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`        If for all real values of x, 4x^2+1 divided by 64x^2-32x sin alpha+29>1/32 then alpha lies in the interval (ans 0,pi/3)`
7 years ago

22 Points
```										Dear Sudha ,
The solution can be found by reorganizing the equation as ( 4x2 + 1 ) * 32 > 64x2 - 32xsin(alpha) + 29
or 128x2 + 32 > 64x2 - 32xsin(alpha) + 29
or 64x2 + 32xsin(alpha) + 3 > 0
The above Identity will hold true for all of x if Determinant is less than 0
because then no solution will exist such that 64x2 + 32xsin(alpha) + 3 = 0
i e D ≤ 0
or ( 32sin(alpha) )2  - 4*64*3 ≤ 0
or 32sin2(alpha) - 24 ≤ 0
or 4sin2(alpha) - 3 ≤ 0
or sin2(alpha) ≤ 3/4
this will give that alpha should lie in the Interval ( 0 , ∏/3 ) ( as sin(∏/3) = sq. root(3) / 2)
Puneet
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
```
7 years ago
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