Consider a function `f (x)` in `[0,2pi]` defined as :
` f(x)=[{:([sinx]+ [cos x],,, 0 le x le pi),( [sin x] -[cos x],,, pi lt x le 2pi):}`
where [.] denotes greatest integer function then.
`lim _(x to ((3pi)/(2))^+), f (x)` equals

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the problem, we need to find the limit of the function \( f(x) \) as \( x \) approaches \( \frac{3\pi}{2} \) from the right. The function is defined piecewise as follows:

\[ f(x) = \begin{cases} [\sin x] + [\cos x] & \text{for } 0 \leq x \leq \pi \\ [\sin x] - [\cos x] & \text{for } \pi < x \leq 2\pi \end{cases} \]

Since we are interested in the limit as \( x \) approaches \( \frac{3\pi}{2} \) from the right, we will use the second case of the function, which applies when \( \pi < x \leq 2\pi \):

\[ f(x) = [\sin x] - [\cos x] \]

### Step 1: Determine the values of \( \sin x \) and \( \cos x \) as \( x \) approaches \( \frac{3\pi}{2} \) from the right.

As \( x \) approaches \( \frac{3\pi}{2} \) from the right (i.e., \( x \to \frac{3\pi}{2}^+ \)):
- \( \sin x \) approaches \( -1 \)
- \( \cos x \) approaches \( 0 \)

### Step 2: Apply the greatest integer function to \( \sin x \) and \( \cos x \).

Now we apply the greatest integer function:
- \( [\sin x] \) approaches \( [-1] = -1 \)
- \( [\cos x] \) approaches \( [0] = 0 \)

### Step 3: Substitute these values into the function.

Substituting these values into the function \( f(x) \):
\[ f(x) = [\sin x] - [\cos x] = -1 - 0 = -1 \]

### Step 4: Write the limit.

Thus, we find:
\[ \lim_{x \to \frac{3\pi}{2}^+} f(x) = -1 \]

### Final Answer:
\[ \lim_{x \to \frac{3\pi}{2}^+} f(x) = -1 \]

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