Question

If $\underset{0}{\overset{11}{\int }}\frac{11^x}{11^{[x]}}dx=\frac{k}{\log 11}$, (where [ ] denotes greatest integer function) then value of $k$ is

• A. 11
• B. 101
• C. 110
• D. 111

Answer 1

Answer: (C)

$I=\underset{0}{\overset{11}{\int }}11^{(x)}dx=\underset{0}{\overset{11\times 1}{\int }}11^{(x)}dx=11\underset{0}{\overset{1}{\int }}11^{\{x\}}dx\{\because \{x\}$ is periodic with period 1$\}$
$=11\underset{0}{\overset{1}{\int }}11^{x}dx=11{\left[\frac{11^x}{\ell n11}\right]}_0^1=11\left[\frac{11}{\ell n11}-\frac{1}{\ell n11}\right]$
$=\frac{110}{\ell n11}=\frac{k}{\ell n11}\Rightarrow k=110$