On which of the following intervals is sin x > √{2}/2} for all the values of (x in the interval?
π/4 < x < 3π/4
In this interval, the sine function is greater than √{2}/2, representing the range where the sine values are positive and exceed this threshold. Specifically, this occurs between the angles of 45° (π/4) and 135° (3π/4), where the sine values rise above √{2}/2.
In this interval, although sin x starts from 0 and increases to 1, it only reaches √{2}/2 at x = π/4. Therefore, sin x is not greater than √{2}/2 for the entire interval, as it is less than this value when x is between 0 and π/4.
This interval correctly captures the range where sin x is greater than √{2}/2. Starting from π/4 where sin x equals √{2}/2 and increasing to its maximum at π/2, then decreasing back to √{2}/2 at 3π/4, this interval fully satisfies the inequality.
While sin x is positive in this interval, it decreases from 1 at π/2 to 0 at π. Thus, it is greater than √{2}/2 only at the very start (π/2) and falls below this value as it approaches π, failing to satisfy the condition for all x in the interval.
In this interval, sin x is negative, as it ranges from √{2}/2 down to -√{2}/2, which does not satisfy the condition of being greater than √{2}/2 for any x values in this range.
Similar to the previous choice, sin x remains negative throughout this interval, starting from -√{2}/2 and going to 0. Therefore, it does not meet the requirement of being greater than √{2}/2.
The interval π/4 < x < 3π/4 is the only range where the sine function consistently exceeds √{2}/2, making it the correct choice. The other intervals either do not maintain the condition for all x or fall into negative sine values, reinforcing the unique nature of this specified range for the given inequality.
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