JEE Main 2024 — Mathematics Calculus

Let, $\displaystyle y=\underset{x\rightarrow \frac{\pi }{2}}{\mathrm{lim}}\frac{{\int }_{{x}^{3}}^{{(\frac{\pi }{2})}^{3}}\mathrm{cos}{t}^{\frac{1}{3}}dt}{{(x-\frac{\pi }{2})}^{2}}$y=x→2π​lim​(x−2π​)2∫x3(2π​)3​cost31​dt​

$$\Rightarrow y=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{{\int }_{{(\frac{\pi }{2}-h)}^{3}}^{{(\frac{\pi }{2})}^{3}}\mathrm{cos}{t}^{\frac{1}{3}}dt}{{(\frac{\pi }{2}-h-\frac{\pi }{2})}^{2}}$$⇒y=h→0lim​(2π​−h−2π​)2∫(2π​−h)3(2π​)3​cost31​dt​

$$\Rightarrow y=\underset{h\rightarrow 0}{\mathrm{lim}}(\frac{{\int }_{{(\frac{\pi }{2}-h)}^{3}}^{{(\frac{\pi }{2})}^{3}}\mathrm{cos}{t}^{\frac{1}{3}}dt}{{h}^{2}})$$⇒y=h→0lim​(h2∫(2π​−h)3(2π​)3​cost31​dt​)

Applying L-Hospital's rule and Newton Leibnitz Theorem we get,

$$\Rightarrow y=\underset{h\rightarrow 0}{\mathrm{lim}}(\frac{0+\mathrm{cos}(\frac{\pi }{2}-h)\times 3\times {(\frac{\pi }{2}-h)}^{2}}{2h})$$⇒y=h→0lim​(2h0+cos(2π​−h)×3×(2π​−h)2​)

$$\Rightarrow y=\underset{h\rightarrow 0}{\mathrm{lim}}(\frac{\mathrm{sin}h\times 3\times {(\frac{\pi }{2}-h)}^{2}}{2h})$$⇒y=h→0lim​(2hsinh×3×(2π​−h)2​)

We know that, $\underset{h\rightarrow 0}{\mathrm{lim}}(\frac{\mathrm{sin}h}{h})=1$h→0lim​(hsinh​)=1

$$\Rightarrow y=\underset{h\rightarrow 0}{\mathrm{lim}}(\frac{3\times {(\frac{\pi }{2}-h)}^{2}}{2})$$⇒y=h→0lim​(23×(2π​−h)2​)

$$\Rightarrow y=\frac{3{\pi }^{2}}{8}$$⇒y=83π2​