GATE 2025 EC Q.37: solution and answer

GATE 2025 · ECQ.37 · 2 marks · MCQ

Consider the polynomial $p (s) = s^{5} + 7 s^{4} + 3 s^{3} - 33 s^{2} + 2 s - 40$ . Let $(L, I, R)$ be defined as follows.

$L$ is the number of roots of $p (s)$ with negative real parts.

$I$ is the number of roots of $p (s)$ that are purely imaginary.

$R$ is the number of roots of $p (s)$ with positive real parts.

Which one of the following options is correct?

(A)
$L = 2, I = 2$ , and $R = 1$
(B)
$L = 3, I = 2$ , and $R = 0$
(C)
$L = 1, I = 2$ , and $R = 2$
(D)
$L = 0, I = 4$ , and $R = 1$

See the question in the paper

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Answer and solution

ANSWER

(A) $L = 2, I = 2$ , and $R = 1$ . The roots of $p (s)$ are $\pm j, 2, - 4, - 5$ .

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

Routh array, first two rows (coefficients taken alternately): $s^{5} : 1, 3, 2$ and $s^{4} : 7, - 33, - 40$ .
Row $s^{3}$ : $\frac{7 ( 3 ) - 1 ( - 33 )}{7} = \frac{54}{7}$ and $\frac{7 ( 2 ) - 1 ( - 40 )}{7} = \frac{54}{7}$ . Dividing a row by a positive number does not change any sign, so divide by $\frac{54}{7}$ to get $1, 1$ .
Row $s^{2}$ (using the scaled $s^{3}$ row): $\frac{1 ( - 33 ) - 7 ( 1 )}{1} = - 40$ and $\frac{1 ( - 40 ) - 7 ( 0 )}{1} = - 40$ , so the row is $- 40, - 40$ .
Row $s^{1}$ : $\frac{( - 40 ) ( 1 ) - ( 1 ) ( - 40 )}{- 40} = 0$ . The whole row is zero. This is the special case where $p (s)$ has roots placed symmetrically about the origin.
Auxiliary polynomial from the $s^{2}$ row: $A (s) = - 40 s^{2} - 40 = - 40 (s^{2} + 1)$ . Its roots are $s = \pm j$ , and they are also roots of $p (s)$ . Replace the zero row with the coefficients of $\frac{d A}{d s} = - 80 s$ , so the $s^{1}$ row is $- 80$ . The $s^{0}$ row is then $- 40$ .
First column: $1, 7, \frac{54}{7} (or 1 after scaling), - 40, - 80, - 40$ . There is exactly one sign change, between the $s^{3}$ and $s^{2}$ entries, so $p (s)$ has one root with positive real part: $R = 1$ .
From the auxiliary row down (the entries $- 40, - 80, - 40$ ) there is no sign change, so $A (s)$ has no root in the right half-plane. Its roots are symmetric about the origin, so it has none in the left half-plane either, and both lie on the $j ω$ axis: $I = 2$ . The remaining roots have negative real parts: $L = 5 - 1 - 2 = 2$ .
Check by factoring: $p (s) = (s^{2} + 1) (s^{3} + 7 s^{2} + 2 s - 40) = (s^{2} + 1) (s - 2) (s + 4) (s + 5)$ . The roots are $\pm j, 2, - 4, - 5$ , which gives $(L, I, R) = (2, 2, 1)$ . This is option (A).

KEY IDEA

A whole row of zeros in the Routh array means $p (s)$ has roots placed symmetrically about the origin, and those roots are the roots of the auxiliary polynomial built from the row above. After replacing the zero row with $d A / d s$ , the total number of sign changes in the first column counts all right-half-plane roots. The sign changes from the auxiliary row downwards count only the auxiliary polynomial's right-half-plane roots. Here that count is zero, so both roots of $A (s) = - 40 (s^{2} + 1)$ lie on the $j ω$ axis.

WATCH OUT FOR

Stopping at the zero row, or putting a small $ϵ$ in place of the whole zero row (the $ϵ$ method is only for a single zero in the first column). That hides the $\pm j$ pair and leads to option (B) or (C). A second trap is reading the one sign change as the only non-left-half-plane root and forgetting to count the imaginary pair: that gives $L = 4$ , which matches no option.

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 15 Routh–Hurwitz questions, with notes

GATE 2025 EC Q.37: solution and answer

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