Question

Let $\underset{0}{\overset{2[x+14]}{\int }}\left\{\frac{x}{2}\right\}dx=\underset{0}{\overset{\left\{x\right\}}{\int }}[x+14]dx$, where $[\cdot ]$ and $\{\cdot \}$ denote the greatest integer and fractional part of $x$ respectively. If $\left[x\right]+14=\lambda \left\{x\right\},$ $\lambda \in R,$ then the value of $\lambda$ is

Answer 1

Given: $\underset{0}{\overset{2[x+14]}{\int }}\left\{\frac{x}{2}\right\}dx=\underset{0}{\overset{\left\{x\right\}}{\int }}[x+14]dx$
Since $\underset{0}{\overset{nT}{\int }}f\left(x\right)dx=n\underset{0}{\overset{T}{\int }}f\left(x\right)dx,$ where $T$ is the period of $f\left(x\right),$ $n\in N$
$\therefore [x+14]\underset{0}{\overset{2}{\int }}\left\{\frac{x}{2}\right\}dx=14\underset{0}{\overset{\left\{x\right\}}{\int }}dx+\underset{0}{\overset{\left\{x\right\}}{\int }}\left[x\right]dx$
$\Rightarrow [x+14]\underset{0}{\overset{2}{\int }}\frac{x}{2}dx=14\left\{x\right\}+0$
$\therefore \left[x\right]+14=14\left\{x\right\}$