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 $\{\cdot \}$ denotes the greatest integer and fractional part of $x$ respectively. Then $\frac{\left[x\right]+14}{14}$ is equal to

• A. $\left\{\frac{x}{2}\right\}$
• B. $\left\{x\right\}$
• C. $x+\left\{x\right\}$
• D. $2\left\{x\right\}$

Answer 1

The correct option is B $\left\{x\right\}$

Given: $\underset{0}{\overset{2[x+14]}{\int }}\left\{\frac{x}{2}\right\}dx=\underset{0}{\overset{\left\{x\right\}}{\int }}[x+14]dx$
Now,
$\Rightarrow \underset{0}{\overset{28+2\left[x\right]}{\int }}\left\{\frac{x}{2}\right\}dx=\underset{0}{\overset{\left\{x\right\}}{\int }}(14+[x\left]\right)dx\Rightarrow \underset{0}{\overset{28}{\int }}\left\{\frac{x}{2}\right\}dx+\underset{28}{\overset{28+2\left[x\right]}{\int }}\left\{\frac{x}{2}\right\}dx=\underset{0}{\overset{\left\{x\right\}}{\int }}14dx+\underset{0}{\overset{\left\{x\right\}}{\int }}\left[x\right]dx\Rightarrow 14\underset{0}{\overset{2}{\int }}\frac{x}{2}dx+\underset{0}{\overset{2\left[x\right]}{\int }}\left\{\frac{x}{2}\right\}dx=14\left\{x\right\}+\underset{0}{\overset{\left\{x\right\}}{\int }}\left[x\right]dx\Rightarrow 14+\underset{0}{\overset{2\left[x\right]}{\int }}\left\{\frac{x}{2}\right\}dx=14\left\{x\right\}+\underset{0}{\overset{\left\{x\right\}}{\int }}\left[x\right]dx\Rightarrow 14+\left[x\right]\underset{0}{\overset{2}{\int }}\frac{x}{2}dx=14\left\{x\right\}+0\Rightarrow 14+\left[x\right]=14\left\{x\right\}\Rightarrow \left[x\right]=14\left(\right\{x\}-1)$
$\Rightarrow \frac{\left[x\right]+14}{14}=\left\{x\right\}$