Let f : R → R be a function defined as f(x) = a sin(π[x]/2) + [2-x], a ∈ R, where [t] is the greates

Q: Why does $[2-x]$ behave differently than $[x]$? When $x \to -1^-$, $x$ is like $-1.01$. So $2-x$ is $2 - (-1.01) = 3.01$. The greatest integer of $3.01$ is $3$. When $x \to -1^+$, $x$ is like $-0.99$. So $2-x$ is $2 - (-0.99) = 2.99$. The greatest integer of $2.99$ is $2$. Always substitute a test value like $-1.01$ or $-0.99$ if you get confused!

You have hit the nail on the head! The key is to treat the expression inside the bracket as a single entity. Always substitute a value slightly to the left or right of the point of interest. It is the most robust way to avoid sign errors.

Q: Can I just integrate $f(x)$ as a single expression? No! The greatest integer function is a step function. It changes its value at every integer. To integrate it, you must partition the interval $[0, 4]$ into unit intervals $[0, 1), [1, 2), [2, 3), [3, 4)$ where the function remains constant.

Exactly. The greatest integer function is not continuous; it is a collection of constant steps. You cannot use standard integration formulas like $\int x^n dx$ directly. You must treat it as a sum of integrals over intervals where the function is constant.

Q: Why did the sine term vanish in the LHL? Because $\sin(\frac{\pi(-2)}{2}) = \sin(-\pi) = 0$. Any integer multiple of $\pi$ inside a sine function is zero. It's a beautiful simplification that rewards you for being precise with your algebra.

That is the beauty of these problems! Often, the most intimidating terms are designed to simplify perfectly if you have correctly identified the integer values. Trust your algebra!

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ be a function defined as $f(x) = a \sin \left( \frac{\pi [x]}{2} ight) + [2-x], a \in \mathbb{R}$, where $[t]$ is the greatest integer less than or equal to $t$. If $\lim_{x 	o -1} f(x)$ exists, then the value of $\int_{0}^{4} f(x) dx$ is equal to :

Options:

(A) -1

(B) -2

(C) 1

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the function $f(x) = a \sin \left( \frac{\pi [x]}{2} ight) + [2-x]$. At first glance, it appears to be a complex mix of trigonometry and step functions.

However, the problem provides a golden key: the limit exists at $x = -1$. This condition is our entry point for determining the constant $a$.

### Phase 1: The Limit at the Edge

To ensure the limit exists at $x = -1$, we must satisfy the condition that the Left Hand Limit (LHL) equals the Right Hand Limit (RHL).

For the LHL, we approach from the left ($x 	o -1^-$). Here, $x$ is slightly less than $-1$ (e.g., $-1.01$), so $[x] = -2$ and $[2-x] = [2 - (-1.01)] = [3.01] = 3$.

Substituting these into the function:
$$f(x) = a \sin \left( \frac{\pi(-2)}{2} ight) + 3 = a \sin(-\pi) + 3$$

Since $\sin(-\pi) = 0$, the trigonometric term vanishes, leaving us with:
$$LHL = 3$$

For the RHL, we approach from the right ($x 	o -1^+$). Here, $x$ is slightly greater than $-1$ (e.g., $-0.99$), so $[x] = -1$ and $[2-x] = [2 - (-0.99)] = [2.99] = 2$.

Substituting these into the function:
$$f(x) = a \sin \left( \frac{\pi(-1)}{2} ight) + 2 = a \sin \left( -\frac{\pi}{2} ight) + 2$$

Since $\sin(-\frac{\pi}{2}) = -1$, we obtain:
$$RHL = -a + 2$$

Equating $LHL = RHL$, we solve for $a$:
$$3 = -a + 2 \Rightarrow a = -1$$

### Phase 2: The Integration Journey

With $a = -1$, our function is fully defined as $f(x) = -\sin \left( \frac{\pi [x]}{2} ight) + [2-x]$. We must evaluate the integral $\int_{0}^{4} f(x) dx$.

Because of the greatest integer function, we partition the interval $[0, 4]$ into four sub-intervals: $[0, 1), [1, 2), [2, 3),$ and $[3, 4)$.

For $x \in [0, 1)$:
$[x] = 0$ and $[2-x] = 1$. Thus, $f(x) = -\sin(0) + 1 = 1$.
$$\int_{0}^{1} 1 dx = 1$$

For $x \in [1, 2)$:
$[x] = 1$ and $[2-x] = 0$. Thus, $f(x) = -\sin \left( \frac{\pi}{2} ight) + 0 = -1$.
$$\int_{1}^{2} -1 dx = -1$$

For $x \in [2, 3)$:
$[x] = 2$ and $[2-x] = -1$. Thus, $f(x) = -\sin(\pi) - 1 = -1$.
$$\int_{2}^{3} -1 dx = -1$$

For $x \in [3, 4)$:
$[x] = 3$ and $[2-x] = -2$. Thus, $f(x) = -\sin \left( \frac{3\pi}{2} ight) - 2 = -(-1) - 2 = -1$.
$$\int_{3}^{4} -1 dx = -1$$

### The Grand Finale

Summing the results of these individual integrals, we obtain:
$$I = 1 + (-1) + (-1) + (-1) = -2$$

The final value of the integral is **$-2$**.

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Calculating LHL at $x = - 1$

Evaluating LHL Expression

Calculating RHL at $x = - 1$

Evaluating RHL Expression

Finding the Constant $a$

Defining the Final Function $f (x)$

Interval 1: $x \in [0, 1)$

Interval 2: $x \in [1, 2)$

Interval 3: $x \in [2, 3)$

Interval 4: $x \in [3, 4)$

Final Summation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

Phase 1

Phase 2

The Grand Finale

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Visualized Solution

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Calculating LHL at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

Evaluating LHL Expression

Calculating RHL at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

Evaluating RHL Expression

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Defining the Final Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Interval 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[0,1)

Interval 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[1,2)

Interval 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[2,3)

Interval 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[3,4)