Question 15

sin x > √{2}/2 for all values of x in the interval (π/4, 3π/4)

The interval π/4 < x < 3π/4 is where the sine function is greater than √{2}/2 for all x values. In this range, sin x increases from √{2}/2 at x = π/4 to a maximum of 1 at x = π/2, and then decreases back to √{2}/2 at x = 3π/4.

A) (0 < x < π/2)

This interval does not satisfy the condition for all values of x. While sin x is greater than √{2}/2 from π/4 to π/2, it is less than √{2}/2 for x values between 0 and π/4, making this option incorrect.

B) π/4 < x < 3π/4

This option is correct because sin x exceeds √{2}/2 throughout the entire interval. Starting from π/4, where sin x equals √{2}/2, it rises to 1 at π/2 and falls back to √{2}/2 at 3π/4, thus meeting the condition for all values of x in this interval.

C) π/2 < x < π

In this interval, sin x decreases from 1 at π/2 to 0 at π. While sin x is greater than √{2}/2 at π/2, it falls below √{2}/2 for all other values leading to π, making this option incorrect.

D) 3π/4 < x < 5π/4

This interval does not meet the condition, as sin x is less than √{2}/2 for values after 3π/4 until it reaches 5π/4. Therefore, this option fails to satisfy the requirement throughout the entire interval.

E) 5π/4 < x < 7π/4

In this interval, sin x is negative, ranging from below 0 at 5π/4 to just above 0 at 7π/4. Consequently, it does not meet the condition of being greater than √{2}/2 at any point.

Conclusion

The correct answer, B, is the only interval where sin x consistently exceeds √{2}/2 for all x values. All other options fail to maintain this condition across their respective intervals, thus confirming B as the definitive solution.