116873
If for all \(x, y \in N\), there exists a function \(f(x)\) satisfying \(f(x+y)=f(x) \times f(y) \quad\) such that \(f(1)=3\) and \(\sum_{x=1}^n f(x)=120\), then value of \(n\) will be
1 4
2 5
3 6
4 None of these
Explanation:
A Given, for \(\mathrm{x}, \mathrm{y} \in \mathrm{N}\), \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) \cdot f(\mathrm{y})\) Then function will be of the form- \( f(\mathrm{x})=\mathrm{a}^{\mathrm{x}}, \text { where } \mathrm{a} \in \mathrm{N} {[\because \mathrm{a} \neq 1]}\) \(\because f(1)=3\) \(\Rightarrow f(1)=\mathrm{a}^1=3\) \(\Rightarrow \mathrm{a}=3\) \(\therefore\) Function is \(f(\mathrm{x})=3^{\mathrm{x}}\) Now, \(\sum_{x=1}^n f(x)=120\) \(\sum_{x=1}^n 3^x=120\) \(3+3^2+3^3+\ldots+3^n=120\) \(\Rightarrow \sum_{x=1}^n 3^x=120\) \(\Rightarrow 3+3^2+3^3+\) \(3+3^2+3^3+\ldots+3^1=120\) \(\Rightarrow \frac{3\left(3^{\mathrm{n}}-1\right)}{3-1}=120\) \(\Rightarrow 3^{\mathrm{n}}=1+\frac{120 \times 2}{3}\) \(\Rightarrow 3^{\mathrm{n}}=81=3^4\) \(\text { So, } \mathrm{n}=4\)
CG PET- 2015
Sets, Relation and Function
116893
Let \(\mathbf{f}=\{(0,-1),(-1,-3),(2,3),(3,5)\}\) be a function from \(Z\) to \(Z\) defined by \(f(x)=a x+b\). Then
1 \(\mathrm{a}=1, \mathrm{~b}=-2\)
2 \(\mathrm{a}=2, \mathrm{~b}=1\)
3 \(a=2, b=-1\)
4 \(\mathrm{a}=1, \mathrm{~b}=2\)
Explanation:
C Given, \(f=\{(0,-1),(-1,-3),(2,3),(3,5)\}\) defined by \(f(x)=a x\) \(+\mathrm{b}\) \(y=a x+b\) For ordered pair \((0,-1)\) \(-1=a(0)+b\) \(\Rightarrow b=-1\) For ordered pair \((-1,-3)\) \(-3=-a+b\) \(-3=-a-1\) \(a=2\)So, \(\mathrm{a}=2\) and \(\mathrm{b}=-1\)
116873
If for all \(x, y \in N\), there exists a function \(f(x)\) satisfying \(f(x+y)=f(x) \times f(y) \quad\) such that \(f(1)=3\) and \(\sum_{x=1}^n f(x)=120\), then value of \(n\) will be
1 4
2 5
3 6
4 None of these
Explanation:
A Given, for \(\mathrm{x}, \mathrm{y} \in \mathrm{N}\), \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) \cdot f(\mathrm{y})\) Then function will be of the form- \( f(\mathrm{x})=\mathrm{a}^{\mathrm{x}}, \text { where } \mathrm{a} \in \mathrm{N} {[\because \mathrm{a} \neq 1]}\) \(\because f(1)=3\) \(\Rightarrow f(1)=\mathrm{a}^1=3\) \(\Rightarrow \mathrm{a}=3\) \(\therefore\) Function is \(f(\mathrm{x})=3^{\mathrm{x}}\) Now, \(\sum_{x=1}^n f(x)=120\) \(\sum_{x=1}^n 3^x=120\) \(3+3^2+3^3+\ldots+3^n=120\) \(\Rightarrow \sum_{x=1}^n 3^x=120\) \(\Rightarrow 3+3^2+3^3+\) \(3+3^2+3^3+\ldots+3^1=120\) \(\Rightarrow \frac{3\left(3^{\mathrm{n}}-1\right)}{3-1}=120\) \(\Rightarrow 3^{\mathrm{n}}=1+\frac{120 \times 2}{3}\) \(\Rightarrow 3^{\mathrm{n}}=81=3^4\) \(\text { So, } \mathrm{n}=4\)
CG PET- 2015
Sets, Relation and Function
116893
Let \(\mathbf{f}=\{(0,-1),(-1,-3),(2,3),(3,5)\}\) be a function from \(Z\) to \(Z\) defined by \(f(x)=a x+b\). Then
1 \(\mathrm{a}=1, \mathrm{~b}=-2\)
2 \(\mathrm{a}=2, \mathrm{~b}=1\)
3 \(a=2, b=-1\)
4 \(\mathrm{a}=1, \mathrm{~b}=2\)
Explanation:
C Given, \(f=\{(0,-1),(-1,-3),(2,3),(3,5)\}\) defined by \(f(x)=a x\) \(+\mathrm{b}\) \(y=a x+b\) For ordered pair \((0,-1)\) \(-1=a(0)+b\) \(\Rightarrow b=-1\) For ordered pair \((-1,-3)\) \(-3=-a+b\) \(-3=-a-1\) \(a=2\)So, \(\mathrm{a}=2\) and \(\mathrm{b}=-1\)
