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)`

Options

A. `-(5)/(2) and 1`
B. `-(1)/(2) and -1`
C. `-(7)/(2) and 2`
D. `-(9)/(2) and 3`

Answer

B

Explanation

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)`.