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GATE 2026 ME – Question 38

Engineering Mathematics · Differential Equations: Initial and boundary value problems, Laplace transforms · 2 marks · Multiple choice

Let $f(t)$ be a function of $t$ defined for all positive values of $t$. The Laplace transform of $f(t)$ denoted by $L\{f(t)\} = \int_0^\infty e^{-st}f(t)\,dt$, provided that the integral exists where $s$ is a parameter which may be a real or complex number.

The Laplace transform of $f(t) = \sin 2t\,\sin 4t$ is

  1. $\frac{16s}{(s^2 + 4)(s^2 + 36)}$
  2. $\frac{32s}{(s^2 + 4)(s^2 + 36)}$
  3. $\frac{16}{(s^2 + 4)(s^2 + 36)}$
  4. $\frac{32}{(s^2 + 4)(s^2 + 36)}$

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Correct answer: (A) $\frac{16s}{(s^2 + 4)(s^2 + 36)}$

Explanation

Use the product formula $\sin 2t\sin 4t = \frac{1}{2}(\cos 2t - \cos 6t)$. The transform is $\frac{1}{2}\left[\frac{s}{s^2 + 4} - \frac{s}{s^2 + 36}\right] = \frac{s}{2}\cdot\frac{32}{(s^2 + 4)(s^2 + 36)} = \frac{16s}{(s^2 + 4)(s^2 + 36)}$.