The range of cos⁡x+42sin⁡x−π4+6 is

f(x)=cos⁡x+42sin⁡x−π4+6 f(x)=cosx+42sinxcosπ4-cosxsinπ4+6 ⇒f(x)=cosx+42sinx-cosx2+6 ⇒f(x)=cosx+4sinx-cosx+6 ⇒f(x)=4sinx-3cosx+6 Range=6-16+9,6+16+9=6-5,6+5=1,11

The range of cos⁡x+42sin⁡x−π4+6 is

f(x)=cos⁡x+42sin⁡x−π4+6 f(x)=cosx+42sinxcosπ4-cosxsinπ4+6 ⇒f(x)=cosx+42sinx-cosx2+6 ⇒f(x)=cosx+4sinx-cosx+6 ⇒f(x)=4sinx-3cosx+6 Range=6-16+9,6+16+9=6-5,6+5=1,11

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$The range of \cos x + 4 \sqrt{2} \sin (x - \frac{π}{4}) + 6 is$

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$[1, 11]$

$[- 1, 11]$

$[2, 10]$

$[10, 11]$

answer is A.

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Detailed Solution

$f (x) = \cos x + 4 \sqrt{2} \sin (x - \frac{π}{4}) + 6 f (x) = \cos x + 4 \sqrt{2} (\sin x \cos \frac{π}{4} - \cos x \sin \frac{π}{4}) + 6 \Rightarrow f (x) = \cos x + 4 \sqrt{2} (\frac{\sin x - \cos x}{\sqrt{2}}) + 6 \Rightarrow f (x) = \cos x + 4 (\sin x - \cos x) + 6 \Rightarrow f (x) = 4 \sin x - 3 \cos x + 6 R a n g e = [6 - \sqrt{16 + 9}, 6 + \sqrt{16 + 9}] = [6 - 5, 6 + 5] = [1, 11]$