Let \(\mathrm{f}(\mathrm{x})=\mathrm{y} \text {, }\) Then, \(y=x+\frac{1}{x}\) \(x y=x^2+1\) \(x^2-x y+1=0\) Then, \(x=\frac{y \pm \sqrt{y^2-4 \times 1 \times 1}}{2 \times 1}\) \(x=\frac{y \pm \sqrt{y^2-4}}{2}\) Since, given \(f\) is positive from \(f:[1, \infty) \rightarrow[2, \infty)\) So, \(\mathrm{x}=\frac{\mathrm{y}+\sqrt{\mathrm{y}^2-4}}{2}=\mathrm{f}^{-1}(\mathrm{y})\)Hence, \(f^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}\)
BCECE-2010
Sets, Relation and Function
117194
Let \(f: R \rightarrow R\) be defined by \(f(x)=2 x+6\) which is a bijective mapping then \(\mathrm{f}^{-1}(x)\) is given by
1 \(\frac{x}{2}-3\)
2 \(2 x+6\)
3 \(\mathrm{x}-3\)
4 \(6 x+2\)
Explanation:
A Given, \(f: R \rightarrow R\) be defined by \(f(x)=2 x+6\) is a bijective mapping mean it is a one-one and onto map. So, let \(\mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=2 x+6\) \(y-6=2 x\) \(2 x=y-6\) \(x=\frac{y-6}{2}\) \(x=\frac{y}{2}-\frac{6}{2}\) \(x=\frac{y}{2}-3=f^{-1}(y)\)Hence, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}}{2}-3\).
117191
In \(Z\), the set of all integers, the inverse of -7 with respect to * defined by \(a * b=a+b+7\) for all \(a, b \in Z\) is
1 -14
2 7
3 14
4 -7
Explanation:
D Given, \([\mathrm{Z}, *]\) is a group Where \(*\) is defined by \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}+7\) \(\Rightarrow \mathrm{a}+\mathrm{e}+7=\mathrm{a} \Rightarrow \mathrm{e}+7=0\) \(\Rightarrow \mathrm{e}+7=0 \Rightarrow \mathrm{e}=-7\) \(\therefore-7\) is the identity element. So, identity element has its own inverse. \(\therefore(-7)^{-1}=-7\)
COMEDK 2011
Sets, Relation and Function
117193
If \(f:[1, \infty) \rightarrow[2, \infty)\) is given by \(f(x)=x+\frac{1}{x}\), then \(f^{-1}(x)\) is equal to
1 \(\frac{x+\sqrt{x^2-4}}{2}\)
2 \(\frac{x}{1+x^2}\)
3 \(\frac{x-\sqrt{x^2-4}}{2}\)
4 \(1+\sqrt{\mathrm{x}-4}\)
Explanation:
A Given, \(\mathrm{f}:[1, \infty) \rightarrow[2, \infty)\) is given by - \(f(x)=x+\frac{1}{x}\) Let \(\mathrm{f}(\mathrm{x})=\mathrm{y} \text {, }\) Then, \(y=x+\frac{1}{x}\) \(x y=x^2+1\) \(x^2-x y+1=0\) Then, \(x=\frac{y \pm \sqrt{y^2-4 \times 1 \times 1}}{2 \times 1}\) \(x=\frac{y \pm \sqrt{y^2-4}}{2}\) Since, given \(f\) is positive from \(f:[1, \infty) \rightarrow[2, \infty)\) So, \(\mathrm{x}=\frac{\mathrm{y}+\sqrt{\mathrm{y}^2-4}}{2}=\mathrm{f}^{-1}(\mathrm{y})\)Hence, \(f^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}\)
BCECE-2010
Sets, Relation and Function
117194
Let \(f: R \rightarrow R\) be defined by \(f(x)=2 x+6\) which is a bijective mapping then \(\mathrm{f}^{-1}(x)\) is given by
1 \(\frac{x}{2}-3\)
2 \(2 x+6\)
3 \(\mathrm{x}-3\)
4 \(6 x+2\)
Explanation:
A Given, \(f: R \rightarrow R\) be defined by \(f(x)=2 x+6\) is a bijective mapping mean it is a one-one and onto map. So, let \(\mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=2 x+6\) \(y-6=2 x\) \(2 x=y-6\) \(x=\frac{y-6}{2}\) \(x=\frac{y}{2}-\frac{6}{2}\) \(x=\frac{y}{2}-3=f^{-1}(y)\)Hence, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}}{2}-3\).