116873
If for all \(x, y \in N\), there exists a function \(f(x)\) satisfying \(f(x+y)=f(x) \times f(y) \quad\) such that \(f(1)=3\) and \(\sum_{x=1}^n f(x)=120\), then value of \(n\) will be
1 4
2 5
3 6
4 None of these
Explanation:
A Given, for \(\mathrm{x}, \mathrm{y} \in \mathrm{N}\), \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) \cdot f(\mathrm{y})\) Then function will be of the form- \( f(\mathrm{x})=\mathrm{a}^{\mathrm{x}}, \text { where } \mathrm{a} \in \mathrm{N} {[\because \mathrm{a} \neq 1]}\) \(\because f(1)=3\) \(\Rightarrow f(1)=\mathrm{a}^1=3\) \(\Rightarrow \mathrm{a}=3\) \(\therefore\) Function is \(f(\mathrm{x})=3^{\mathrm{x}}\) Now, \(\sum_{x=1}^n f(x)=120\) \(\sum_{x=1}^n 3^x=120\) \(3+3^2+3^3+\ldots+3^n=120\) \(\Rightarrow \sum_{x=1}^n 3^x=120\) \(\Rightarrow 3+3^2+3^3+\) \(3+3^2+3^3+\ldots+3^1=120\) \(\Rightarrow \frac{3\left(3^{\mathrm{n}}-1\right)}{3-1}=120\) \(\Rightarrow 3^{\mathrm{n}}=1+\frac{120 \times 2}{3}\) \(\Rightarrow 3^{\mathrm{n}}=81=3^4\) \(\text { So, } \mathrm{n}=4\)
CG PET- 2015
Sets, Relation and Function
116893
Let \(\mathbf{f}=\{(0,-1),(-1,-3),(2,3),(3,5)\}\) be a function from \(Z\) to \(Z\) defined by \(f(x)=a x+b\). Then
1 \(\mathrm{a}=1, \mathrm{~b}=-2\)
2 \(\mathrm{a}=2, \mathrm{~b}=1\)
3 \(a=2, b=-1\)
4 \(\mathrm{a}=1, \mathrm{~b}=2\)
Explanation:
C Given, \(f=\{(0,-1),(-1,-3),(2,3),(3,5)\}\) defined by \(f(x)=a x\) \(+\mathrm{b}\) \(y=a x+b\) For ordered pair \((0,-1)\) \(-1=a(0)+b\) \(\Rightarrow b=-1\) For ordered pair \((-1,-3)\) \(-3=-a+b\) \(-3=-a-1\) \(a=2\)So, \(\mathrm{a}=2\) and \(\mathrm{b}=-1\)
116873
If for all \(x, y \in N\), there exists a function \(f(x)\) satisfying \(f(x+y)=f(x) \times f(y) \quad\) such that \(f(1)=3\) and \(\sum_{x=1}^n f(x)=120\), then value of \(n\) will be
1 4
2 5
3 6
4 None of these
Explanation:
A Given, for \(\mathrm{x}, \mathrm{y} \in \mathrm{N}\), \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) \cdot f(\mathrm{y})\) Then function will be of the form- \( f(\mathrm{x})=\mathrm{a}^{\mathrm{x}}, \text { where } \mathrm{a} \in \mathrm{N} {[\because \mathrm{a} \neq 1]}\) \(\because f(1)=3\) \(\Rightarrow f(1)=\mathrm{a}^1=3\) \(\Rightarrow \mathrm{a}=3\) \(\therefore\) Function is \(f(\mathrm{x})=3^{\mathrm{x}}\) Now, \(\sum_{x=1}^n f(x)=120\) \(\sum_{x=1}^n 3^x=120\) \(3+3^2+3^3+\ldots+3^n=120\) \(\Rightarrow \sum_{x=1}^n 3^x=120\) \(\Rightarrow 3+3^2+3^3+\) \(3+3^2+3^3+\ldots+3^1=120\) \(\Rightarrow \frac{3\left(3^{\mathrm{n}}-1\right)}{3-1}=120\) \(\Rightarrow 3^{\mathrm{n}}=1+\frac{120 \times 2}{3}\) \(\Rightarrow 3^{\mathrm{n}}=81=3^4\) \(\text { So, } \mathrm{n}=4\)
CG PET- 2015
Sets, Relation and Function
116893
Let \(\mathbf{f}=\{(0,-1),(-1,-3),(2,3),(3,5)\}\) be a function from \(Z\) to \(Z\) defined by \(f(x)=a x+b\). Then
1 \(\mathrm{a}=1, \mathrm{~b}=-2\)
2 \(\mathrm{a}=2, \mathrm{~b}=1\)
3 \(a=2, b=-1\)
4 \(\mathrm{a}=1, \mathrm{~b}=2\)
Explanation:
C Given, \(f=\{(0,-1),(-1,-3),(2,3),(3,5)\}\) defined by \(f(x)=a x\) \(+\mathrm{b}\) \(y=a x+b\) For ordered pair \((0,-1)\) \(-1=a(0)+b\) \(\Rightarrow b=-1\) For ordered pair \((-1,-3)\) \(-3=-a+b\) \(-3=-a-1\) \(a=2\)So, \(\mathrm{a}=2\) and \(\mathrm{b}=-1\)
