116895
Let \(A=\{-1,0,1,2\}, B=\{-4,-2,0,2\}\) and \(f, g: A \rightarrow B\) be functions defined by \(f(x)=x^2\) \(-x\) and \(g(x)=2\left|x-\frac{1}{2}\right|-1\). Then
1 \(f=g\)
2 \(f=2 g\)
3 \(\mathrm{g}=2 \mathrm{f}\)
4 none of these
Explanation:
A Since, \(f(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) has same domain and codomain A and B and \(f(1)=(1)^2-1=0\) \(\mathrm{g}(1)=2\left|1-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0\) \(f(1)=0=\mathrm{g}(1), f(0)=0=\mathrm{g}(0)\) \(f(-1)=2=\mathrm{g}(-1), f(2)=2=\mathrm{g}(2)\) \(\mathrm{A}=\{-1,0,1,2\}\) \(\mathrm{B}=\{-4,-2,0,2\}\) So, the functions are equal \((f=\mathrm{g})\)
AMU-2010
Sets, Relation and Function
116896
If \(1^4+2^4+3^4+\ldots \ldots+n^4=f(n)\left(1^2+2^2+\ldots \ldots+n^2\right)\), \(\forall \mathbf{n} \in \mathbf{N}\) then \(\mathbf{f}(\mathbf{4})=\)
116901
Assuming \(|x|\) to be so small, that \(x^2\) and higher powers of \(x\) can be neglected, then \(\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\)
1 \(1+\frac{5 x}{4}\)
2 \(1-\frac{5 x}{4}\)
3 \(1+\frac{4 \mathrm{x}}{5}\)
4 \(1-\frac{4 \mathrm{x}}{5}\)
Explanation:
B Given, \(|x|\) is very small, \(\mathrm{x}^2\) is negligible - \(\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\frac{1+\frac{1}{2} x+1-\frac{3}{2} x}{1+x+1+\frac{x}{2}}\) \(=\frac{2-x}{2+\frac{3 x}{2}}=\frac{4-2 x}{4+3 x}=\frac{(4-2 x)(4-3 x)}{\left(16-9 x^2\right)}\) \(=\frac{16-20 x+6 x^2}{16-9 x^2}=\frac{16-20 x}{16}\left(\because x^2 \text { negligible }\right)\) \(=1-\frac{5 x}{4}\)
116895
Let \(A=\{-1,0,1,2\}, B=\{-4,-2,0,2\}\) and \(f, g: A \rightarrow B\) be functions defined by \(f(x)=x^2\) \(-x\) and \(g(x)=2\left|x-\frac{1}{2}\right|-1\). Then
1 \(f=g\)
2 \(f=2 g\)
3 \(\mathrm{g}=2 \mathrm{f}\)
4 none of these
Explanation:
A Since, \(f(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) has same domain and codomain A and B and \(f(1)=(1)^2-1=0\) \(\mathrm{g}(1)=2\left|1-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0\) \(f(1)=0=\mathrm{g}(1), f(0)=0=\mathrm{g}(0)\) \(f(-1)=2=\mathrm{g}(-1), f(2)=2=\mathrm{g}(2)\) \(\mathrm{A}=\{-1,0,1,2\}\) \(\mathrm{B}=\{-4,-2,0,2\}\) So, the functions are equal \((f=\mathrm{g})\)
AMU-2010
Sets, Relation and Function
116896
If \(1^4+2^4+3^4+\ldots \ldots+n^4=f(n)\left(1^2+2^2+\ldots \ldots+n^2\right)\), \(\forall \mathbf{n} \in \mathbf{N}\) then \(\mathbf{f}(\mathbf{4})=\)
116901
Assuming \(|x|\) to be so small, that \(x^2\) and higher powers of \(x\) can be neglected, then \(\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\)
