If $\alpha$ and $\beta$ are the zeros of the polynomial $f(x)=x^{2}+p x+q$, then a polynomial having $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is its zero is
(a) $x^{2}+q x+p$
(b) $x^{2}-p x+q$
(c) $q x^{2}+p x+1$
(d) $p x^{2}+q x+1$
Let $\alpha$ and $\beta$ be the zeros of the polynomial $f(x)=x^{2}+p x+q$. Then,
$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$
$=-\frac{p}{1}$
$=-p$
And
$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$
$=\frac{q}{1}$
$=q$
Let S and R denote respectively the sum and product of the zeros of a polynomial
Whose zeros are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.then
$S=\frac{1}{\alpha}+\frac{1}{\beta}$
$=\frac{\alpha+\beta}{\alpha \beta}$
$=\frac{-P}{q}$
$R=\frac{1}{\alpha} \times \frac{1}{\beta}$
$=\frac{1}{\alpha \beta}$
$=\frac{1}{q}$
Hence, the required polynomial $g(x)$ whose sum and product of zeros are $\mathrm{S}$ and $\mathrm{R}$ is given by
$x^{2}-S x+R=0$
$x^{2}+\frac{P}{q} x+\frac{1}{q}=0$
$\frac{q x^{2}+P x+1}{q}=0$
$\Rightarrow \quad q x^{2}+P x+1$
So $g(x)=q x^{2}+P x+1$
Hence, the correct choice is $(c)$