Question

If the value of definite integral $\underset{1}{\overset{a}{\int }}x\cdot a^{-\left[{\log }_{a}x\right]}dx$ where $a>1$, and $[x]$ denotes the greatest integer, is $\frac{e-1}{2}$ then the value of 'a' equals

• A. $\sqrt{e}$
• B. $e$
• C. $\sqrt{e+1}$
• D. $e-1$

Answer 1

Answer: (A)

$\underset{1}{\overset{a}{\int }}x\cdot a^{-\left[{\log }_{a}x\right]}dx$ $\text{ put }{\log }_{a}x=t\Rightarrow a^t=x$ $I=\ln a\cdot \underset{0}{\overset{1}{\int }}\left(a^t\cdot a^{-[t]}\cdot a^t\right)dt=\ln a\cdot \underset{0}{\overset{1}{\int }}\left(a^{t-[t]}\cdot a^t\right)dt=\ln a\cdot {\int }_0^1\left(a^{\{t\}}\cdot a^t\right)dt=\ln a\cdot \underset{0}{\overset{1}{\int }}{a}^{2t}dt$ ${=\frac{\ln a}{2}\cdot \frac{a^{2t}}{\ln a}]}_0^1=\frac{1}{2}\left(a^2-1\right)$ $\text{ (as }\{t\}=t\text{ if }t\in (0,1)\text{ ) }$ $\therefore \frac{1}{2}\left(a^2-1\right)=\frac{c-1}{2}\Rightarrow a=\sqrt{e}$