JEE Main 2023 — Mathematics Calculus

Given,

$$\phi (x)=\frac{1}{\sqrt{x}}{\int }_{\pi /4}^{x}(4\sqrt{2}\mathrm{sin}t-3{\phi }^{'}(t))dt$$ϕ(x)=x​1​∫π/4x​(42​sint−3ϕ′(t))dt

Differentiate both sides w.r.t. $x$x we get,

$${\phi }^{'}(x)=\frac{-1}{2{x}^{3/2}}{\int }_{\pi /4}^{x}(4\sqrt{2}\mathrm{sin}t-3{\phi }^{'}(t))dt+\frac{1}{\sqrt{x}}(4\sqrt{2}\mathrm{sin}x-3{\phi }^{'}(x))$$ϕ′(x)=2x3/2−1​∫π/4x​(42​sint−3ϕ′(t))dt+x​1​(42​sinx−3ϕ′(x))

Put $x=\frac{\pi }{4}$x=4π​

$${\phi }^{'}(\frac{\pi }{4})=\frac{-1}{2{(\frac{\pi }{4})}^{3/2}}\times 0+\sqrt{\frac{4}{\pi }}(4\sqrt{2}\times \frac{1}{\sqrt{2}}-3{\phi }^{'}(\frac{\pi }{4}))$$ϕ′(4π​)=2(4π​)3/2−1​×0+π4​​(42​×2​1​−3ϕ′(4π​))

$$\Rightarrow \sqrt{\frac{\pi }{4}}{\phi }^{'}(\frac{\pi }{4})+3{\phi }^{'}(\frac{\pi }{4})=4$$⇒4π​​ϕ′(4π​)+3ϕ′(4π​)=4

$$\Rightarrow {\phi }^{'}(\frac{\pi }{4})(3+\sqrt{\frac{\pi }{4}})=4$$⇒ϕ′(4π​)(3+4π​​)=4

$$\Rightarrow {\phi }^{'}(\frac{\pi }{4})=\frac{8}{6+\sqrt{\pi }}$$⇒ϕ′(4π​)=6+π​8​