117191
In \(Z\), the set of all integers, the inverse of -7 with respect to * defined by \(a * b=a+b+7\) for all \(a, b \in Z\) is
1 -14
2 7
3 14
4 -7
Explanation:
D Given, \([\mathrm{Z}, *]\) is a group Where \(*\) is defined by \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}+7\) \(\Rightarrow \mathrm{a}+\mathrm{e}+7=\mathrm{a} \Rightarrow \mathrm{e}+7=0\) \(\Rightarrow \mathrm{e}+7=0 \Rightarrow \mathrm{e}=-7\) \(\therefore-7\) is the identity element. So, identity element has its own inverse. \(\therefore(-7)^{-1}=-7\)
COMEDK 2011
Sets, Relation and Function
117193
If \(f:[1, \infty) \rightarrow[2, \infty)\) is given by \(f(x)=x+\frac{1}{x}\), then \(f^{-1}(x)\) is equal to
1 \(\frac{x+\sqrt{x^2-4}}{2}\)
2 \(\frac{x}{1+x^2}\)
3 \(\frac{x-\sqrt{x^2-4}}{2}\)
4 \(1+\sqrt{\mathrm{x}-4}\)
Explanation:
A Given, \(\mathrm{f}:[1, \infty) \rightarrow[2, \infty)\) is given by - \(f(x)=x+\frac{1}{x}\) Let \(\mathrm{f}(\mathrm{x})=\mathrm{y} \text {, }\) Then, \(y=x+\frac{1}{x}\) \(x y=x^2+1\) \(x^2-x y+1=0\) Then, \(x=\frac{y \pm \sqrt{y^2-4 \times 1 \times 1}}{2 \times 1}\) \(x=\frac{y \pm \sqrt{y^2-4}}{2}\) Since, given \(f\) is positive from \(f:[1, \infty) \rightarrow[2, \infty)\) So, \(\mathrm{x}=\frac{\mathrm{y}+\sqrt{\mathrm{y}^2-4}}{2}=\mathrm{f}^{-1}(\mathrm{y})\)Hence, \(f^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}\)
BCECE-2010
Sets, Relation and Function
117194
Let \(f: R \rightarrow R\) be defined by \(f(x)=2 x+6\) which is a bijective mapping then \(\mathrm{f}^{-1}(x)\) is given by
1 \(\frac{x}{2}-3\)
2 \(2 x+6\)
3 \(\mathrm{x}-3\)
4 \(6 x+2\)
Explanation:
A Given, \(f: R \rightarrow R\) be defined by \(f(x)=2 x+6\) is a bijective mapping mean it is a one-one and onto map. So, let \(\mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=2 x+6\) \(y-6=2 x\) \(2 x=y-6\) \(x=\frac{y-6}{2}\) \(x=\frac{y}{2}-\frac{6}{2}\) \(x=\frac{y}{2}-3=f^{-1}(y)\)Hence, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}}{2}-3\).
117191
In \(Z\), the set of all integers, the inverse of -7 with respect to * defined by \(a * b=a+b+7\) for all \(a, b \in Z\) is
1 -14
2 7
3 14
4 -7
Explanation:
D Given, \([\mathrm{Z}, *]\) is a group Where \(*\) is defined by \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}+7\) \(\Rightarrow \mathrm{a}+\mathrm{e}+7=\mathrm{a} \Rightarrow \mathrm{e}+7=0\) \(\Rightarrow \mathrm{e}+7=0 \Rightarrow \mathrm{e}=-7\) \(\therefore-7\) is the identity element. So, identity element has its own inverse. \(\therefore(-7)^{-1}=-7\)
COMEDK 2011
Sets, Relation and Function
117193
If \(f:[1, \infty) \rightarrow[2, \infty)\) is given by \(f(x)=x+\frac{1}{x}\), then \(f^{-1}(x)\) is equal to
1 \(\frac{x+\sqrt{x^2-4}}{2}\)
2 \(\frac{x}{1+x^2}\)
3 \(\frac{x-\sqrt{x^2-4}}{2}\)
4 \(1+\sqrt{\mathrm{x}-4}\)
Explanation:
