Question

If $f(\alpha )=\underset{1}{\overset{\alpha }{\int }}\frac{{\log }_{10}t}{1+t}dt,\alpha >0$, then $f\left(e^3\right)+f\left(e^{-3}\right)$ is equal to :

• A. 9
• B. $\frac{9}{2}$
• C. $\frac{9}{{\log }_c(10)}$
• D. $\frac{9}{2{\log }_c(10)}$

Answer 1

Answer: (D)

$f\left(e^3\right)={\int }_1^{c^3}\frac{\ell nt}{\ell n10(1+t})dt\dots ..(1)$
$f(\alpha )={\int }_1^{\alpha }\frac{\ell nt}{(\ell n10)(1+t}dt$
$t=\frac{1}{x}\Rightarrow x=\frac{1}{t}$
$dt=\frac{-1}{x^2}dx$
$={\int }_1^{\frac{1}{\alpha }}\frac{-\ell nx}{\left(\ell n10()+\frac{1}{x}\right)}\left(-\frac{1}{x^2}\right)dx$
$f(\alpha )=\frac{1}{\ell n10}{\int }_1^{\frac{1}{\alpha }}\frac{\ell nx}{x(x+1}dx$
$f\left(e^{-3}\right)=\frac{1}{\ell n10}{\int }_1^{e^3}\frac{\ell nt}{t(t+1}dt\dots \dots \dots (2)$
$Add(1)\&(2)$
$f\left(e^3\right)+f\left(e^{-3}\right)$
$=\left(\frac{1}{\ell n10}\right){\int }_1^{c^3}\frac{\ell nt}{(1+t}\left[1+\frac{1}{t}\right]dt$
$=\left(\frac{1}{\ell n10}\right){\int }_1^3\frac{\ell nt}{t}dt$
$\ell nt=r$
$\frac{dt}{t}=dr$
$=\frac{1}{\ell n10}{\int }_0^{3}rdr$
$={\left(\frac{1}{\ell n10}\right)\left(\frac{r^2}{2}\right)|}_0^3$
$=\left(\frac{1}{\log 10}\right)\left(\frac{9}{2}\right)$
$=\frac{9}{2{\log }_{c}10}$