`lim_(x to 0) (A "sin" x + B log (1 + x^(2)) + C (1 - cos x))/(x^(2)) = 2` then
Statement-1 : A = 1
Statement-2 : 2B + C = 4
Statement-3 A + B + C = 0

Options

A. TTT
B. TTF
C. FTF
D. FFF

Answer

To solve the limit problem, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{A \sin x + B \log(1 + x^2) + C(1 - \cos x)}{x^2} = 2 \]

### Step 1: Expand the terms in the limit

1. **Expand \(\sin x\)** using its Taylor series:
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]

2. **Expand \(\log(1 + x^2)\)** using its Taylor series:
\[ \log(1 + x^2) = x^2 - \frac{x^4}{2} + O(x^6) \]

3. **Expand \(1 - \cos x\)** using its Taylor series:
\[ 1 - \cos x = \frac{x^2}{2} - \frac{x^4}{24} + O(x^6) \]

### Step 2: Substitute the expansions into the limit

Substituting these expansions into the limit, we get:

\[ \lim_{x \to 0} \frac{A\left(x - \frac{x^3}{6} + O(x^5)\right) + B\left(x^2 - \frac{x^4}{2} + O(x^6)\right) + C\left(\frac{x^2}{2} - \frac{x^4}{24} + O(x^6)\right)}{x^2} \]

### Step 3: Simplify the expression

This simplifies to:

\[ \lim_{x \to 0} \frac{Ax - \frac{Ax^3}{6} + Bx^2 - \frac{Bx^4}{2} + \frac{Cx^2}{2} - \frac{Cx^4}{24}}{x^2} \]

Breaking this down, we can separate the terms:

\[ = \lim_{x \to 0} \left( \frac{A}{x} - \frac{A}{6}x + B + \frac{C}{2} - \frac{B}{2}x^2 - \frac{C}{24}x^2 \right) \]

### Step 4: Evaluate the limit as \(x \to 0\)

As \(x \to 0\), the term \(\frac{A}{x}\) will diverge unless \(A = 0\). Therefore, we set \(A = 0\).

Now our limit simplifies to:

\[ \lim_{x \to 0} \left( B + \frac{C}{2} \right) = 2 \]

This gives us the equation:

\[ B + \frac{C}{2} = 2 \quad \text{(1)} \]

### Step 5: Analyze the second term

From the expansion of \(\log(1 + x^2)\):

\[ \lim_{x \to 0} \frac{B \log(1 + x^2)}{x^2} = B \]

And for \(1 - \cos x\):

\[ \lim_{x \to 0} \frac{C(1 - \cos x)}{x^2} = \frac{C}{2} \]

Thus, we have:

\[ B + \frac{C}{2} = 2 \]

### Step 6: Analyze the third term

From the expansion of \(1 - \cos x\):

\[ \lim_{x \to 0} \frac{C(1 - \cos x)}{x^2} = \frac{C}{2} \]

### Step 7: Solve for \(C\)

From the limit, we also have:

\[ 2B + C = 4 \quad \text{(2)} \]

### Step 8: Solve the equations

Now we have two equations:

1. \(B + \frac{C}{2} = 2\)
2. \(2B + C = 4\)

From equation (1), we can express \(C\) in terms of \(B\):

\[ C = 4 - 2B \]

Substituting this into equation (1):

\[ B + \frac{4 - 2B}{2} = 2 \]
\[ B + 2 - B = 2 \quad \Rightarrow \quad 2 = 2 \]

This is always true, so we need another condition.

### Step 9: Find \(A + B + C = 0\)

Since \(A = 0\), we can conclude:

\[ 0 + B + C = 0 \quad \Rightarrow \quad B + C = 0 \quad \text{(3)} \]

### Step 10: Solve for \(B\) and \(C\)

From equation (2):

\[ C = 4 - 2B \]

Substituting into equation (3):

\[ B + (4 - 2B) = 0 \quad \Rightarrow \quad 4 - B = 0 \quad \Rightarrow \quad B = 4 \]

Then substituting back to find \(C\):

\[ C = 4 - 2(4) = -4 \]

### Final Values

Thus we have:

- \(A = 0\)
- \(B = 4\)
- \(C = -4\)

### Conclusion

Now we can check the statements:

1. **Statement 1**: \(A = 1\) → False
2. **Statement 2**: \(2B + C = 4\) → True (since \(2(4) - 4 = 4\))
3. **Statement 3**: \(A + B + C = 0\) → True (since \(0 + 4 - 4 = 0\))