Question

If $f\left(x\right)=3\left[x\right]+5$ and $f\left(x\right)=5[x-2]+7$ then ${\int }_1^{2}x[x+f(x\left)\right]dx$ equals, where [.] respresents greatest integer function

• A. $63$
• B. $\frac{63}{2}$
• C. $126$
• D. None of these

Answer 1

Answer: (B)

$\frac{63}{2}$

$f\left(x\right)=3\left[x\right]+5=5[x-2]+7\Rightarrow 3\left[x\right]+5=5\left[x\right]-10+7\Rightarrow 2\left[x\right]=8\Rightarrow \left[x\right]=4\Rightarrow xϵ[4,5)\Rightarrow 4\le x<5$
$\therefore x=4+$ fraction part
$\therefore f\left(x\right)=3\left[x\right]+5=5[x-2]+7\therefore f\left(4\right)=12+5=17$
$\therefore [x+f(x\left)\right]=$[4 + fraction part +17]=21
Now ${\int }_1^{2}x[x+f(x\left)\right]dx={\int }_1^{2}21xdx=\frac{63}{2}$