UPSC NDA 2 2017 Mathematics Solved Paper

Concept:If the sum of coefficients of a quadratic equation $ax^{2}+bx+c=0$ is zero ($a+b+c=0$), then one root is $1$ and the other is $\frac{c}{a}$.Explanation:Given equation: $(q-r)x^{2}+(r-p)x+(p-q)=0$.Sum of coefficients: $(q-r)+(r-p)+(p-q)=0$.Thus, $x=1$ is a root.Product of roots $= \frac{c}{a} = \frac{p-q}{q-r}$.Since one root is $1$, the other root is $\frac{p-q}{q-r}$.Alternatively, using the quadratic formula: $a=(q-r)$, $b=(r-p)$, $c=(p-q)$.Discriminant: $(r-p)^{2}-4(q-r)(p-q) = (r+p-2q)^{2}$.Roots: $x = \frac{-(r-p)\pm(r+p-2q)}{2(q-r)}$.This gives $x = \frac{p-q}{q-r}$ and $x=1$.Answer:The roots are $\frac{p-q}{q-r}$ and $1$, which corresponds to option B.