The minimum and maximum values of cosx+32sinx+π4+6 are

cosx+32sinx+π4+6=cosx+32sinx.12+cosx12+6=4cosx+3sinx+6Minimeu value=c−a2+b2=6-5=1 Maximum value =c+a2+b2=6+5=11

The minimum and maximum values of cosx+32sinx+π4+6 are

cosx+32sinx+π4+6=cosx+32sinx.12+cosx12+6=4cosx+3sinx+6Minimeu value=c−a2+b2=6-5=1 Maximum value =c+a2+b2=6+5=11

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The minimum and maximum values of $\cos x + 3 \sqrt{2} \sin (x + \frac{π}{4}) + 6$ are

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11, 1

6, 5

5, 6

1, 11

answer is D.

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Detailed Solution

$\cos x + 3 \sqrt{2} \sin (x + \frac{π}{4}) + 6$ $= \cos x + 3 \sqrt{2} (\sin x . \frac{1}{\sqrt{2}} + \cos x \frac{1}{\sqrt{2}}) + 6$ $= 4 \cos x + 3 \sin x + 6$

$M i n i m e u v a l u e = c - \sqrt{a^{2} + b^{2}} = 6 - 5 = 1 M a x i m u m v a l u e = c + \sqrt{a^{2} + b^{2}} = 6 + 5 = 11$

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The minimum and maximum values of $\cos x + 3 \sqrt{2} \sin (x + \frac{π}{4}) + 6$ are