The range of fx-8-2sin-216-x2 is

The correct option is A [0,8] Value inside square root can not be negative thus. π/4≤x≤π/4⇒ #160; 0≤√(π^2/16 x^2)≤π/4 So the value of function wil ...

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Standard XII

Mathematics

Theorems for Continuity

The range of ...

Question

The range of $f (x) = 8 \sqrt{2} sin \sqrt{\frac{π^{2}}{16} - x^{2}}$ is

[- 1, 1]

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[0, 1]

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[0, 8]

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[0, 4]

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Solution

The correct option is A $[0, 8]$
Value inside square root can not be negative thus
. $- \frac{π}{4} \leq x \leq \frac{π}{4} \Rightarrow 0 \leq \sqrt{\frac{π^{2}}{16} - x^{2}} \leq \frac{π}{4}$
So the value of function will belong to
$\Rightarrow 8 \sqrt{2} sin \sqrt{\frac{π^{2}}{16} - x^{2}} \in [0, 8]$

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The range of f(x)=8√2sin√π216−x2 is

The correct option is A [0,8]Value inside square root can not be negative thus. −π4≤x≤π4⇒0≤√π216−x2≤π4So the value of function will belong to⇒8√2sin√π216−x2∈[0,8]

The range of $f (x) = 8 \sqrt{2} sin \sqrt{\frac{π^{2}}{16} - x^{2}}$ is