GATE EC 2022 Set 1 — Question 44

Given the even polynomial $p(s) = s^4 + 5s^2 + 4 + K$.
For all roots of $p(s)$ to lie on the imaginary axis, we can substitute $s^2 = y$. The polynomial becomes a quadratic in $y$:
$q(y) = y^2 + 5y + (4+K) = 0$If $s$ is purely imaginary, say $s = j\omega$, then $s^2 = (j\omega)^2 = -\omega^2$. This means that the roots of $q(y)$ must be real and negative.Let the roots of $q(y)$ be $y_1$ and $y_2$. For $y_1$ and $y_2$ to be real and negative, two conditions must be met: The discriminant $\Delta = b^2 - 4ac$ must be $\ge 0$.
Here, $a=1, b=5, c=(4+K)$.
$\Delta = 5^2 - 4(1)(4+K) \ge 0$
$25 - 16 - 4K \ge 0$
$9 - 4K \ge 0$
$9 \ge 4K \implies K \le \frac{9}{4}$ From Vieta's formulas, the product of roots $y_1 y_2 = c/a = 4+K$.
For both roots to be negative (or one zero), their product must be $\ge 0$.
$4+K \ge 0 \implies K \ge -4$ From Vieta's formulas, the sum of roots $y_1 + y_2 = -b/a = -5$.
This condition is already satisfied, as $-5 < 0$.Combining the conditions $K \le 9/4$ and $K \ge -4$, we get:
$-4 \le K \le 9/4$Let's verify the boundary cases:

• If $K = -4$, then $q(y) = y^2 + 5y = y(y+5) = 0$. The roots are $y=0$ and $y=-5$.

If $y=0$, then $s^2=0 \implies s=0$ (on imaginary axis).
If $y=-5$, then $s^2=-5 \implies s=\pm j\sqrt{5}$ (on imaginary axis).
So, $K=-4$ is included.

• If $K = 9/4$, then $q(y) = y^2 + 5y + (4+9/4) = y^2 + 5y + (16/4+9/4) = y^2 + 5y + 25/4 = (y+5/2)^2 = 0$.

The roots are $y = -5/2$ (repeated).
If $y=-5/2$, then $s^2=-5/2 \implies s=\pm j\sqrt{5/2}$ (on imaginary axis).
So, $K=9/4$ is included.Thus, the complete range for $K$ is $-4 \le K \le 9/4$.The final answer is $\boxed{A}$