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

The Correct Option is D

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): \(\frac{9}{2}log_e10\)