A Given, \(\mathrm{f}:[1, \infty) \rightarrow[2, \infty)\) is given by - \(f(x)=x+\frac{1}{x}\) Let \(\mathrm{f}(\mathrm{x})=\mathrm{y} \text {, }\) Then, \(y=x+\frac{1}{x}\) \(x y=x^2+1\) \(x^2-x y+1=0\) Then, \(x=\frac{y \pm \sqrt{y^2-4 \times 1 \times 1}}{2 \times 1}\) \(x=\frac{y \pm \sqrt{y^2-4}}{2}\) Since, given \(f\) is positive from \(f:[1, \infty) \rightarrow[2, \infty)\) So, \(\mathrm{x}=\frac{\mathrm{y}+\sqrt{\mathrm{y}^2-4}}{2}=\mathrm{f}^{-1}(\mathrm{y})\)Hence, \(f^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}\)
BCECE-2010
Sets, Relation and Function
117194
Let \(f: R \rightarrow R\) be defined by \(f(x)=2 x+6\) which is a bijective mapping then \(\mathrm{f}^{-1}(x)\) is given by
1 \(\frac{x}{2}-3\)
2 \(2 x+6\)
3 \(\mathrm{x}-3\)
4 \(6 x+2\)
Explanation:
A Given, \(f: R \rightarrow R\) be defined by \(f(x)=2 x+6\) is a bijective mapping mean it is a one-one and onto map. So, let \(\mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=2 x+6\) \(y-6=2 x\) \(2 x=y-6\) \(x=\frac{y-6}{2}\) \(x=\frac{y}{2}-\frac{6}{2}\) \(x=\frac{y}{2}-3=f^{-1}(y)\)Hence, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}}{2}-3\).
117191
In \(Z\), the set of all integers, the inverse of -7 with respect to * defined by \(a * b=a+b+7\) for all \(a, b \in Z\) is
1 -14
2 7
3 14
4 -7
Explanation:
D Given, \([\mathrm{Z}, *]\) is a group Where \(*\) is defined by \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}+7\) \(\Rightarrow \mathrm{a}+\mathrm{e}+7=\mathrm{a} \Rightarrow \mathrm{e}+7=0\) \(\Rightarrow \mathrm{e}+7=0 \Rightarrow \mathrm{e}=-7\) \(\therefore-7\) is the identity element. So, identity element has its own inverse. \(\therefore(-7)^{-1}=-7\)
COMEDK 2011
Sets, Relation and Function
117193
If \(f:[1, \infty) \rightarrow[2, \infty)\) is given by \(f(x)=x+\frac{1}{x}\), then \(f^{-1}(x)\) is equal to
1 \(\frac{x+\sqrt{x^2-4}}{2}\)
2 \(\frac{x}{1+x^2}\)
3 \(\frac{x-\sqrt{x^2-4}}{2}\)
4 \(1+\sqrt{\mathrm{x}-4}\)
Explanation:
A Given, \(\mathrm{f}:[1, \infty) \rightarrow[2, \infty)\) is given by - \(f(x)=x+\frac{1}{x}\) Let \(\mathrm{f}(\mathrm{x})=\mathrm{y} \text {, }\) Then, \(y=x+\frac{1}{x}\) \(x y=x^2+1\) \(x^2-x y+1=0\) Then, \(x=\frac{y \pm \sqrt{y^2-4 \times 1 \times 1}}{2 \times 1}\) \(x=\frac{y \pm \sqrt{y^2-4}}{2}\) Since, given \(f\) is positive from \(f:[1, \infty) \rightarrow[2, \infty)\) So, \(\mathrm{x}=\frac{\mathrm{y}+\sqrt{\mathrm{y}^2-4}}{2}=\mathrm{f}^{-1}(\mathrm{y})\)Hence, \(f^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}\)
BCECE-2010
Sets, Relation and Function
117194
Let \(f: R \rightarrow R\) be defined by \(f(x)=2 x+6\) which is a bijective mapping then \(\mathrm{f}^{-1}(x)\) is given by
1 \(\frac{x}{2}-3\)
2 \(2 x+6\)
3 \(\mathrm{x}-3\)
4 \(6 x+2\)
Explanation:
A Given, \(f: R \rightarrow R\) be defined by \(f(x)=2 x+6\) is a bijective mapping mean it is a one-one and onto map. So, let \(\mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=2 x+6\) \(y-6=2 x\) \(2 x=y-6\) \(x=\frac{y-6}{2}\) \(x=\frac{y}{2}-\frac{6}{2}\) \(x=\frac{y}{2}-3=f^{-1}(y)\)Hence, \(\mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}}{2}-3\).
