117158
Let \(f: R \rightarrow R\) be the function defined by \(f(x)=\sqrt{\mathbf{x}-3}, \forall \mathbf{x} \in \mathbf{R}\) then, \(\boldsymbol{f}^{-1}(\mathbf{x})=\)
1 \(x+3\)
2 \(x^2+3\)
3 \(\frac{x+3}{2}\)
4 \(\frac{x^2+3}{2}\)
Explanation:
Exp: (B) : Given, \(f(x)=\sqrt{x-3} \forall x \in R\) Let \(f(x)=y\) \(y=\sqrt{x-3}\) \(y^2=x-3\) \(x=y^2+3=f^{-1}(y)\)Hence, \(\mathrm{f}^{-1}(\mathrm{x})=\mathrm{x}^2+3\)
[CG PET- 2017]
Sets, Relation and Function
117159
Which one of the following binary operations * is associative on the set of real numbers?
Exp: (B) : Let \(\mathrm{a}=1, \mathrm{~b}=2, \mathrm{c}=3\) are real no. For associative- \((a * b) * c=a *(b * c)\) By option (b) \(-\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1\) Then, \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=(1 * 2) * 3=(1+2-1) * 3=(2 * 3)=2+3-1=4\) \(=\mathrm{a} *(\mathrm{~b} * \mathrm{c})=1 *(2 * 3)=1 *(2+3-1)=(1 * 4)=1+4-1=4\) \(\text { So, }\) \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1 \text { is associative on the set of real numbers. }\)
[SCRA-2015]
Sets, Relation and Function
117160
Let \(f: R-\left\{\frac{5}{4}\right\} \rightarrow R\) be a function defined as \(f(x)=\frac{5 x}{4 x+5}\). The inverse of \(f\) is the map \(g\) : Range \(\mathrm{f} \rightarrow \mathrm{R}-\left\{\frac{5}{4}\right\}\) given by
1 \(g(y)=\frac{y}{5-4 y}\)
2 \(g(y)=\frac{5 y}{5+4 y}\)
3 \(g(y)=\frac{5 y}{5-4 y}\)
4 none of these
Explanation:
Exp: (C) : We have, \(f(\mathrm{x})=\frac{5 \mathrm{x}}{4 \mathrm{x}+5}, \mathrm{x} \in \mathrm{R}-\left\{\frac{5}{4}\right\}\) Let \(\mathrm{f}(\mathrm{x})=\mathrm{y} \Rightarrow \mathrm{x}=f^{-1}(\mathrm{y})\) \(y=\frac{5 x}{4 x+5} \Rightarrow 4 x y+5 y=5 x\) \(5 y=5 x-4 x y=x(5-4 y)\) \(\Rightarrow x=\frac{5 y}{5-4 y}\) \(\therefore \quad g(y)=f^{-1}(y)=\frac{5 y}{5-4 y}, x \in R-\left\{\frac{5}{4}\right\}\)
[AMU-2010]
Sets, Relation and Function
117162
Let * be a binary operation on the set \(R\) of real numbers defined by \(a * b=\frac{3 a b}{7}\), then the identity element in \(\mathbf{R}\) for ' \(*\) ' is
117158
Let \(f: R \rightarrow R\) be the function defined by \(f(x)=\sqrt{\mathbf{x}-3}, \forall \mathbf{x} \in \mathbf{R}\) then, \(\boldsymbol{f}^{-1}(\mathbf{x})=\)
1 \(x+3\)
2 \(x^2+3\)
3 \(\frac{x+3}{2}\)
4 \(\frac{x^2+3}{2}\)
Explanation:
Exp: (B) : Given, \(f(x)=\sqrt{x-3} \forall x \in R\) Let \(f(x)=y\) \(y=\sqrt{x-3}\) \(y^2=x-3\) \(x=y^2+3=f^{-1}(y)\)Hence, \(\mathrm{f}^{-1}(\mathrm{x})=\mathrm{x}^2+3\)
[CG PET- 2017]
Sets, Relation and Function
117159
Which one of the following binary operations * is associative on the set of real numbers?
