GATE 2018 EC Q.6: solution and answer

Official answer: BSource: Official answer key, GATE 2018 EC (IIT Guwahati)

1. Differentiate: $p^{\prime }(s)=3s^2+2a_{2}s+a_1$. This is a real quadratic with leading coefficient $3>0$.

2. It has no real roots, so its discriminant is negative: $(2a_2{)}^2-4(3)(a_1)=4a_2^2-12a_1<0$. A quadratic with a positive leading coefficient and no real roots never changes sign, so it stays positive: $p^{\prime }(s)>0$ for every real $s$.

3. Since $p^{\prime }(s)>0$ everywhere, $p(s)$ is strictly increasing on the whole real line.

4. Check the ends. The leading term $s^3$ dominates, so $p(s)\to -\infty$ as $s\to -\infty$ and $p(s)\to +\infty$ as $s\to +\infty$. $p$ is continuous, so by the intermediate value theorem it crosses zero at least once.

5. A strictly increasing function can take the value $0$ at most once, so $p(s)$ has exactly one real root $s=r$.

6. That root is simple, because $p^{\prime }(r)>0\ne 0$. Write $p(s)=(s-r)q(s)$, where $q$ is a real quadratic. If $q$ had a real root, $p$ would have a second real zero (or a repeated one at $r$, which would force $p^{\prime }(r)=0$). So the other two roots are a complex-conjugate pair $\sigma \pm j\omega$ with $\omega \ne 0$, which is what real coefficients require.

7. Check with an example: $p(s)=s^3+s$ gives $p^{\prime }(s)=3s^2+1$, which has no real roots. Then $p(s)=s(s^2+1)$ has roots $0$ and $\pm j$, so exactly one root is real.

8. The number of real roots of $p(s)$ is $1$, which is option (B).