JEE Main 2022 — Mathematics Calculus

Given,

$$f(\alpha )={\int }_{1}^{\alpha }\frac{{\mathrm{log}}_{10}t}{1+t}dt$$f(α)=∫1α​1+tlog10​t​dt

So, $\displaystyle f({e}^{3})={\int }_{1}^{{e}^{3}}\frac{\mathrm{ln}t}{\mathrm{ln}10(1+t)}\mathrm{dt}\cdots (1)$f(e3)=∫1e3​ln10(1+t)lnt​dt⋯(1)

Again $\displaystyle f(\alpha )={\int }_{1}^{\alpha }\frac{\mathrm{ln}t}{(\mathrm{ln}10)(1+t)}dt$f(α)=∫1α​(ln10)(1+t)lnt​dt

Now let $t=\frac{1}{x}\Rightarrow x=\frac{1}{t}$t=x1​⇒x=t1​$\Rightarrow dt=\frac{-1}{{x}^{2}}dx$⇒dt=x2−1​dx

So, $\displaystyle f(\alpha )={\int }_{1}^{\frac{1}{a}}\frac{-\mathrm{ln}nx}{(\mathrm{ln}10)(1+\frac{1}{x})}(-\frac{1}{{x}^{2}})dx$f(α)=∫1a1​​(ln10)(1+x1​)−lnnx​(−x21​)dx

$$\Rightarrow f(\alpha )=\frac{1}{ln10}{\int }_{1}^{\frac{1}{\alpha }}\frac{lnx}{x(x+1)}dx$$⇒f(α)=ln101​∫1α1​​x(x+1)lnx​dx

So, $\displaystyle f({e}^{-3})=\frac{1}{ln10}{\int }_{1}^{{e}^{3}}\frac{lnt}{t(t+1)}dt\ldots (2)$f(e−3)=ln101​∫1e3​t(t+1)lnt​dt…(2)

Now adding equation $(1)$(1) and $(2)$(2) we get,

$f({e}^{3})+f({e}^{-3})$f(e3)+f(e−3)$\displaystyle =(\frac{1}{\mathrm{ln}n10}){\int }_{1}^{{c}^{3}}\frac{lnt}{(1+t)}[1+\frac{1}{t}]dt$=(lnn101​)∫1c3​(1+t)lnt​[1+t1​]dt

$$\Rightarrow f({e}^{3})+f({e}^{-3})=(\frac{1}{\mathrm{ln}10}){\int }_{1}^{3}\frac{\mathrm{ln}t}{t}dt$$⇒f(e3)+f(e−3)=(ln101​)∫13​tlnt​dt

Now let $\mathrm{ln}t=r$lnt=r$\Rightarrow \frac{dt}{t}=dr$⇒tdt​=dr

So, $\displaystyle f({e}^{3})+f({e}^{-3})=\frac{1}{\mathrm{ln}10}{\int }_{0}^{3}rdr$f(e3)+f(e−3)=ln101​∫03​rdr

$$\Rightarrow f({e}^{3})+f({e}^{-3}){=(\frac{1}{\mathrm{ln}}10)(\frac{{r}^{2}}{2})}_{0}^{3}$$⇒f(e3)+f(e−3)=(ln1​10)(2r2​)03​

$$\Rightarrow f({e}^{3})+f({e}^{-3})=(\frac{1}{\mathrm{log}10})(\frac{9}{2})$$⇒f(e3)+f(e−3)=(log101​)(29​)

$$\Rightarrow f({e}^{3})+f({e}^{-3})=\frac{9}{2{\mathrm{log}}_{e}10}$$⇒f(e3)+f(e−3)=2loge​109​