GATE 2020 EE Q.2: solution and answer

GATE 2020 · EEQ.2 · 1 mark · MCQ

Which of the following is true for all possible non-zero choices of integers $m, n; m \neq = n$ , or all possible non-zero choices of real numbers $p, q; p \neq = q$ , as applicable?

(A)
$\frac{1}{π} \int_{0}^{π} sin m θ sin n θ d θ = 0$
(B)
$\frac{1}{2 π} \int_{- π /2}^{π /2} sin pθ sin q θ d θ = 0$
(C)
$\frac{1}{2 π} \int_{- π}^{π} sin pθ cos q θ d θ = 0$
(D)
$α \to \infty lim \frac{1}{2 α} \int_{- α}^{α} sin pθ sin q θ d θ = 0$

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Answer and solution

ANSWER

(C): $\frac{1}{2 π} \int_{- π}^{π} sin pθ cos q θ d θ = 0$ for all non-zero real $p \neq = q$ , because the integrand is odd and the interval is symmetric.

Official answer: CSource: Official answer key, GATE 2020 EE (IIT Delhi)

The statement has to hold for every allowed choice, so one counter-example rules an option out. Note that $m \neq = n$ still allows $m = - n$ , and $p \neq = q$ still allows $p = - q$ .
Product-to-sum identity: $sin a θ sin b θ = \frac{1}{2} [cos (a - b) θ - cos (a + b) θ]$ .
(A): take $m = 1, n = - 1$ . Both are non-zero and unequal, so the choice is allowed. Then $sin θ sin (- θ) = - sin^{2} θ$ , so $\frac{1}{π} \int_{0}^{π} - sin^{2} θ d θ = \frac{1}{π} (- \frac{π}{2}) = - \frac{1}{2} \neq = 0.$ The orthogonality result also needs $m + n \neq = 0$ , so (A) is not always true.
(B): $[- π /2, π /2]$ is not a full period, so orthogonality does not apply. For $p = 1, q = 2$ the integrand is $sin θ sin 2 θ = 2 sin^{2} θ cos θ$ , which is even, so $\frac{1}{2 π} \cdot 2 \int_{0}^{π /2} 2 sin^{2} θ cos θ d θ = \frac{1}{π} [\frac{2 sin ^{3} θ}{3}]_{0}^{π /2} = \frac{2}{3 π} \neq = 0.$ (B) is false.
(C): odd-function property. $sin pθ$ is odd and $cos q θ$ is even, so $f (θ) = sin pθ cos q θ$ satisfies $f (- θ) = - f (θ)$ . The integral of an odd function over the symmetric interval $[- π, π]$ is $0$ , and this holds for every real $p, q$ .
(D): for $p + q \neq = 0$ the identity gives $\frac{1}{2 α} \int_{- α}^{α} sin pθ sin q θ d θ = \frac{1}{2 α} [\frac{sin ( p - q ) α}{p - q} - \frac{sin ( p + q ) α}{p + q}] \to 0.$ But $q = - p$ is allowed. Then the integrand is $- sin^{2} pθ$ and $\frac{1}{2 α} \int_{- α}^{α} - sin^{2} pθ d θ = - \frac{1}{2} + \frac{sin 2 p α}{4 p α} \to - \frac{1}{2} \neq = 0.$ (D) fails.
Only (C) is true for every allowed choice. Answer: (C).

KEY IDEA

The only property that holds for every allowed choice is symmetry: $sin pθ cos q θ$ is odd, so its integral over $[- π, π]$ is zero whatever $p$ and $q$ are. The sine-sine orthogonality results in (A), (B) and (D) need $m + n \neq = 0$ (or $p + q \neq = 0$ ), and (B) also needs a full period, so each one has a counter-example.

WATCH OUT FOR

Choosing (A) or (D) from the memorised orthogonality relation "different frequencies integrate to zero". The question only requires $m \neq = n$ (or $p \neq = q$ ) and non-zero values, so negative values are allowed. With $n = - m$ or $q = - p$ the product becomes $- sin^{2}$ , and its integral (or long-run average) is $- \frac{1}{2}$ , not $0$ .

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GATE 2020 EE Q.2: solution and answer

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