Question

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Answer 1

Step-by-step explanation:

We have,

$\tt{f(x) = \sqrt{ cos \left( \dfrac{\pi}{ {x}^{2} } \right) } }$

$\tt{ \implies \: cos \left( \dfrac{\pi}{ {x}^{2} } \right) \ge0}$

$\tt{ \implies \:(4n - 1) \dfrac{\pi}{2} \leqslant \dfrac{\pi}{ {x}^{2} } \leqslant (4n + 1) \dfrac{\pi}{2} \:, \: \: \: \: \: \: \: \: \: \forall \: n \in \mathbb{Z}}$

$\tt{ \implies \:\dfrac{4n - 1}{2} \leqslant \dfrac{1}{ {x}^{2} } \leqslant \dfrac{4n + 1}{2} \:, \: \: \: \: \: \: \: \: \: \forall \: n \in \mathbb{Z}}$

$\tt{ \implies \:\dfrac{2}{4n - 1} \geqslant {x}^{2} \geqslant \dfrac{2}{4n + 1} \:, \: \: \: \: \: \: \: \: \: \forall \: n \in \mathbb{Z}}$

$\tt{ \implies \: \sqrt{\dfrac{2}{4n + 1} }\leqslant x \leqslant \sqrt{\dfrac{2}{4n - 1}} \:, \: \: \: \: \: \: \: \: \: \forall \: n \in \mathbb{Z^{+}}}$

So,

$\bigstar \: \: \green{ \bf{dom(f) \in \left[ \sqrt{ \dfrac{2}{4n + 1} } \: \: , \: \: \sqrt{ \dfrac{2}{4n - 1} } \right] \: \: \: \: \: \: \: \: \forall \: n \in \mathbb{ Z^{+}}}}$

Answer 2

Answer:

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