If f(x)=cos2x1+sin2x, then f(π4)-3f'(π4)=

f(x)=cos2x1+sen2x f(x)=1-sin2x1+sin2x f1(x)=-2sin 2x(1+xin2x)2 fπ4=1-121+12=13 f1π4=-21+122=-89 fπ4-3f1π4=13+83=3

If f(x)=cos2x1+sin2x, then f(π4)-3f'(π4)=

f(x)=cos2x1+sen2x f(x)=1-sin2x1+sin2x f1(x)=-2sin 2x(1+xin2x)2 fπ4=1-121+12=13 f1π4=-21+122=-89 fπ4-3f1π4=13+83=3

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If f(x)= $\frac{\cos^{2} x}{1 + \sin^{2} x}$ , then f( $\frac{π}{4})$ -3f' $(\frac{π}{4})$ =

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$\frac{5}{3}$

$\frac{11}{3}$

$\frac{13}{9}$

answer is D.

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Detailed Solution

$f (x) = \frac{\cos^{2} x}{1 + {sen}^{2} x} f (x) = \frac{1 - \sin^{2} x}{1 + \sin^{2} x} f^{1} (x) = \frac{- 2 \sin 2 x}{{(1 + {xin}^{2} x)}^{2}} f (\frac{π}{4}) = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} = \frac{1}{3} f^{1} (\frac{π}{4}) = \frac{- 2}{{(1 + \frac{1}{2})}^{2}} = \frac{- 8}{9} f (\frac{π}{4}) - 3 f^{1} (\frac{π}{4}) = \frac{1}{3} + \frac{8}{3} = 3$