If `f(x)` be a cubic polynomial and `lim_(x->0)(sin^2x)/(f(x))=1/3` then `f(1)` can not be equal to :

Options

A. 0
B. `-5`
C. `3`
D. `-2`

Answer

To solve the problem, we need to analyze the limit given and apply L'Hôpital's rule appropriately. Let's go through the steps systematically.

### Step-by-Step Solution

1. **Understanding the Limit**: We are given that
\[ \lim_{x \to 0} \frac{\sin^2 x}{f(x)} = \frac{1}{3} \]
where \( f(x) \) is a cubic polynomial.

2. **Form of the Polynomial**: A cubic polynomial can be expressed as:
\[ f(x) = ax^3 + bx^2 + cx + d \]

3. **Evaluating the Limit**: As \( x \to 0 \), we know that \( \sin^2 x \) approaches \( 0 \). Therefore, for the limit to exist and equal \( \frac{1}{3} \), \( f(x) \) must also approach \( 0 \) as \( x \to 0 \). This implies that \( f(0) = d = 0 \).

4. **Substituting into the Limit**: We can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\sin^2 x}{ax^3 + bx^2 + cx} = \frac{1}{3} \]

5. **Applying L'Hôpital's Rule**: Since both the numerator and denominator approach \( 0 \) as \( x \to 0 \), we can apply L'Hôpital's rule:
\[ \lim_{x \to 0} \frac{2\sin x \cos x}{3ax^2 + 2bx + c} \]

6. **Evaluating the Derivative**: As \( x \to 0 \), \( \sin x \to 0 \) and \( \cos x \to 1 \), so:
\[ \lim_{x \to 0} \frac{2\sin x \cos x}{3ax^2 + 2bx + c} = \lim_{x \to 0} \frac{0}{c} = 0 \]
This means \( c \) must also be \( 0 \) for the limit to exist.

7. **Reapplying L'Hôpital's Rule**: Now, we have:
\[ f(x) = ax^3 + bx^2 \]
We apply L'Hôpital's rule again:
\[ \lim_{x \to 0} \frac{2\sin x \cos x}{3ax^2 + 2bx} \]

8. **Evaluating Again**: As \( x \to 0 \):
\[ \lim_{x \to 0} \frac{2\sin x \cos x}{3ax^2 + 2bx} = \lim_{x \to 0} \frac{2 \cdot 0 \cdot 1}{0} = \text{still } 0 \]
This implies \( b \) must also be \( 0 \).

9. **Final Form of f(x)**: Therefore, we have:
\[ f(x) = ax^3 \]

10. **Setting the Limit**: Now we can set the limit:
\[ \lim_{x \to 0} \frac{\sin^2 x}{ax^3} = \frac{1}{3} \]
Using the fact that \( \sin^2 x \sim x^2 \) as \( x \to 0 \):
\[ \lim_{x \to 0} \frac{x^2}{ax^3} = \lim_{x \to 0} \frac{1}{ax} = \frac{1}{3} \]
For this limit to equal \( \frac{1}{3} \), we need \( a = 3 \).

11. **Finding f(1)**: Thus, we have:
\[ f(x) = 3x^3 \]
Therefore,
\[ f(1) = 3 \cdot 1^3 = 3 \]

12. **Conclusion**: The question asks for the value that \( f(1) \) cannot be equal to. Since \( f(1) = 3 \), the answer is:
\[ \text{f(1) cannot be equal to } 3. \]