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If `alpha, beta` are the roots of the equation `lambda(x^(2)-x)+x+5=0` and if `lambda_(1) and lambda_(2)` are two values of `lambda` obtained from `(alpha)/(beta)+(beta)/(alpha)=(4)/(5)`, then `(lambda_(1))/(lambda_(2)^(2))+(lambda_(2))/(lambda_(1)^(2))` equalsA. 4192B. 4144C. 4096D. 4048 |
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Answer» Correct Answer - D Since `alpha and beta` are the roots of the equation `lambda x^(2) + x(1-lambda)+5 = 0`. `therefore" "alpha + beta = (lambda -1)/(lambda) and alpha beta =(5)/(lambda)` Now, `(alpha)/(beta)+(beta)/(alpha)=(4)/(5)` `rArr" "5(alpha + beta)^(2) = 14 alpha beta` `5(lambda -1)^(2) = 70 lambda rArr lambda^(2) - 16 lambda + 1 = 0` Since `lambda_(1) and lambda_(2)` are roots of this equation. `therefore" "lambda_(1) + lambda_(2) = 16 and lambda_(1) lambda_(2) = 1` Hence, `(lambda_(1))/(lambda_(2)^(2))+(lambda_(2))/(lambda_(1)^(2))=(lambda_(1)^(3)+lambda_(2)^(3))/((lambda_(1)lambda_(2)))=((lambda_(1)+ lambda_(2))^(3)-3lambda_(1) lambda_(2)(lambda_(1)+lambda_(2)))/((lambda_(1) lambda_(2))^(2))=16^(3) - 3 xx 16 = 4048` |
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