If $\alpha ,\beta$ are zeroes of polynomial $f\left(x\right)=x^2+px+q$ then polynomial having $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ as its zeroes is :

The correct option is B $q{x}^2+px+1$

$\alpha$ and $\beta$ are the roots of $x^2+px+q=0$

So,$\alpha +\beta =\frac{-p}{1}=-p$

and $\alpha \beta =\frac{q}{1}=q$

Let $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ be the roots of new polynomial $g\left(x\right)$

So, sum of roots $=\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-p}{q}$

and product of roots $\frac{1}{\alpha \beta }=\frac{1}{q}$

So, $g\left(x\right)=x^2-$(sum of roots)$x+$(product of roots)

So, $g\left(x\right)=x^2-\left(\frac{-p}{q}\right)x+\frac{1}{q}$

So, $g\left(x\right)=q{x}^2+px+1$

The answer is option (C)