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

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(D) Given the inequality: $\frac{4x^2 + 1}{64x^2 - 96x \sin \alpha + 5}  0$,we

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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} Assuming the denominator $64x^2 - 96x \sin \alpha + 5 > 0$,we cross-multiply:$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 here we require the inequality to hold for some $x$. However,the condition "for all real values of $x$" implies the quadratic $64x^2 + 96x \sin \alpha + 27$ must be negative,which is impossible for a parabola opening upwards. Re-evaluating the condition,the inequality holds if the discriminant $D > 0$ and the quadratic is negative between its roots.$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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