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The range of real constant `t` for which `(1-tan^2 t)sin theta^2+tan^2 t*tan theta^2 >= theta^2;` always holds `AA theta in (0,pi/2)` is `[alpha,beta)` then `beta/alpha` is equal to |
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Answer» Correct Answer - 3 `tan^(2)ge (theta^(2)-sintheta^(2))/(tantheta^(2)-sintheta^(2)) AA theta in(0,(pi)/2)` So, `tan^(2)t ge ((theta^(2)-sintheta^(2))/(tantheta^(2)-sintheta^(2)))_("max")AA theta in (0,(pi)/2)` Since, in `(0,(pi)/2):tantheta^(2)gt theta^(2)` and the same is subtracted from `N^(-r)` and `D^(-r)` both So , maximum value occurs at `theta to O^(+)` Therefore `tan^(2)t ge 1/3, t in ((pi)/6, (pi)/2)` |
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