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प्रश्न
Show that the function given by f(x) = sin x is
- strictly increasing in `(0, pi/2)`
- strictly decreasing in `(pi/2, pi)`
- neither increasing nor decreasing in (0, π)
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उत्तर
The given function is f(x) = sin x.
f'(x) = cos x
a. Since for each `x in (0, pi/2)`, cos x > 0, we have f'(x) > 0
Hence, f is strictly increasing in `(0. pi/2)`
b. Since for each `x in (pi/2 , pi), cos x < 0` we have f'(x) < 0
Hence, f is strictly decreasing in `(pi/2, pi)`
c. From the results obtained in (a) and (b), it is clear that f is neither increasing nor decreasing in (0, π).
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