**Question:**

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) $p x^{2}+q x+1$

**Solution:**

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 $(c)$

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