Question

If the value of definite integral $\int\limits_1^a x \cdot a ^{-\left[\log _{ a } x \right]} dx$ 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)

$ \int\limits_1^a x \cdot a^{-\left[\log _a x\right]} d x$ $\text { put } \log _{ a } x = t \Rightarrow a ^{ t }= x $ $I =\ln a \cdot \int\limits_0^1\left( a ^{ t } \cdot a ^{-[ t ]} \cdot a ^{ t }\right) dt =\ln a \cdot \int\limits_0^1\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 \int\limits_0^1 a ^{2 t } dt $ $\left.=\frac{\ln a }{2} \cdot \frac{ a ^{2 t }}{\ln a }\right]_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 } $