117037
If \(f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in R\) then \(\mathbf{f}\left(\frac{1}{2023}\right)+\mathbf{f}\left(\frac{2}{2023}\right)+\mathbf{f}\left(\frac{3}{2023}\right)+\ldots \ldots+\mathbf{f}\left(\frac{2022}{2023}\right)\) is equal to
1 1011
2 2010
3 1012
4 2011
Explanation:
A Given, \(f(x)=\frac{2^{2 x}}{2^{2 x}+2}\) Put, \(\quad \mathrm{x} \rightarrow 1-\mathrm{x}\) then we get- \(f(1-x)=\frac{2^{2(1-x)}}{2^{2(1-x)}+2}\) \(\Rightarrow \quad \frac{4^{1-x}}{4^{1-x}+2}\) Then, adding we get- \(f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2}\) \(\Rightarrow \quad \frac{4^x}{4^x+2}+\frac{\frac{4}{4^x}}{\frac{4}{4^x}+2}\) \(\Rightarrow \quad \frac{4^x}{4^x+2}+\frac{2}{4^x+2}\) \(\Rightarrow \quad \frac{4^x+2}{4^x+2}=1\) Now, \(\mathrm{f}\left(\frac{1}{2023}\right) +\mathrm{f}\left(\frac{2}{2023}\right)+\mathrm{f}\left(\frac{3}{2023}\right)+\ldots . .+\mathrm{f}\left(\frac{2022}{2023}\right)\) \(=(1+1+1+1+\ldots \ldots+1,1011 \text { times })\) \(=1011\)
Shift-II
Sets, Relation and Function
117038
Function \(f: R \rightarrow R\) defined by \(f(x)=x^2+5\) is
1 many-one and onto
2 one-one and onto
3 one-one and into
4 many-one and into
Explanation:
D Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\), defined as \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2+5\) Polynomial function of even degree is many-one. \(\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}^2+5\) is many-one. There are some elements in co-domain of \(f\) which has no pre-image in its domain. Hence, \(f(x)\) is into function. Thus, \(f(x)=x^2+5\) is many one and into function.
MHT-CET 2019
Sets, Relation and Function
117039
Let \(A=\{x: x \in R ; x\) is not a positive integer \(\}\) Define \(f: A \rightarrow R\) as \(f(x)=\frac{2 x}{x-1}\), then \(f\) is
1 injective but not surjective
2 surjective but not injective
3 bijective
4 neither injective nor surjective
Explanation:
A Given, \(f(x) =\frac{2 x}{x-1}\) \(f^{\prime}(x) =\frac{(x-1) 2-2 x(1)}{(x-1)^2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{2 \mathrm{x}-2-2 \mathrm{x}}{(\mathrm{x}-1)^2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-2}{(\mathrm{x}-1)^2}, \forall \mathrm{x} \in \mathrm{A} .\) We see that \(f\) is decreasing in its domain So, \(\mathrm{f}\) is one-one (injective) Let, \(y=f(x)\) \(y=\frac{2 x}{x-1}\) \(x y-y=2 x\) \(x y-2 x=y\) \(x(y-2)=y\) \(x=\frac{y}{y-2}\) Consider \(\mathrm{y}=3\), then \(\mathrm{x}=\frac{3}{3-2}=3>0\) Since, \(\mathrm{x}\) is not a positive integer. So, \(\mathrm{f}\) is not onto (Surjective).
