InterviewSolution
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InterviewSolution
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let `alpha(a)` and `beta(a)` be the roots of the equation `((1+a)^(1/3)-1)x^2 +((1+a)^(1/2)-1)x+((1+a)^(1/6)-1)=0` where `agt-1` then, `lim_(a->0^+)alpha(a)` and `lim_(a->0^+)beta(a)`A. `-(5)/(2) and 1`B. `-(1)/(2) and -1`C. `-(7)/(2) and 2`D. `-(9)/(2) and 3` |
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Answer» Correct Answer - B Let 1 + a = y. Then, the given equation becomes `(y^(1//3)-1)x^(2)+(y^(1//2)-1)x+(y^(1//6)-1)=0` `rArr" "((y^(1//3)-1)/(y-1))x^(2)+((y^(1//2)-1)/(y-1))x+((y^(1//6)-1)/(y-1))=0" "...(i)` Now, `a rarr 0^(+) rArr 1 + a rarr 1^(+) and 1 + a gt 1 rArr y rarr 1^(+)` Taking `underset(y rarr 1^(+))("lim")` on both sides of (i), we get `underset(y rarr 1^(+))("lim")((y^(1//3)-1^(1//3))/(y-1))x^(2)+underset(y rarr 1^(+))("lim")((y^(1//2)-1^(1//2))/(y-1))x + underset(y rarr 1^(+))("lim")((y^(1//6)-1^(6))/(y-1))=0` `rArr" "(1)/(3) x^(2) +(1)/(2)x +(1)/(6)=0` `rArr" "2x^(2)+3x + 1 = 0` `rArr" "(2x+1)(x+1)=0` `rArr" "x = -1, -(1)/(2)` `rArr" "underset(a rarr o^(+))("lim") beta(a) = -1 and underset(a rarr o^(+))("lim") alpha(a) = -(1)/(2)`. |
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