Question

If α and β are the zeros of the polynomial f(x) = x² + px + q, then a polynomial having \[\frac{1}{\alpha} \text{and}\frac{1}{\beta}\]  is its zero is 

Options

•  x² + qx + p

• x² − px + q

• qx² + px + 1

• px² + qx + 1

Answer 1

Let `alpha` and `beta` be the zeros of the polynomial f(x) = x² + px + q,.Then,

`alpha + ß = - (text{coefficient of x})/(text{coefficient of } x^2)`

`= -p/1`

`=-p`

`alpha  ß = (\text{constat term})/(text{coefficient of} x^2)`

`= q/1`

`= q`

Let S and R denote respectively the sum and product of the zeros of a polynomial

Whose zeros are  `1/alpha`  and  `1/beta` .then

`s = 1/alpha + 1/beta`

`=(alpha + beta)/(alpha beta)`

`= (-p)/q`

` R = 1/alpha xx1/beta`

`=1/(alpha beta)`

`= 1/q`

Hence, the required polynomial `g(x)` whose sum and product of zeros are S and R is given by

`x^2 - Sx + R = 0`

`x^2 + p/qx+ 1/q =0`

`(qx^2+px+1)/q=0`

`⇒ qx^2 + px + 1`

So `g(x) = qx^2 + px + 1`

Hence, the correct choice is `(C)`