The range of frac{1}{sin^2 x + 3 sin x cos x | vedclass

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(D) Let $f(x) = \frac{1}{\sin^2 x + 3 \sin x \cos x + 5 \cos^2 x}$.Divide numerator and denominator by $\cos^2 x$ (assuming $\cos x 
eq 0$):$f(x) = \

Vedclass · Accepted Answer

$\left[\frac{2}{11}, 2\right]$

Vedclass · Answer

(D) Let $f(x) = \frac{1}{\sin^2 x + 3 \sin x \cos x + 5 \cos^2 x}$.Divide numerator and denominator by $\cos^2 x$ (assuming $\cos x 
eq 0$):$f(x) = \frac{\sec^2 x}{	an^2 x + 3 	an x + 5} = \frac{1 + 	an^2 x}{	an^2 x + 3 	an x + 5}$.Let $t = 	an x$. Then $f(t) = \frac{1 + t^2}{t^2 + 3t + 5}$.Let $y = \frac{1 + t^2}{t^2 + 3t + 5}$.$y(t^2 + 3t + 5) = 1 + t^2 \implies (y-1)t^2 + 3yt + (5y-1) = 0$.For $t$ to be real,the discriminant $D \geq 0$:$(3y)^2 - 4(y-1)(5y-1) \geq 0$.$9y^2 - 4(5y^2 - 6y + 1) \geq 0$.$9y^2 - 20y^2 + 24y - 4 \geq 0$.$-11y^2 + 24y - 4 \geq 0 \implies 11y^2 - 24y + 4 \leq 0$.Solving $11y^2 - 24y + 4 = 0$ using the quadratic formula:$y = \frac{24 \pm \sqrt{576 - 176}}{22} = \frac{24 \pm \sqrt{400}}{22} = \frac{24 \pm 20}{22}$.$y_1 = \frac{4}{22} = \frac{2}{11}$ and $y_2 = \frac{44}{22} = 2$.Thus,the range is $\left[\frac{2}{11}, 2ight]$.

The range of $\frac{1}{\sin^2 x + 3 \sin x \cos x + 5 \cos^2 x}$ is

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