| Column I | Column II | ||
| A. | \[\frac{1}{\alpha }+\frac{1}{\beta }\] | 1. | - 6 |
| B. | \[{{\left( \alpha -\beta \right)}^{2}}\] | 2. | \[\frac{-4}{25}\] |
| C. | \[\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}\] | 3. | \[\frac{-2}{5}\] |
| D. | \[\frac{\alpha }{\beta }+\frac{\beta }{\alpha }\] | 4. | \[\frac{4}{5}\] |
A) A-4, B-1, C-2, D-3
B) A-4, B-2, C-1, D-3
C) A-1, B-2, C-3, D-4
D) A-1, B-4, C-2, D-3
Correct Answer: A
Solution :
| \[\alpha +\beta =-\left( -\frac{4}{2} \right)=2\] and \[\alpha \,\,.\,\,\beta =\frac{5}{2}\] |
| [A] \[\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\beta +\alpha }{\alpha \,.\,\beta }=\frac{2\times 2}{5}=\frac{4}{5}\] |
| [B] \[{{\left( \alpha -\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \,.\,\beta \] |
| \[={{\left( \alpha \,+\beta \right)}^{2}}-4\alpha \beta \] |
| \[={{\left( 2 \right)}^{2}}-4\times \frac{5}{2}\] |
| \[=4-10=-6\] |
| [C] \[\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}=\frac{{{\beta }^{2}}+{{\alpha }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}}\] |
| \[=\frac{{{\left( \alpha +\beta \right)}^{2}}-2\alpha \,.\,\beta }{{{\left( \alpha \,.\,\beta \right)}^{2}}}\] |
| \[=\frac{{{\left( 2 \right)}^{2}}-2\times \frac{5}{2}}{{{\left( \frac{5}{2} \right)}^{2}}}\] |
| \[=\frac{4-5}{\frac{25}{4}}=-\frac{4}{25}\] |
| [D] \[\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \,.\,\beta }\] |
| \[=\frac{{{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta }{\alpha \beta }\] |
| \[=\frac{{{\left( 2 \right)}^{2}}-2\times \frac{5}{2}}{\frac{5}{2}}\] |
| \[=\frac{4-5}{\frac{5}{2}}=\frac{-2}{5}\] |
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