JEE Main 2022 — Mathematics Calculus

Let, $f(x)=8\mathrm{sin}x-\mathrm{sin}2x$f(x)=8sinx−sin2x

$$\Rightarrow {f}^{'}(x)=8\mathrm{cos}x-2\mathrm{cos}2x$$⇒f′(x)=8cosx−2cos2x

$$\Rightarrow {f}^{''}(x)=-8\mathrm{sin}x+4\mathrm{sin}2x$$⇒f′′(x)=−8sinx+4sin2x

$$\Rightarrow {f}^{''}(x)=-8\mathrm{sin}x(1-\mathrm{cos}x)$$⇒f′′(x)=−8sinx(1−cosx)

So, ${f}^{'}(x)<0\text{when }x\in (\frac{\pi }{4},\frac{\pi }{3})$f′(x)<0when x∈(4π​,3π​)

Hence ${f}^{'}(x)$f′(x) is decreasing function in given domain,

Now, ${f}^{'}(\frac{\pi }{3})<{f}^{'}(x)<{f}^{'}(\frac{\pi }{4})$f′(3π​)<f′(x)<f′(4π​)

$$\Rightarrow 5<{f}^{'}(x)<\frac{8}{\sqrt{2}}$$⇒5<f′(x)<2​8​

$$\Rightarrow 5<{f}^{'}(x)<4\sqrt{2}$$⇒5<f′(x)<42​

Now integrating all with respect to $dx$dx

$$\int 5dx<\int {f}^{'}(x)dx<\int 4\sqrt{2}dx$$∫5dx<∫f′(x)dx<∫42​dx

$$\Rightarrow 5x<f(x)<4\sqrt{2}x$$⇒5x<f(x)<42​x

$$\Rightarrow 5<\frac{f(x)}{x}<4\sqrt{2}$$⇒5<xf(x)​<42​

Again integrating with $\displaystyle {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}$∫4π​3π​​

We get, $\displaystyle {\int }_{\frac{\pi }{4}}^{\frac{\pi }{4}}5<\int \frac{f(x)}{x}<{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}4\sqrt{2}$∫4π​4π​​5<∫xf(x)​<∫4π​3π​​42​

$$\Rightarrow {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}5<\int \frac{8\mathrm{sin}x-\mathrm{sin}2x}{x}<{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}4\sqrt{2}$$⇒∫4π​3π​​5<∫x8sinx−sin2x​<∫4π​3π​​42​

$$\Rightarrow \frac{5\pi }{12}<I<\frac{\sqrt{2}\pi }{3}$$⇒125π​<I<32​π​