Let `A ={x in RR : [X+3] + [X+4],le 3}`
`B ={x in RR :3^x(sum_(r=1)^oo3/(10^r))^(x-3) gt 3^(-3x)}` where [t] denotes greatest integer function. Then,

Options

A. `B sub C ,A!= B`
B. ` A nn B ,= phi`
C. `A sub B ,A!= B`
D. `A= B`

Answer

To solve the problem, we need to analyze the two sets \( A \) and \( B \) defined in the question.

### Step 1: Analyze Set \( A \)

Set \( A \) is defined as:
\[ A = \{ x \in \mathbb{R} : [x + 3] + [x + 4] \leq 3 \} \]
where \([t]\) denotes the greatest integer function.

We can rewrite the condition:
\[ [x + 3] + [x + 4] \leq 3 \]

### Step 2: Simplify the Condition for Set \( A \)

The greatest integer function has the property that:
\[ [x + 4] = [x + 3] + 1 \quad \text{if } x \text{ is not an integer} \]
\[ [x + 4] = [x + 3] \quad \text{if } x \text{ is an integer} \]

Thus, we can analyze two cases:

1. **Case 1:** \( x \) is an integer.
\[ [x + 3] + [x + 4] = [x + 3] + [x + 3] = 2[x + 3] \leq 3 \]
This gives:
\[ [x + 3] \leq 1 \implies x + 3 \leq 1 \implies x \leq -2 \]

2. **Case 2:** \( x \) is not an integer.
\[ [x + 3] + [x + 4] = [x + 3] + [x + 3] + 1 = 2[x + 3] + 1 \leq 3 \]
This gives:
\[ 2[x + 3] \leq 2 \implies [x + 3] \leq 1 \implies x + 3 \leq 1 \implies x \leq -2 \]

In both cases, we find that:
\[ x \leq -2 \]

Thus, the set \( A \) can be expressed as:
\[ A = (-\infty, -2] \]

### Step 3: Analyze Set \( B \)

Set \( B \) is defined as:
\[ B = \{ x \in \mathbb{R} : 3^x \left( \sum_{r=1}^{\infty} \frac{3}{10^r} \right)^{x-3} > 3^{-3x} \} \]

### Step 4: Simplify the Condition for Set \( B \)

The series \( \sum_{r=1}^{\infty} \frac{3}{10^r} \) is a geometric series with:
- First term \( a = \frac{3}{10} \)
- Common ratio \( r = \frac{1}{10} \)

The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{\frac{3}{10}}{1 - \frac{1}{10}} = \frac{\frac{3}{10}}{\frac{9}{10}} = \frac{3}{9} = \frac{1}{3} \]

Thus, we can rewrite the condition for set \( B \):
\[ 3^x \left( \frac{1}{3} \right)^{x-3} > 3^{-3x} \]

### Step 5: Further Simplification

This simplifies to:
\[ 3^x \cdot 3^{-(x-3)} > 3^{-3x} \]
which means:
\[ 3^x \cdot 3^{-x + 3} > 3^{-3x} \]
or:
\[ 3^3 > 3^{-3x} \]
This implies:
\[ 27 > 3^{-3x} \]

Taking logarithm base 3 on both sides:
\[ 3 > -3x \implies x > -1 \]

Thus, the set \( B \) can be expressed as:
\[ B = (-1, \infty) \]

### Step 6: Conclusion

Now we have:
- \( A = (-\infty, -2] \)
- \( B = (-1, \infty) \)

Since there is no overlap between the two sets, we conclude that:
\[ A \neq B \]

### Final Answer

The answer is that \( A \) and \( B \) are not equal.