If f be the greatest integer function and g be the moduls function, then `(gof) (-(5)/(2)) - (fog ) (-(5)/(3))=`

Options

A. 1
B. `-1`
C. 2
D. 4

Answer

To solve the problem, we need to evaluate the expression \((g \circ f) \left(-\frac{5}{2}\right) - (f \circ g) \left(-\frac{5}{3}\right)\).

### Step 1: Evaluate \(f\left(-\frac{5}{2}\right)\)
The function \(f\) is the greatest integer function, which gives the largest integer less than or equal to a given number.

\[ f\left(-\frac{5}{2}\right) = f\left(-2.5\right) = -3 \]

**Hint:** The greatest integer function rounds down to the nearest integer.

### Step 2: Evaluate \(g(f(-\frac{5}{2}))\)
Now we need to apply the modulus function \(g\) to the result from step 1.

\[ g(f(-\frac{5}{2})) = g(-3) = | -3 | = 3 \]

**Hint:** The modulus function converts negative numbers to positive.

### Step 3: Evaluate \(g\left(-\frac{5}{3}\right)\)
Next, we evaluate \(g\) at \(-\frac{5}{3}\).

\[ g\left(-\frac{5}{3}\right) = | -\frac{5}{3} | = \frac{5}{3} \]

**Hint:** Again, apply the modulus function to get the positive value.

### Step 4: Evaluate \(f(g(-\frac{5}{3}))\)
Now we apply the greatest integer function \(f\) to the result from step 3.

\[ f(g(-\frac{5}{3})) = f\left(\frac{5}{3}\right) = 1 \]

**Hint:** The greatest integer function rounds down to the nearest integer.

### Step 5: Combine the results
Now we can substitute the results back into the original expression:

\[ (g \circ f) \left(-\frac{5}{2}\right) - (f \circ g) \left(-\frac{5}{3}\right) = 3 - 1 = 2 \]

### Final Answer
\[ \text{The final result is } 2. \]

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