A) A = 3, B = 8
B) A = -3, B = 8
C) A = 3, B = - 8
D) None of these
Correct Answer: A
Solution :
| [a] \[\alpha ,\beta ,\gamma \] and \[\delta \]are in H.P. \[\Rightarrow \,\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{\delta }\] are in A.P. |
| Let d be the common difference of this A.P. |
| Now, \[\alpha .\gamma \] are roots of \[A{{x}^{2}}-4x+1=0\] |
| \[\therefore \frac{\alpha +\gamma }{\alpha \gamma }=\frac{4/A}{1/A}=4\] or \[\frac{1}{\alpha }+\frac{1}{\gamma }=4\] i.e. |
| \[\frac{1}{\alpha }+\frac{1}{\alpha }+2d=4\] |
| or \[\frac{1}{\alpha }+d=2\] ?. (i) |
| \[\beta ,\delta \] are roots of \[B{{x}^{2}}-6x+1=0\] |
| \[\therefore \frac{\beta +\delta }{\beta \delta }=\frac{1}{\beta }+\frac{1}{\delta }=\frac{6/B}{1/B}=6\]or \[\frac{1}{\alpha }+d+\frac{1}{\alpha }+3d=6\] |
| \[\frac{1}{\alpha }+2d=3\] ?. (ii) |
| From (i) and (ii), on solving, we get |
| \[\frac{1}{\alpha }=1,\,\,d=1.\therefore \,\frac{1}{\alpha }=1,\frac{1}{\beta }=2,\frac{1}{\gamma }=3,\frac{1}{\delta }=4\] |
| Since, \[\frac{1}{\alpha \gamma }=A\,\,\therefore \,\,A=3.\]Also, \[\frac{1}{\beta \delta }=B,\,\,\therefore B=8\] |
| Hence \[A=3\] and \[B=8\]. |
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