If for all real values of x,frac{4x^2 + 1}{64 | vedclass

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(D) Given the inequality: $\frac{4x^2 + 1}{64x^2 - 96x \sin \alpha + 5} < \frac{1}{32}$.Since the expression must hold for all real $x$,the denominato

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$b$ or $c$ both

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(D) Given the inequality: $\frac{4x^2 + 1}{64x^2 - 96x \sin \alpha + 5} Since the expression must hold for all real $x$,the denominator $64x^2 - 96x \sin \alpha + 5$ must be positive for all $x$ (as the numerator is always positive).Cross-multiplying: $32(4x^2 + 1) $128x^2 + 32 $64x^2 + 96x \sin \alpha + 27 For this quadratic in $x$ to be negative for all $x$,the discriminant $D$ must be positive,but wait,the inequality must hold for all $x$. Actually,for the quadratic $ax^2 + bx + c  0$. Re-evaluating: the expression must be negative for all $x$. This implies the quadratic $64x^2 + 96x \sin \alpha + 27$ must be less than $0$ for all $x$,which is impossible for a parabola opening upwards. The condition implies the range of the function must be less than $1/32$. The discriminant of $64x^2 + 96x \sin \alpha + 27 = 0$ must be positive for real roots to exist,allowing the expression to be negative between the roots. Thus,$D = (96 \sin \alpha)^2 - 4(64)(27) > 0$.$9216 \sin^2 \alpha - 6912 > 0$.$\sin^2 \alpha > \frac{6912}{9216} = \frac{3}{4}$.$|\sin \alpha| > \frac{\sqrt{3}}{2}$.This implies $\sin \alpha > \frac{\sqrt{3}}{2}$ or $\sin \alpha For $\sin \alpha > \frac{\sqrt{3}}{2}$,$\alpha \in (\pi/3, 2\pi/3)$.For $\sin \alpha < -\frac{\sqrt{3}}{2}$,$\alpha \in (4\pi/3, 5\pi/3)$.

If for all real values of $x$,$\frac{4x^2 + 1}{64x^2 - 96x \sin \alpha + 5} < \frac{1}{32}$,then $\alpha$ lies in the interval

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