1.

If ` 1/ sqrt(alpha) and 1/sqrt(beta)` are the roots of equation `a x^2 + bx + 1 = 0`(`a!=0, (a,b in R)`), then the equation `x(x + b^3) + (a^3 - 3abx) = 0` has roots -

Answer» eqn is `ax^2 + bx +1 =0`
so,roots will be `1/sqrt alpha & 1/sqrt beta`
can be written in the form `1/sqrt alpha + 1/sqrt beta = -b/a`
`(sqrt alpha + sqrt beta)/(sqrt alpha * sqrt beta)= -b/a`
`=1/sqrt(alpha beta) = 1/a`
now, `x(x+b^3) + (a^3 - 3abx) = 0`
`x^2 + b^3x - 3abx+a^3 = 0`
`x^2 + x(b^3 - 3ab) + alpha^(3/2) beta^(3/2) = 0`
now,` x^2 + x(-(sqrt alpha + sqrt beta)^3 + 3 sqrt(alpha beta) (sqrt alpha + sqrt beta)`
`x^2 -x[(sqrt alpha + sqrt beta)^3 - 3 sqrt(alpha beta) (sqrt alpha + sqrt beta)] + alpha^(3/2)beta^(3/2)`
`x^2 - x[alpha^3/2 + beta^(3/2) + 3 sqrt(alpha beta)(sqrt alpha + sqrt beta) - 3 sqrt (alpha beta) ( sqrt alpha + sqrt beta) ] + alpha^(3/2)* beta^(3/2) = 0`
`x^2 - x[alpha^(3/2) + beta^(3/2)] + alpha^(3/2) beta^(3/2) = 0`
option A is correct


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