Q61 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Mathematics

Let $\alpha$ be the period of $3 \sin \frac{\pi x}{3}-\cos \frac{\pi x}{2}+\tan \frac{\pi x}{4}, \beta$ be the period of $\sin ^2\left(\frac{\pi}{7}+\frac{x}{4}\right)-\sin ^2\left(\frac{\pi}{7}-\frac{x}{4}\right)$, and $\gamma$ be the period of $\cos ^4 x+\sin ^4 x$. Then $\frac{\alpha \gamma}{\beta}=$

  • Correct answer
    A
    $\frac{3}{2}$
  • B
    $\frac{3}{4}$
  • C
    3
  • D
    6

Solution

Period of $3 \sin \frac{\pi x}{3}-\cos \frac{\pi x}{2}+\tan \frac{\pi x}{4}$ is $12(\alpha)$, period of $\sin ^2\left(\frac{\pi}{7}+\frac{x}{4}\right)-\sin ^2 \frac{\pi}{4}-\frac{x}{4}$ is $4 \times \pi(\beta)$ and period of $\cos ^4(x)$ $+\sin ^4(\mathrm{x})$ is $\frac{\pi}{2}(\gamma)$ Hence, $\frac{\alpha \gamma}{\beta}=\frac{12 \times \frac{\pi}{2}}{4 \pi}=\frac{3}{2}$

Hence, the correct answer is option: 'A'.

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