Question

Let $f:R\to R$ be a function defined as
$f(x)=a\sin \left(\frac{\pi [x]}{2}\right)+[2-x],a\in R,$
where [t] is the greatest integer less than or equal to $t$. If $\underset{x\to -1}{\lim }f(x)$ exists, then the value of ${\int }_0^{4}f(x)dx$ is equal to :

• A. $-1$
• B. $-2$
• C. 1
• D. 2

Answer 1

Answer: (B)

$\underset{x\to -1^{+}}{\lim }a\sin \left(\pi \frac{[x]}{2}\right)+[2-x]=-a+2$
$\underset{x\to -1^{-}}{\lim }a\sin \left(\pi \frac{[x]}{2}\right)+[2-x]=$
$\underset{x\to -1}{\lim }f(x)\text{ exist when }a=-1$
Now,
$\underset{0}{\overset{4}{\int }}f(x)dx=\underset{0}{\overset{1}{\int }}f(x)dx+\underset{1}{\overset{2}{\int }}f(x)dx+\underset{2}{\overset{3}{\int }}f(x)dx+\underset{3}{\overset{4}{\int }}f(x)dx$
$=\underset{0}{\overset{1}{\int }}(0+1)dx+\underset{1}{\overset{2}{\int }}(-1+0)dx+\underset{2}{\overset{3}{\int }}(0-1)dx+$
$\underset{3}{\overset{4}{\int }}(1-2)dx$
$=1-1-1-1=-2$