To find \( f(2.5) \) where \( f(x) = \lfloor x \rfloor + \lfloor -x \rfloor \), let's evaluate each part of the function:
Step 1: Calculate \( \lfloor 2.5 \rfloor \)
The floor function \( \lfloor x \rfloor \) returns the greatest integer less than or equal to \( x \). Thus, \( \lfloor 2.5 \rfloor = 2 \).
Step 2: Calculate \( \lfloor -2.5 \rfloor \)
For \( \lfloor -x \rfloor \), we apply the floor function to \( -2.5 \). Since \( -2.5 \) is between \(-3\) and \(-2\), the greatest integer less than or equal to \(-2.5\) is \(-3\). Hence, \( \lfloor -2.5 \rfloor = -3 \).
Step 3: Compute \( f(2.5) \)
Substitute the results from the previous steps into the function:
\( f(2.5) = \lfloor 2.5 \rfloor + \lfloor -2.5 \rfloor = 2 + (-3) = -1 \).
Therefore, the value of \( f(2.5) \) is \(-1\).
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |