The maximum value of f (x) = sin x in the interval [π,2π] isa) 6b) 0c...
f(x) = sin x
f’(x) =cosx
f”(x) = -sin x
f”(3pi/2) = -sin(3pi/2)
= -(-1)
=> 1 > 0 (local minima)
f(pi) = sin(pi) = 0
f(2pi) = sin(2pi) = 0
Hence, 0 is the maxima.
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The maximum value of f (x) = sin x in the interval [π,2π] isa) 6b) 0c...
Solution:
To find the maximum value of f(x) = sin x in the interval [π, 2π], we need to find the value of x in the given interval at which sin x is maximum.
The maximum value of sin x is 1, which occurs when x = π/2.
However, π/2 is not in the given interval [π, 2π].
The next maximum value of sin x occurs at x = 3π/2, which is also not in the given interval.
Therefore, the maximum value of sin x in the interval [π, 2π] is 0, which occurs at x = π and x = 2π.
Hence, the correct answer is option B, 0.
The maximum value of f (x) = sin x in the interval [π,2π] isa) 6b) 0c...
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