Functions Ques 24

The values of $f(x)=3 \sin (\sqrt{\frac{\pi^{2}}{16}-x^{2}})$ lie in the interval…

(1983, 2M)

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Answer:

Correct Answer: 24.$[0, \frac{3}{\sqrt{2}}]$

Solution:

Formula:

Domain and Range:

  1. Given, $f(x)=3 \sin \sqrt{\frac{\pi^{2}}{16}-x^{2}}$

$\Rightarrow$ Domain $\in[-\frac{\pi}{4}, \frac{\pi}{4}]$

$\therefore$ For range, $f^{\prime}(x)=3 \cos (\sqrt{\frac{\pi^{2}}{16}-x^{2}}) \cdot \frac{1(-2 x)}{2 \sqrt{\frac{\pi^{2}}{16}-x^{2}}}=0$

Where, $\quad \cos (\sqrt{\frac{\pi^{2}}{16}-x^{2}})=0 \quad$ or $\quad x=0$

neglecting $\cos (\sqrt{\frac{\pi^{2}}{16}-x^{2}})=0 \Rightarrow \frac{\pi^{2}}{16}-x^{2}=\frac{\pi^{2}}{4}$

$\Rightarrow x^{2}=-\frac{3 \pi^{2}}{16}$, never possible

$\Rightarrow \quad x=0$

Thus,

$ f(0)=3 \sin \frac{\pi}{4}=\frac{3}{\sqrt{2}} $

and

$ f(-\frac{\pi}{4})=f (\frac{\pi}{4})=0 $

Hence, $\quad$ range $\in [0, \frac{3}{\sqrt{2}}]$



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