shift-II
Sets, Relation and Function
117040
For any natural number \(n,\left(15 \times 5^{2 n}\right)+(2 \times\) \(2^{3 \mathrm{n}}\) ) is divisible by
1 7
2 11
3 13
4 17
Explanation:
D \(\left(15 \times 5^{2 \mathrm{n}}\right)+\left(2 \times 2^{3 \mathrm{n}}\right)\) Put, \(\mathrm{n}=1\) \(=15 \times 5^{2 \times 1}+2 \times 2^{3 \times 1}\) \(=15 \times 5^2+2 \times 2^3\) \(=15 \times 25+16\) \(=375+16=391\) Which is divisible by 17
117037
If \(f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in R\) then \(\mathbf{f}\left(\frac{1}{2023}\right)+\mathbf{f}\left(\frac{2}{2023}\right)+\mathbf{f}\left(\frac{3}{2023}\right)+\ldots \ldots+\mathbf{f}\left(\frac{2022}{2023}\right)\) is equal to
1 1011
2 2010
3 1012
4 2011
Explanation:
A Given, \(f(x)=\frac{2^{2 x}}{2^{2 x}+2}\) Put, \(\quad \mathrm{x} \rightarrow 1-\mathrm{x}\) then we get- \(f(1-x)=\frac{2^{2(1-x)}}{2^{2(1-x)}+2}\) \(\Rightarrow \quad \frac{4^{1-x}}{4^{1-x}+2}\) Then, adding we get- \(f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2}\) \(\Rightarrow \quad \frac{4^x}{4^x+2}+\frac{\frac{4}{4^x}}{\frac{4}{4^x}+2}\) \(\Rightarrow \quad \frac{4^x}{4^x+2}+\frac{2}{4^x+2}\) \(\Rightarrow \quad \frac{4^x+2}{4^x+2}=1\) Now, \(\mathrm{f}\left(\frac{1}{2023}\right) +\mathrm{f}\left(\frac{2}{2023}\right)+\mathrm{f}\left(\frac{3}{2023}\right)+\ldots . .+\mathrm{f}\left(\frac{2022}{2023}\right)\) \(=(1+1+1+1+\ldots \ldots+1,1011 \text { times })\) \(=1011\)
Shift-II
Sets, Relation and Function
117038
Function \(f: R \rightarrow R\) defined by \(f(x)=x^2+5\) is
1 many-one and onto
2 one-one and onto
3 one-one and into
4 many-one and into
Explanation:
D Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\), defined as \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2+5\) Polynomial function of even degree is many-one. \(\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}^2+5\) is many-one. There are some elements in co-domain of \(f\) which has no pre-image in its domain. Hence, \(f(x)\) is into function. Thus, \(f(x)=x^2+5\) is many one and into function.
MHT-CET 2019
Sets, Relation and Function
117039
Let \(A=\{x: x \in R ; x\) is not a positive integer \(\}\) Define \(f: A \rightarrow R\) as \(f(x)=\frac{2 x}{x-1}\), then \(f\) is
1 injective but not surjective
2 surjective but not injective
3 bijective
4 neither injective nor surjective
Explanation:
A Given, \(f(x) =\frac{2 x}{x-1}\) \(f^{\prime}(x) =\frac{(x-1) 2-2 x(1)}{(x-1)^2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{2 \mathrm{x}-2-2 \mathrm{x}}{(\mathrm{x}-1)^2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-2}{(\mathrm{x}-1)^2}, \forall \mathrm{x} \in \mathrm{A} .\) We see that \(f\) is decreasing in its domain So, \(\mathrm{f}\) is one-one (injective) Let, \(y=f(x)\) \(y=\frac{2 x}{x-1}\) \(x y-y=2 x\) \(x y-2 x=y\) \(x(y-2)=y\) \(x=\frac{y}{y-2}\) Consider \(\mathrm{y}=3\), then \(\mathrm{x}=\frac{3}{3-2}=3>0\) Since, \(\mathrm{x}\) is not a positive integer. So, \(\mathrm{f}\) is not onto (Surjective).
