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$\sf \: 5x \leqslant 8sinx - sin2x \leqslant 6x \: for \: all \: 0 \leqslant x \leqslant \frac{\pi}{3}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\red{\rm :\longmapsto\: f(x) \: = \: 8sinx - sin2x \: \: \: \: \: \: \forall \: \: \: 0 \leqslant x \leqslant \dfrac{\pi}{3}}$

Now,

$\rm :\longmapsto\: f(x) \: = \: 8sinx - sin2x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} f(x) \: = \dfrac{d}{dx}\: (8sinx - sin2x)$

$\rm :\longmapsto\:f'(x) = 8cosx - 2cos2x - - - - (1)$

Again on differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f'(x) =\dfrac{d}{dx}( 8cosx - 2cos2x)$

$\rm :\longmapsto\:f''(x) = - 8sinx + 4sin2x$

$\rm :\longmapsto\:f''(x) = - 8sinx + 4(2 \: sinx \: cosx)$

$\rm :\longmapsto\:f''(x) = - 8sinx + 8 \: sinx \: cosx$

$\rm :\longmapsto\:f''(x) = - 8sinx (1 - cosx)$

As,

$\boxed{ \rm \:0 \leqslant x \leqslant \dfrac{\pi}{3}}$

$\bf\implies \:f''(x) = - 8sinx(1 - cosx) \leqslant 0$

$\bf\implies \:f'(x) \: is \: decreasing \: function \: on \: 0 \leqslant x \leqslant \dfrac{\pi}{3}$

So, by definition of decreasing function,

$\rm :\longmapsto\:f'\bigg[\dfrac{\pi}{3} \bigg] \leqslant f'(x) \leqslant f'(0)$

$\rm :\longmapsto\:8cos\bigg[\dfrac{\pi}{3} \bigg] - 2cos\bigg[\dfrac{2\pi}{3} \bigg] \leqslant f'(x) \leqslant 8cos0 - 2cos0$

$\rm :\longmapsto\:8 \times \dfrac{1}{2} - 2 \times \dfrac{( - 1)}{2} \leqslant f'(x) \leqslant 8 - 2$

$\rm :\longmapsto\:4 + 1 \leqslant f'(x) \leqslant 6$

$\rm :\longmapsto\:5 \leqslant f'(x) \leqslant 6$

So, on integrating with respect to x, between 0 to x, we get

$\rm :\longmapsto\:\displaystyle\int_0^x 5 dx\leqslant \displaystyle\int_0^x f'(x)dx \leqslant \displaystyle\int_0^x 6dx$

$\rm :\longmapsto\:\bigg[5x\bigg]_0^x \leqslant \bigg[f(x)\bigg]_0^x \leqslant \bigg[6x\bigg]_0^x$

$\rm :\longmapsto\:5x \leqslant f(x) - f(0) \leqslant 6x$

$\rm :\longmapsto\:5x \leqslant (8sinx - sin2x) - (8sin0 - sin0) \leqslant 6x$

$\rm :\longmapsto\:5x \leqslant (8sinx - sin2x) - (0 - 0) \leqslant 6x$

$\rm :\longmapsto\:5x \leqslant 8sinx - sin2x \leqslant 6x$

Hence, Proved