Let [f: R R] be a function defined as [f(x)=a (([x])/(2))+[2-x], a R] where [[t]] is the g…

Question

Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a function defined as

$$f(x)=a \sin \left(\frac{\pi[x]}{2}\right)+[2-x], a \in \mathbb{R}$$ where $$[t]$$ is the greatest integer less than or equal to $t$. If $$\mathop {\lim }\limits_{x \to -1 } f(x)$$ exists, then the value of $$\int\limits_{0}^{4} f(x) d x$$ is equal to

Accepted Answer

$$f(x) = a\sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]\,a \in R$$

Now,

$$\because$$ $$\mathop {\lim }\limits_{x \to  - 1} f(x)$$ exist

$$\therefore$$ $$\mathop {\lim }\limits_{x \to  - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to  - {1^ + }} f(x)$$

$$ \Rightarrow a\sin \left( {{{ - 2\pi } \over 2}} \right) + 3 = a\sin \left( {{{ - \pi } \over 2}} \right) + 2$$

$$ \Rightarrow  - a = 1 \Rightarrow a =  - 1$$

Now, $$\int_0^4 {f(x)dx = \int_0^4 {\left( { - \sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]} \right)dx} } $$

$$ = \int_0^1 {1dx + \int_1^2 { - 1dx + \int_2^3 { - 1dx + \int_3^4 {(1 - 2)dx} } } } $$

$= 1 - 1 - 1 - 1 =  - 2$

Solutions · 1