First slide

Functions (XII)

Question

The range of the function $y = 3 \sin \sqrt{\frac{π^{2}}{16} - x^{2}} is$

Moderate

A

$[0, \frac{3}{\sqrt{2}}]$
B

$[- \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}]$
C

$[- \frac{3}{\sqrt{2}}, 0]$
D

None of these

Solution

For y to be defined, $\frac{π^{2}}{16} - x^{2} \geq 0$
$\begin{array}{l} \Rightarrow (\frac{π}{4} - x) (\frac{π}{4} + x) \geq 0 \Rightarrow (x - \frac{π}{4}) (x + \frac{π}{4}) \leq 0 \\ \Rightarrow - \frac{π}{4} \leq x \leq \frac{π}{4} \end{array}$
$∴$ Domain of y $= [- \frac{π}{4}, \frac{π}{4}]$
Clearly, for $x \in [- \frac{π}{4}, \frac{π}{4}], \sqrt{\frac{π^{2}}{16} - x^{2}} \in [0, \frac{π}{4}]$
Since sin x is an increasing function on $[0, \frac{π}{4}]$
Therefore, $\sin 0 \leq \sin \sqrt{\frac{π^{2}}{16} - x^{2}} \leq \sin \frac{π}{4}$
$\Rightarrow 0 \leq 3 \sin \sqrt{\frac{π^{2}}{16} - x^{2}} \leq \frac{3}{\sqrt{2}} \Rightarrow 0 \leq y \leq \frac{3}{\sqrt{2}}$
$∴$ Range of y $= [0, \frac{3}{\sqrt{2}}]$

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