Question

If $f(\alpha)$ =$\int_{1}^{\alpha}\frac{log_{10}t}{1+t}dt$ , $\alpha>0$, then f(e3) + f(e–3) is equal to :

• A. 9
• B. $\frac{9}{2}$
• C. $\frac{9}{log_e{10}}$
• D. \frac{9}{2}$$log_e10

Answer 1

Answer: (D)

\frac{9}{2}$$log_e10

Answer 2

f(α)=\int_{1}^{\alpha}$$\frac{log 10t}{1+t} dt⋯(i)
f($\frac{1}{\alpha}$)=$\int_{1}^{\frac{1}{\alpha}}\frac{log_{10}t}{1+t}dt$
Substituting t = $\frac{1}{p}$
f($\frac{1}{\alpha}$)=$\int_{1}^{\frac{1}{\alpha}}\frac{log_{10}(\frac{1}{p})}dt$
=$\int_{1}^{\alpha} \frac{log_{10}p}{p(p+1)}dp$=$\int_{1}^{\alpha}(log \frac{10t }{t}-\frac{log\,10t}{t}+1)$dt ⋯(ii)
By (i) + (ii)
f(α)+f($\frac{1}{\alpha}$)=$\int_{1}^{\alpha}$ $\frac{log\,10t}{t}$ dt=$\int_{1}^{\alpha}$ $\frac{ln\,t}{t.log_{10}e}$dt
α=e3⇒f(e3)+f(e−3)=$\frac{9}{2}log_e10$
So, the correct option is (D): $\frac{9}{2}log_e10$