Exp: (B) : Let \(\mathrm{a}=1, \mathrm{~b}=2, \mathrm{c}=3\) are real no. For associative- \((a * b) * c=a *(b * c)\) By option (b) \(-\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1\) Then, \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=(1 * 2) * 3=(1+2-1) * 3=(2 * 3)=2+3-1=4\) \(=\mathrm{a} *(\mathrm{~b} * \mathrm{c})=1 *(2 * 3)=1 *(2+3-1)=(1 * 4)=1+4-1=4\) \(\text { So, }\) \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1 \text { is associative on the set of real numbers. }\)
[SCRA-2015]
Sets, Relation and Function
117160
Let \(f: R-\left\{\frac{5}{4}\right\} \rightarrow R\) be a function defined as \(f(x)=\frac{5 x}{4 x+5}\). The inverse of \(f\) is the map \(g\) : Range \(\mathrm{f} \rightarrow \mathrm{R}-\left\{\frac{5}{4}\right\}\) given by
1 \(g(y)=\frac{y}{5-4 y}\)
2 \(g(y)=\frac{5 y}{5+4 y}\)
3 \(g(y)=\frac{5 y}{5-4 y}\)
4 none of these
Explanation:
Exp: (C) : We have, \(f(\mathrm{x})=\frac{5 \mathrm{x}}{4 \mathrm{x}+5}, \mathrm{x} \in \mathrm{R}-\left\{\frac{5}{4}\right\}\) Let \(\mathrm{f}(\mathrm{x})=\mathrm{y} \Rightarrow \mathrm{x}=f^{-1}(\mathrm{y})\) \(y=\frac{5 x}{4 x+5} \Rightarrow 4 x y+5 y=5 x\) \(5 y=5 x-4 x y=x(5-4 y)\) \(\Rightarrow x=\frac{5 y}{5-4 y}\) \(\therefore \quad g(y)=f^{-1}(y)=\frac{5 y}{5-4 y}, x \in R-\left\{\frac{5}{4}\right\}\)
[AMU-2010]
Sets, Relation and Function
117162
Let * be a binary operation on the set \(R\) of real numbers defined by \(a * b=\frac{3 a b}{7}\), then the identity element in \(\mathbf{R}\) for ' \(*\) ' is
117158
Let \(f: R \rightarrow R\) be the function defined by \(f(x)=\sqrt{\mathbf{x}-3}, \forall \mathbf{x} \in \mathbf{R}\) then, \(\boldsymbol{f}^{-1}(\mathbf{x})=\)
1 \(x+3\)
2 \(x^2+3\)
3 \(\frac{x+3}{2}\)
4 \(\frac{x^2+3}{2}\)
Explanation:
Exp: (B) : Given, \(f(x)=\sqrt{x-3} \forall x \in R\) Let \(f(x)=y\) \(y=\sqrt{x-3}\) \(y^2=x-3\) \(x=y^2+3=f^{-1}(y)\)Hence, \(\mathrm{f}^{-1}(\mathrm{x})=\mathrm{x}^2+3\)
[CG PET- 2017]
Sets, Relation and Function
117159
Which one of the following binary operations * is associative on the set of real numbers?
