If `f(x)=x^(3)+3x^(2)+4x+asinx+bcosx, forall x in R` is a one-one fuction, then the greatest value of `(a^(2)+b^(2))` is

Options

A. (a)1
B. (b)2
C. (c)`sqrt(2)`
D. (d)None of these

Answer

To solve the problem, we need to determine the conditions under which the function \( f(x) = x^3 + 3x^2 + 4x + a \sin x + b \cos x \) is a one-to-one function. This can be done by analyzing the derivative of the function.

### Step-by-step Solution:

1. **Find the Derivative of \( f(x) \)**:
\[ f'(x) = \frac{d}{dx}(x^3 + 3x^2 + 4x + a \sin x + b \cos x) \]
Using the power rule and the derivatives of sine and cosine, we get:
\[ f'(x) = 3x^2 + 6x + 4 + a \cos x - b \sin x \]

2. **Condition for One-to-One Function**:
For \( f(x) \) to be a one-to-one function, \( f'(x) \) must be greater than or equal to zero for all \( x \in \mathbb{R} \):
\[ f'(x) \geq 0 \]
This implies:
\[ 3x^2 + 6x + 4 + a \cos x - b \sin x \geq 0 \]

3. **Analyze the Quadratic Part**:
The quadratic part \( 3x^2 + 6x + 4 \) can be analyzed for its minimum value. The vertex of the quadratic \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \):
\[ x = -\frac{6}{2 \cdot 3} = -1 \]
Now, evaluate the quadratic at \( x = -1 \):
\[ 3(-1)^2 + 6(-1) + 4 = 3 - 6 + 4 = 1 \]
Therefore, the minimum value of \( 3x^2 + 6x + 4 \) is 1.

4. **Bound the Trigonometric Terms**:
The expression \( a \cos x - b \sin x \) can be bounded using the Cauchy-Schwarz inequality:
\[ |a \cos x - b \sin x| \leq \sqrt{a^2 + b^2} \]

5. **Combine the Inequalities**:
From the previous steps, we have:
\[ 1 + |a \cos x - b \sin x| \geq 0 \]
Thus:
\[ 1 - \sqrt{a^2 + b^2} \geq 0 \]
This leads to:
\[ \sqrt{a^2 + b^2} \leq 1 \]
Squaring both sides gives:
\[ a^2 + b^2 \leq 1 \]

6. **Find the Greatest Value**:
The greatest value of \( a^2 + b^2 \) that satisfies this inequality is:
\[ a^2 + b^2 = 1 \]

### Final Answer:
The greatest value of \( a^2 + b^2 \) is \( \boxed{1} \).