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Question Number 159046 by mnjuly1970 last updated on 12/Nov/21

Commented by cortano last updated on 13/Nov/21

$$f\left(x\right)=\frac{3{(1-\cos {}^{2}x)}^2+2}{6\cos {}^{2}x+1}$$$$f\left(x\right)=\frac{3(\cos {}^{4}x-2\cos {}^{2}x+1)+2}{6\cos {}^{2}x+1}$$$$f\left(x\right)=\frac{3\cos {}^{4}x-6\cos {}^{2}x+5}{6\cos {}^{2}x+1}$$$$let\cos {}^{2}x=u\wedge 0⩽u⩽1$$$$\Rightarrow y=\frac{3u^2-6u+5}{6u+1}$$$$whenu=0\Rightarrow y=5$$$$whenu=1\Rightarrow y=\frac{2}{7}$$$$R_f:\frac{2}{7}⩽f\left(x\right)⩽5$$

Commented by mnjuly1970 last updated on 13/Nov/21

$$thankyousomuchmaster$$

Answered by MJS_new last updated on 12/Nov/21

$$[\frac{2}{7};5]$$

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