JEE Main 2024 — Mathematics Calculus

Given: $\displaystyle f(x)={\int }_{0}^{x}g(t)\mathrm{log}(\frac{1-t}{1+t})\mathrm{dt}$f(x)=∫0x​g(t)log(1+t1−t​)dt

$$\Rightarrow f(-x)={\int }_{0}^{-x}g(t)\mathrm{log}(\frac{1-t}{1+t})\mathrm{dt}$$⇒f(−x)=∫0−x​g(t)log(1+t1−t​)dt

$$\Rightarrow f(-x)=-{\int }_{0}^{x}g(-y)\mathrm{log}(\frac{1+y}{1-y})\mathrm{dy}$$⇒f(−x)=−∫0x​g(−y)log(1−y1+y​)dy

$\displaystyle =-{\int }_{0}^{x}g(y)\mathrm{ln}(\frac{1-y}{1+y})\mathrm{dy}$=−∫0x​g(y)ln(1+y1−y​)dy ($g$g is odd)

$f(-x)=-f(x)\Rightarrow f$f(−x)=−f(x)⇒f is also odd.

Now,

$$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f(x)+\frac{{x}^{2}\mathrm{cosx}}{1+{e}^{x}})\mathrm{dx}...(i)$$I=∫−2π​2π​​(f(x)+1+exx2cosx​)dx...(i)

$$\Rightarrow I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f(-x)+\frac{{x}^{2}{e}^{x}\mathrm{cosx}}{1+{e}^{x}})\mathrm{dx}...(\mathrm{ii})$$⇒I=∫−2π​2π​​(f(−x)+1+exx2excosx​)dx...(ii)

$$\Rightarrow 2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{x}^{2}\mathrm{cosxdx}$$⇒2I=∫−2π​2π​​x2cosxdx

$$\Rightarrow 2I=2{\int }_{0}^{\frac{\pi }{2}}{x}^{2}\mathrm{cosxdx}$$⇒2I=2∫02π​​x2cosxdx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}{x}^{2}\mathrm{cosxdx}$$⇒I=∫02π​​x2cosxdx

$$\Rightarrow I={({x}^{2}\mathrm{sinx})}_{0}^{\frac{\pi }{2}}-{\int }_{0}^{\frac{\pi }{2}}2\mathrm{xsinxdx}$$⇒I=(x2sinx)02π​​−∫02π​​2xsinxdx

$$\Rightarrow I=\frac{{\pi }^{2}}{4}-2{(-\mathrm{xcosx}+\mathrm{sinx})}_{0}^{\frac{\pi }{2}}$$⇒I=4π2​−2(−xcosx+sinx)02π​​

$$\Rightarrow I=\frac{{\pi }^{2}}{4}-2(0+1)$$⇒I=4π2​−2(0+1)

$$\Rightarrow I=\frac{{\pi }^{2}}{4}-2$$⇒I=4π2​−2

$$\Rightarrow I={(\frac{\pi }{2})}^{2}-2$$⇒I=(2π​)2−2

$$\therefore \alpha =2$$∴α=2