1 \(1+\frac{5 x}{4}\)
2 \(1-\frac{5 x}{4}\)
3 \(1+\frac{4 \mathrm{x}}{5}\)
4 \(1-\frac{4 \mathrm{x}}{5}\)
Explanation:
B Given, \(|x|\) is very small, \(\mathrm{x}^2\) is negligible - \(\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\frac{1+\frac{1}{2} x+1-\frac{3}{2} x}{1+x+1+\frac{x}{2}}\) \(=\frac{2-x}{2+\frac{3 x}{2}}=\frac{4-2 x}{4+3 x}=\frac{(4-2 x)(4-3 x)}{\left(16-9 x^2\right)}\) \(=\frac{16-20 x+6 x^2}{16-9 x^2}=\frac{16-20 x}{16}\left(\because x^2 \text { negligible }\right)\) \(=1-\frac{5 x}{4}\)
116895
Let \(A=\{-1,0,1,2\}, B=\{-4,-2,0,2\}\) and \(f, g: A \rightarrow B\) be functions defined by \(f(x)=x^2\) \(-x\) and \(g(x)=2\left|x-\frac{1}{2}\right|-1\). Then
1 \(f=g\)
2 \(f=2 g\)
3 \(\mathrm{g}=2 \mathrm{f}\)
4 none of these
Explanation:
A Since, \(f(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) has same domain and codomain A and B and \(f(1)=(1)^2-1=0\) \(\mathrm{g}(1)=2\left|1-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0\) \(f(1)=0=\mathrm{g}(1), f(0)=0=\mathrm{g}(0)\) \(f(-1)=2=\mathrm{g}(-1), f(2)=2=\mathrm{g}(2)\) \(\mathrm{A}=\{-1,0,1,2\}\) \(\mathrm{B}=\{-4,-2,0,2\}\) So, the functions are equal \((f=\mathrm{g})\)
AMU-2010
Sets, Relation and Function
116896
If \(1^4+2^4+3^4+\ldots \ldots+n^4=f(n)\left(1^2+2^2+\ldots \ldots+n^2\right)\), \(\forall \mathbf{n} \in \mathbf{N}\) then \(\mathbf{f}(\mathbf{4})=\)
116901
Assuming \(|x|\) to be so small, that \(x^2\) and higher powers of \(x\) can be neglected, then \(\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\)
1 \(1+\frac{5 x}{4}\)
2 \(1-\frac{5 x}{4}\)
3 \(1+\frac{4 \mathrm{x}}{5}\)
4 \(1-\frac{4 \mathrm{x}}{5}\)
Explanation:
B Given, \(|x|\) is very small, \(\mathrm{x}^2\) is negligible - \(\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\frac{1+\frac{1}{2} x+1-\frac{3}{2} x}{1+x+1+\frac{x}{2}}\) \(=\frac{2-x}{2+\frac{3 x}{2}}=\frac{4-2 x}{4+3 x}=\frac{(4-2 x)(4-3 x)}{\left(16-9 x^2\right)}\) \(=\frac{16-20 x+6 x^2}{16-9 x^2}=\frac{16-20 x}{16}\left(\because x^2 \text { negligible }\right)\) \(=1-\frac{5 x}{4}\)
116895
Let \(A=\{-1,0,1,2\}, B=\{-4,-2,0,2\}\) and \(f, g: A \rightarrow B\) be functions defined by \(f(x)=x^2\) \(-x\) and \(g(x)=2\left|x-\frac{1}{2}\right|-1\). Then
1 \(f=g\)
2 \(f=2 g\)
3 \(\mathrm{g}=2 \mathrm{f}\)
4 none of these
Explanation:
A Since, \(f(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) has same domain and codomain A and B and \(f(1)=(1)^2-1=0\) \(\mathrm{g}(1)=2\left|1-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0\) \(f(1)=0=\mathrm{g}(1), f(0)=0=\mathrm{g}(0)\) \(f(-1)=2=\mathrm{g}(-1), f(2)=2=\mathrm{g}(2)\) \(\mathrm{A}=\{-1,0,1,2\}\) \(\mathrm{B}=\{-4,-2,0,2\}\) So, the functions are equal \((f=\mathrm{g})\)