shift-II
Sets, Relation and Function
117040
For any natural number \(n,\left(15 \times 5^{2 n}\right)+(2 \times\) \(2^{3 \mathrm{n}}\) ) is divisible by
1 7
2 11
3 13
4 17
Explanation:
D \(\left(15 \times 5^{2 \mathrm{n}}\right)+\left(2 \times 2^{3 \mathrm{n}}\right)\) Put, \(\mathrm{n}=1\) \(=15 \times 5^{2 \times 1}+2 \times 2^{3 \times 1}\) \(=15 \times 5^2+2 \times 2^3\) \(=15 \times 25+16\) \(=375+16=391\) Which is divisible by 17
117037
If \(f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in R\) then \(\mathbf{f}\left(\frac{1}{2023}\right)+\mathbf{f}\left(\frac{2}{2023}\right)+\mathbf{f}\left(\frac{3}{2023}\right)+\ldots \ldots+\mathbf{f}\left(\frac{2022}{2023}\right)\) is equal to
1 1011
2 2010
3 1012
4 2011
Explanation:
A Given, \(f(x)=\frac{2^{2 x}}{2^{2 x}+2}\) Put, \(\quad \mathrm{x} \rightarrow 1-\mathrm{x}\) then we get- \(f(1-x)=\frac{2^{2(1-x)}}{2^{2(1-x)}+2}\) \(\Rightarrow \quad \frac{4^{1-x}}{4^{1-x}+2}\) Then, adding we get- \(f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2}\) \(\Rightarrow \quad \frac{4^x}{4^x+2}+\frac{\frac{4}{4^x}}{\frac{4}{4^x}+2}\) \(\Rightarrow \quad \frac{4^x}{4^x+2}+\frac{2}{4^x+2}\) \(\Rightarrow \quad \frac{4^x+2}{4^x+2}=1\) Now, \(\mathrm{f}\left(\frac{1}{2023}\right) +\mathrm{f}\left(\frac{2}{2023}\right)+\mathrm{f}\left(\frac{3}{2023}\right)+\ldots . .+\mathrm{f}\left(\frac{2022}{2023}\right)\) \(=(1+1+1+1+\ldots \ldots+1,1011 \text { times })\) \(=1011\)
Shift-II
Sets, Relation and Function
117038
Function \(f: R \rightarrow R\) defined by \(f(x)=x^2+5\) is
1 many-one and onto
2 one-one and onto
3 one-one and into
4 many-one and into
Explanation:
D Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\), defined as \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2+5\) Polynomial function of even degree is many-one. \(\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}^2+5\) is many-one. There are some elements in co-domain of \(f\) which has no pre-image in its domain. Hence, \(f(x)\) is into function. Thus, \(f(x)=x^2+5\) is many one and into function.
MHT-CET 2019
Sets, Relation and Function
117039
Let \(A=\{x: x \in R ; x\) is not a positive integer \(\}\) Define \(f: A \rightarrow R\) as \(f(x)=\frac{2 x}{x-1}\), then \(f\) is
1 injective but not surjective
2 surjective but not injective
3 bijective
4 neither injective nor surjective
Explanation:
A Given, \(f(x) =\frac{2 x}{x-1}\) \(f^{\prime}(x) =\frac{(x-1) 2-2 x(1)}{(x-1)^2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{2 \mathrm{x}-2-2 \mathrm{x}}{(\mathrm{x}-1)^2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-2}{(\mathrm{x}-1)^2}, \forall \mathrm{x} \in \mathrm{A} .\) We see that \(f\) is decreasing in its domain So, \(\mathrm{f}\) is one-one (injective) Let, \(y=f(x)\) \(y=\frac{2 x}{x-1}\) \(x y-y=2 x\) \(x y-2 x=y\) \(x(y-2)=y\) \(x=\frac{y}{y-2}\) Consider \(\mathrm{y}=3\), then \(\mathrm{x}=\frac{3}{3-2}=3>0\) Since, \(\mathrm{x}\) is not a positive integer. So, \(\mathrm{f}\) is not onto (Surjective).