Exp: (B) : Let \(\mathrm{a}=1, \mathrm{~b}=2, \mathrm{c}=3\) are real no. For associative- \((a * b) * c=a *(b * c)\) By option (b) \(-\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1\) Then, \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=(1 * 2) * 3=(1+2-1) * 3=(2 * 3)=2+3-1=4\) \(=\mathrm{a} *(\mathrm{~b} * \mathrm{c})=1 *(2 * 3)=1 *(2+3-1)=(1 * 4)=1+4-1=4\) \(\text { So, }\) \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1 \text { is associative on the set of real numbers. }\)
[SCRA-2015]
Sets, Relation and Function
117160
Let \(f: R-\left\{\frac{5}{4}\right\} \rightarrow R\) be a function defined as \(f(x)=\frac{5 x}{4 x+5}\). The inverse of \(f\) is the map \(g\) : Range \(\mathrm{f} \rightarrow \mathrm{R}-\left\{\frac{5}{4}\right\}\) given by
1 \(g(y)=\frac{y}{5-4 y}\)
2 \(g(y)=\frac{5 y}{5+4 y}\)
3 \(g(y)=\frac{5 y}{5-4 y}\)
4 none of these
Explanation:
Exp: (C) : We have, \(f(\mathrm{x})=\frac{5 \mathrm{x}}{4 \mathrm{x}+5}, \mathrm{x} \in \mathrm{R}-\left\{\frac{5}{4}\right\}\) Let \(\mathrm{f}(\mathrm{x})=\mathrm{y} \Rightarrow \mathrm{x}=f^{-1}(\mathrm{y})\) \(y=\frac{5 x}{4 x+5} \Rightarrow 4 x y+5 y=5 x\) \(5 y=5 x-4 x y=x(5-4 y)\) \(\Rightarrow x=\frac{5 y}{5-4 y}\) \(\therefore \quad g(y)=f^{-1}(y)=\frac{5 y}{5-4 y}, x \in R-\left\{\frac{5}{4}\right\}\)
[AMU-2010]
Sets, Relation and Function
117162
Let * be a binary operation on the set \(R\) of real numbers defined by \(a * b=\frac{3 a b}{7}\), then the identity element in \(\mathbf{R}\) for ' \(*\) ' is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117158
Let \(f: R \rightarrow R\) be the function defined by \(f(x)=\sqrt{\mathbf{x}-3}, \forall \mathbf{x} \in \mathbf{R}\) then, \(\boldsymbol{f}^{-1}(\mathbf{x})=\)
1 \(x+3\)
2 \(x^2+3\)
3 \(\frac{x+3}{2}\)
4 \(\frac{x^2+3}{2}\)
Explanation:
Exp: (B) : Given, \(f(x)=\sqrt{x-3} \forall x \in R\) Let \(f(x)=y\) \(y=\sqrt{x-3}\) \(y^2=x-3\) \(x=y^2+3=f^{-1}(y)\)Hence, \(\mathrm{f}^{-1}(\mathrm{x})=\mathrm{x}^2+3\)
[CG PET- 2017]
Sets, Relation and Function
117159
Which one of the following binary operations * is associative on the set of real numbers?
Exp: (B) : Let \(\mathrm{a}=1, \mathrm{~b}=2, \mathrm{c}=3\) are real no. For associative- \((a * b) * c=a *(b * c)\) By option (b) \(-\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1\) Then, \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=(1 * 2) * 3=(1+2-1) * 3=(2 * 3)=2+3-1=4\) \(=\mathrm{a} *(\mathrm{~b} * \mathrm{c})=1 *(2 * 3)=1 *(2+3-1)=(1 * 4)=1+4-1=4\) \(\text { So, }\) \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1 \text { is associative on the set of real numbers. }\)
[SCRA-2015]
Sets, Relation and Function
117160
Let \(f: R-\left\{\frac{5}{4}\right\} \rightarrow R\) be a function defined as \(f(x)=\frac{5 x}{4 x+5}\). The inverse of \(f\) is the map \(g\) : Range \(\mathrm{f} \rightarrow \mathrm{R}-\left\{\frac{5}{4}\right\}\) given by
