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(B) For the limit $\lim_{x \rightarrow -1} f(x)$ to exist,the left-hand limit and right-hand limit must be equal.Right-hand limit: $\lim_{x \rightarro

Vedclass · Accepted Answer

$-2$

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(B) For the limit $\lim_{x \rightarrow -1} f(x)$ to exist,the left-hand limit and right-hand limit must be equal.Right-hand limit: $\lim_{x \rightarrow -1^+} f(x) = a \sin \left(\frac{\pi(-1)}{2}\right) + [2 - (-1^+)] = a \sin(-\pi/2) + [3^-] = -a + 2$.Left-hand limit: $\lim_{x \rightarrow -1^-} f(x) = a \sin \left(\frac{\pi(-2)}{2}\right) + [2 - (-1^-)] = a \sin(-\pi) + [3^+] = 0 + 3 = 3$.Equating the two: $-a + 2 = 3 \implies a = -1$.Now,we calculate $\int_{0}^{4} f(x) dx$ with $a = -1$,so $f(x) = -\sin \left(\frac{\pi[x]}{2}\right) + [2-x]$.$\int_{0}^{4} f(x) dx = \int_{0}^{1} f(x) dx + \int_{1}^{2} f(x) dx + \int_{2}^{3} f(x) dx + \int_{3}^{4} f(x) dx$.For $x \in [0, 1)$,$[x] = 0$,$[2-x] = 1 \implies f(x) = 1$.For $x \in [1, 2)$,$[x] = 1$,$[2-x] = 0 \implies f(x) = -\sin(\pi/2) + 0 = -1$.For $x \in [2, 3)$,$[x] = 2$,$[2-x] = -1 \implies f(x) = -\sin(\pi) - 1 = -1$.For $x \in [3, 4)$,$[x] = 3$,$[2-x] = -2 \implies f(x) = -\sin(3\pi/2) - 2 = 1 - 2 = -1$.Integral = $\int_{0}^{1} (1) dx + \int_{1}^{2} (-1) dx + \int_{2}^{3} (-1) dx + \int_{3}^{4} (-1) dx = 1 - 1 - 1 - 1 = -2$.

Let $f: R \rightarrow R$ 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 $\lim_{x \rightarrow -1} f(x)$ exists,then the value of $\int_{0}^{4} f(x) dx$ is equal to.

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