AMU-2010
Sets, Relation and Function
116896
If \(1^4+2^4+3^4+\ldots \ldots+n^4=f(n)\left(1^2+2^2+\ldots \ldots+n^2\right)\), \(\forall \mathbf{n} \in \mathbf{N}\) then \(\mathbf{f}(\mathbf{4})=\)
116901
Assuming \(|x|\) to be so small, that \(x^2\) and higher powers of \(x\) can be neglected, then \(\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\)
1 \(1+\frac{5 x}{4}\)
2 \(1-\frac{5 x}{4}\)
3 \(1+\frac{4 \mathrm{x}}{5}\)
4 \(1-\frac{4 \mathrm{x}}{5}\)
Explanation:
B Given, \(|x|\) is very small, \(\mathrm{x}^2\) is negligible - \(\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\frac{1+\frac{1}{2} x+1-\frac{3}{2} x}{1+x+1+\frac{x}{2}}\) \(=\frac{2-x}{2+\frac{3 x}{2}}=\frac{4-2 x}{4+3 x}=\frac{(4-2 x)(4-3 x)}{\left(16-9 x^2\right)}\) \(=\frac{16-20 x+6 x^2}{16-9 x^2}=\frac{16-20 x}{16}\left(\because x^2 \text { negligible }\right)\) \(=1-\frac{5 x}{4}\)
116895
Let \(A=\{-1,0,1,2\}, B=\{-4,-2,0,2\}\) and \(f, g: A \rightarrow B\) be functions defined by \(f(x)=x^2\) \(-x\) and \(g(x)=2\left|x-\frac{1}{2}\right|-1\). Then
1 \(f=g\)
2 \(f=2 g\)
3 \(\mathrm{g}=2 \mathrm{f}\)
4 none of these
Explanation:
A Since, \(f(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) has same domain and codomain A and B and \(f(1)=(1)^2-1=0\) \(\mathrm{g}(1)=2\left|1-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0\) \(f(1)=0=\mathrm{g}(1), f(0)=0=\mathrm{g}(0)\) \(f(-1)=2=\mathrm{g}(-1), f(2)=2=\mathrm{g}(2)\) \(\mathrm{A}=\{-1,0,1,2\}\) \(\mathrm{B}=\{-4,-2,0,2\}\) So, the functions are equal \((f=\mathrm{g})\)
AMU-2010
Sets, Relation and Function
116896
If \(1^4+2^4+3^4+\ldots \ldots+n^4=f(n)\left(1^2+2^2+\ldots \ldots+n^2\right)\), \(\forall \mathbf{n} \in \mathbf{N}\) then \(\mathbf{f}(\mathbf{4})=\)
116901
Assuming \(|x|\) to be so small, that \(x^2\) and higher powers of \(x\) can be neglected, then \(\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\)
1 \(1+\frac{5 x}{4}\)
2 \(1-\frac{5 x}{4}\)
3 \(1+\frac{4 \mathrm{x}}{5}\)
4 \(1-\frac{4 \mathrm{x}}{5}\)
Explanation:
B Given, \(|x|\) is very small, \(\mathrm{x}^2\) is negligible - \(\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\frac{1+\frac{1}{2} x+1-\frac{3}{2} x}{1+x+1+\frac{x}{2}}\) \(=\frac{2-x}{2+\frac{3 x}{2}}=\frac{4-2 x}{4+3 x}=\frac{(4-2 x)(4-3 x)}{\left(16-9 x^2\right)}\) \(=\frac{16-20 x+6 x^2}{16-9 x^2}=\frac{16-20 x}{16}\left(\because x^2 \text { negligible }\right)\) \(=1-\frac{5 x}{4}\)