1 \(g(y)=\frac{y}{5-4 y}\)
2 \(g(y)=\frac{5 y}{5+4 y}\)
3 \(g(y)=\frac{5 y}{5-4 y}\)
4 none of these
Explanation:
Exp: (C) : We have, \(f(\mathrm{x})=\frac{5 \mathrm{x}}{4 \mathrm{x}+5}, \mathrm{x} \in \mathrm{R}-\left\{\frac{5}{4}\right\}\) Let \(\mathrm{f}(\mathrm{x})=\mathrm{y} \Rightarrow \mathrm{x}=f^{-1}(\mathrm{y})\) \(y=\frac{5 x}{4 x+5} \Rightarrow 4 x y+5 y=5 x\) \(5 y=5 x-4 x y=x(5-4 y)\) \(\Rightarrow x=\frac{5 y}{5-4 y}\) \(\therefore \quad g(y)=f^{-1}(y)=\frac{5 y}{5-4 y}, x \in R-\left\{\frac{5}{4}\right\}\)
[AMU-2010]
Sets, Relation and Function
117162
Let * be a binary operation on the set \(R\) of real numbers defined by \(a * b=\frac{3 a b}{7}\), then the identity element in \(\mathbf{R}\) for ' \(*\) ' is
117158
Let \(f: R \rightarrow R\) be the function defined by \(f(x)=\sqrt{\mathbf{x}-3}, \forall \mathbf{x} \in \mathbf{R}\) then, \(\boldsymbol{f}^{-1}(\mathbf{x})=\)
1 \(x+3\)
2 \(x^2+3\)
3 \(\frac{x+3}{2}\)
4 \(\frac{x^2+3}{2}\)
Explanation:
Exp: (B) : Given, \(f(x)=\sqrt{x-3} \forall x \in R\) Let \(f(x)=y\) \(y=\sqrt{x-3}\) \(y^2=x-3\) \(x=y^2+3=f^{-1}(y)\)Hence, \(\mathrm{f}^{-1}(\mathrm{x})=\mathrm{x}^2+3\)
[CG PET- 2017]
Sets, Relation and Function
117159
Which one of the following binary operations * is associative on the set of real numbers?
Exp: (B) : Let \(\mathrm{a}=1, \mathrm{~b}=2, \mathrm{c}=3\) are real no. For associative- \((a * b) * c=a *(b * c)\) By option (b) \(-\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1\) Then, \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=(1 * 2) * 3=(1+2-1) * 3=(2 * 3)=2+3-1=4\) \(=\mathrm{a} *(\mathrm{~b} * \mathrm{c})=1 *(2 * 3)=1 *(2+3-1)=(1 * 4)=1+4-1=4\) \(\text { So, }\) \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1 \text { is associative on the set of real numbers. }\)
[SCRA-2015]
Sets, Relation and Function
117160
Let \(f: R-\left\{\frac{5}{4}\right\} \rightarrow R\) be a function defined as \(f(x)=\frac{5 x}{4 x+5}\). The inverse of \(f\) is the map \(g\) : Range \(\mathrm{f} \rightarrow \mathrm{R}-\left\{\frac{5}{4}\right\}\) given by
1 \(g(y)=\frac{y}{5-4 y}\)
2 \(g(y)=\frac{5 y}{5+4 y}\)
3 \(g(y)=\frac{5 y}{5-4 y}\)
4 none of these
Explanation:
Exp: (C) : We have, \(f(\mathrm{x})=\frac{5 \mathrm{x}}{4 \mathrm{x}+5}, \mathrm{x} \in \mathrm{R}-\left\{\frac{5}{4}\right\}\) Let \(\mathrm{f}(\mathrm{x})=\mathrm{y} \Rightarrow \mathrm{x}=f^{-1}(\mathrm{y})\) \(y=\frac{5 x}{4 x+5} \Rightarrow 4 x y+5 y=5 x\) \(5 y=5 x-4 x y=x(5-4 y)\) \(\Rightarrow x=\frac{5 y}{5-4 y}\) \(\therefore \quad g(y)=f^{-1}(y)=\frac{5 y}{5-4 y}, x \in R-\left\{\frac{5}{4}\right\}\)
[AMU-2010]
Sets, Relation and Function
117162
Let * be a binary operation on the set \(R\) of real numbers defined by \(a * b=\frac{3 a b}{7}\), then the identity element in \(\mathbf{R}\) for ' \(*\) ' is