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WB JEE 2008

Subjective

Show that sin x is a monotonic increasing function of x in 0 < x < $$\pi$$/2.

Answer

.

Explanation

Let x1, x2 be two values of x in the given interval of 0 < x < $$\pi$$/2, where x1 < x2

$$\therefore$$ $${{{x_2} - {x_1}} \over 2}$$ and $${{{x_2} + {x_1}} \over 2}$$ both will lie between 0 and $$\pi$$/2

$$\therefore$$ $$f({x_2}) - f({x_1}) = \sin {x_2} - \sin {x_1}$$

$$ = 2\cos {{{x_1} + {x_2}} \over 2}.\sin {{{x_2} - {x_1}} \over 2}$$

Now, $${{{x_1} + {x_2}} \over 2}$$ and $${{{x_2} - {x_1}} \over 2}$$ both being acute angles, both sine and cosine of them are +ve.

$$\therefore$$ $$f({x_2}) - f({x_1}) > 0$$

$$\therefore$$ $$f({x_2}) > f({x_1})$$, where x2 > x1.

$$\therefore$$ sin x is a monotonic increasing function in 0 to $$\pi$$/2.

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