A root of the equation, $sin x + x - 1 = 0$ , lies in the interval

(0, π / 2)

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(- π / 2, 0)

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(π / 2, π)

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(- π, - π / 2)

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Solution

The correct option is A $(0, π / 2)$
$f (x) = s i n x + x - 1$
$f (0) = s i n 0 + 0 - 1$
$f (0) = - 1$
$f (π / 2) = s i n \frac{π}{2} + \frac{π}{2} - 1 = 1 + \frac{π}{2} - 1 = \frac{π}{2} f (\frac{π}{2}) = \frac{π}{2} f (0) . f (\frac{π}{2}) < 0$ one root lies [IMVT] Theorem between $(0, \frac{π}{2})$

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