Let `f(x)=a x^3+b x^2+c x+1` has exterma at `x=alpha,beta` such that `alphabeta<0a n df(alpha)f(beta)<0` . Then the equation `f(x)=0` has three equal real roots one negative root if `f(alpha)<<0a n df(beta)>>0` one positive root if `f(alpha)<<0a n df(beta)>>0` none of these

Options

A. three distinct real roots
B. one positive root if `f(alpha)lt 0 and f(beta) gt 0 `
C. on negative root if `f(alpha) gt 0 ` and ` (beta) lt 0 `
D. all the above

Answer

D

Explanation

We have ,
`alpha beta lt 0`
`rArr alpha and beta` are of opposite signs
Let `alpha lt 0 and beta gt 0 `
If is given that f(x) has extrema at `x=alpha, beta` Therefore `alpha and beta ` are two distinct real roots of f' (x)=0 .But we know that between two distinct roots a polynominal there is at least one real root of its derivative. Therefore,f(x) has three distinct real roots `gamma, mu ` and v (say) such that
`gamma lt alpha lt mu lt beta lt v `

Thus,option (a)is correct.
If f(x)=0 has exactly one positive root, then it is evident from Fig .23 that `v gt 0 and gamma ,mu lt 0 `

Therefore
`alpha lt mu lt 0`
`rArr f(alpha)f(0)lt 0 " "[because mu` lies between alpha and 0]
`rArr f(alpha) lt 0 " "[because f(0) = 1 lt 0 ]`
`rArr (alpha) gt 0 " "[ because f(alpha ) f(beta ) lt 0 ] ` ltbtgt Thus option (b) is correct
If f(x) =0 has exactly one negative real root , then from Fig , 24 we have

`gamma lt 0 and mu v gt 0 `
`rArr 0 lt mu lt beta`
`