Question

${\int }_1^{a}x.a^{-\left[lo{g}_{a}x\right]}dx=\frac{e-1}{2},\text{where}a>1\text{and}[.]$ denotes the greatest integer function, then the value of $a^2$is

• A. $e-1$
• B. $e$
• C. $e+1$
• D. $e^2-1$

Answer 1

The correct option is B

$e$

$\because 1<x<a\Rightarrow 0<lo{g}_{a}x<1\therefore \left[lo{g}_{a}x\right]=0\Rightarrow {\int }_1^{a}x.a^{\left[lo{g}_{a}x\right]}dx={\int }_1^{a}xdx={\left[\frac{x^2}{2}\right]}_1^a=\frac{a^2-1}{2}\Rightarrow \frac{a^2-1}{2}=\frac{e-1}{2}\therefore a^2=e$