shift-II
Sets, Relation and Function
117040
For any natural number \(n,\left(15 \times 5^{2 n}\right)+(2 \times\) \(2^{3 \mathrm{n}}\) ) is divisible by
1 7
2 11
3 13
4 17
Explanation:
D \(\left(15 \times 5^{2 \mathrm{n}}\right)+\left(2 \times 2^{3 \mathrm{n}}\right)\) Put, \(\mathrm{n}=1\) \(=15 \times 5^{2 \times 1}+2 \times 2^{3 \times 1}\) \(=15 \times 5^2+2 \times 2^3\) \(=15 \times 25+16\) \(=375+16=391\) Which is divisible by 17
117037
If \(f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in R\) then \(\mathbf{f}\left(\frac{1}{2023}\right)+\mathbf{f}\left(\frac{2}{2023}\right)+\mathbf{f}\left(\frac{3}{2023}\right)+\ldots \ldots+\mathbf{f}\left(\frac{2022}{2023}\right)\) is equal to
1 1011
2 2010
3 1012
4 2011
Explanation:
A Given, \(f(x)=\frac{2^{2 x}}{2^{2 x}+2}\) Put, \(\quad \mathrm{x} \rightarrow 1-\mathrm{x}\) then we get- \(f(1-x)=\frac{2^{2(1-x)}}{2^{2(1-x)}+2}\) \(\Rightarrow \quad \frac{4^{1-x}}{4^{1-x}+2}\) Then, adding we get- \(f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2}\) \(\Rightarrow \quad \frac{4^x}{4^x+2}+\frac{\frac{4}{4^x}}{\frac{4}{4^x}+2}\) \(\Rightarrow \quad \frac{4^x}{4^x+2}+\frac{2}{4^x+2}\) \(\Rightarrow \quad \frac{4^x+2}{4^x+2}=1\) Now, \(\mathrm{f}\left(\frac{1}{2023}\right) +\mathrm{f}\left(\frac{2}{2023}\right)+\mathrm{f}\left(\frac{3}{2023}\right)+\ldots . .+\mathrm{f}\left(\frac{2022}{2023}\right)\) \(=(1+1+1+1+\ldots \ldots+1,1011 \text { times })\) \(=1011\)
Shift-II
Sets, Relation and Function
117038
Function \(f: R \rightarrow R\) defined by \(f(x)=x^2+5\) is
1 many-one and onto
2 one-one and onto
3 one-one and into
4 many-one and into
Explanation:
D Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\), defined as \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2+5\) Polynomial function of even degree is many-one. \(\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}^2+5\) is many-one. There are some elements in co-domain of \(f\) which has no pre-image in its domain. Hence, \(f(x)\) is into function. Thus, \(f(x)=x^2+5\) is many one and into function.
MHT-CET 2019
Sets, Relation and Function
117039
Let \(A=\{x: x \in R ; x\) is not a positive integer \(\}\) Define \(f: A \rightarrow R\) as \(f(x)=\frac{2 x}{x-1}\), then \(f\) is
1 injective but not surjective
2 surjective but not injective
3 bijective
4 neither injective nor surjective
Explanation:
A Given, \(f(x) =\frac{2 x}{x-1}\) \(f^{\prime}(x) =\frac{(x-1) 2-2 x(1)}{(x-1)^2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{2 \mathrm{x}-2-2 \mathrm{x}}{(\mathrm{x}-1)^2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-2}{(\mathrm{x}-1)^2}, \forall \mathrm{x} \in \mathrm{A} .\) We see that \(f\) is decreasing in its domain So, \(\mathrm{f}\) is one-one (injective) Let, \(y=f(x)\) \(y=\frac{2 x}{x-1}\) \(x y-y=2 x\) \(x y-2 x=y\) \(x(y-2)=y\) \(x=\frac{y}{y-2}\) Consider \(\mathrm{y}=3\), then \(\mathrm{x}=\frac{3}{3-2}=3>0\) Since, \(\mathrm{x}\) is not a positive integer. So, \(\mathrm{f}\) is not onto (Surjective).
shift-II
Sets, Relation and Function
117040
For any natural number \(n,\left(15 \times 5^{2 n}\right)+(2 \times\) \(2^{3 \mathrm{n}}\) ) is divisible by
1 7
2 11
3 13
4 17
Explanation:
D \(\left(15 \times 5^{2 \mathrm{n}}\right)+\left(2 \times 2^{3 \mathrm{n}}\right)\) Put, \(\mathrm{n}=1\) \(=15 \times 5^{2 \times 1}+2 \times 2^{3 \times 1}\) \(=15 \times 5^2+2 \times 2^3\) \(=15 \times 25+16\) \(=375+16=391\) Which is divisible by 17
