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If the roots of the equation `x^3+3ax^2+3bx+c=0` are in `H.P.`, then(i) `2b^2=c(3ab-c)`(ii) `2b^3=c(3ab-c)`(iii) `2b^3=c^(2)(3ab-c)`(iv) `2b^2=c^(2)(3ab-c)`A. `beta = (1)/(alpha)`B. `beta = b`C. `beta = (c)/(b)`D. `beta = (b)/(c)` |
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Answer» Correct Answer - C Clearly, `(1)/(alpha), (1)/(beta), (1)/(gamma)` are the roots of the equation `-cx^(3) + 3bx^(2) - 3ax + 1 = 0` and are in A.P. Now, `(1)/(alpha), (1)/(beta), (1)/(gamma) = (3b)/(c)` `rArr" "(3)/(beta) = (3b)/(c)" "[therefore (1)/(alpha)+(1)/(gamma)=(2)/(beta)]` `rArr" "beta = (c)/(b)` |
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