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 $[\mathrm{t}]$ is the greatest integer less than or equal to $t$. If $\lim _{x \rightarrow-1} f(x)$ exists, then the value of $\int_{0}^{4} f(x) d x$ is equal to :

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Continuity and Differentiability

Let

f : R \to R

be a function defined as

f (x) = a sin (\frac{π [ x ]}{2}) + [2 - x], a \in R

, where

[t]

is the greatest integer less than or equal to

t

. If

lim_{x \to - 1} f (x)

exists, then the value of

\int_{0}^{4} f (x) d x

is equal to :

Answer: (B)

For the limit to exist at

x = - 1

lim_{x \to - 1^{+}} f (x) = a sin (\frac{π ( - 1 )}{2}) + [2 - (- 1)] = - a + 2

lim_{x \to - 1^{-}} f (x) = a sin (\frac{π ( - 2 )}{2}) + [2 - (- 1)] = 0 + 3 = 3

For the limit to exist:

- a + 2 = 3

, so

a = - 1

.

Now compute

\int_{0}^{4} f (x) d x

with

a = - 1

\int_{0}^{1} (0 + 1) d x + \int_{1}^{2} (- 1 + 0) d x + \int_{2}^{3} (0 - 1) d x + \int_{3}^{4} (1 - 2) d x

= 1 - 1 - 1 - 1 = - 2

. Answer: (B)

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option