If for all real values of x- 4x2-164x2-96x sin-5 - 132 then - lies in the interval-

The correct options are A (π3, 2π3) D (4π3, 5π3) . 128x^2+32 lt;64x^2 96sin #10;α+5 #10; 64x^2+96sinα+27 lt;0 #10;Thus x is real. ...

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Standard XII

Mathematics

Expansions to Remove Indeterminate Form

If for all re...

Question

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

(0, \frac{π}{3})

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(\frac{π}{3}, \frac{2 π}{3})

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(\frac{2 π}{3}, π)

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(\frac{4 π}{3}, \frac{5 π}{3})

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Solution

The correct options are
A $(\frac{π}{3}, \frac{2 π}{3})$
D $(\frac{4 π}{3}, \frac{5 π}{3})$ .
$128 x^{2} + 32 < 64 x^{2} - 96 sin α + 5$
$64 x^{2} + 96 sin α + 27 < 0$
Thus x is real.
For real x,
$B^{2} - 4 A C > 0$
Or
$(96 sin α)^{2} - 4 (64 \times 27) > 0$
Or
${sin}^{2} α > \frac{3}{4}$
Or
$sin α > \frac{\sqrt{3}}{2}$ or $sin α < \frac{- \sqrt{3}}{2}$ .
Or
$α ϵ (\frac{π}{3}, \frac{2 π}{3}) \cup (\frac{4 π}{3}, \frac{5 π}{3})$

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If for all real values of x, 4x2+164x2−96xsinα+5<132 then α lies in the interval-

The correct options are A (π3,2π3) D (4π3,5π3).128x2+32<64x2−96sinα+5 64x2+96sinα+27<0 Thus x is real. For real x, B2−4AC>0 Or (96sinα)2−4(64×27)>0 Or sin2α>34 Or sinα>√32 or sinα<−√32. Or αϵ(π3,2π3)∪(4π3,5π3)

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