If `f(x)=x^3+3x^2+4x+ bsinx+ c cosx AA x in R` is a one-one function then the value of `b^2 + c^2` is

Options

A. `ge 1`
B. `ge 2`
C. `le 1`
D. None of these

Answer

To solve the problem, we need to determine the value of \( b^2 + c^2 \) such that the function \( f(x) = x^3 + 3x^2 + 4x + b \sin x + c \cos x \) is a one-one function. A function is one-one if its derivative is always positive or always negative.

### Step-by-Step Solution:

1. **Differentiate the Function:**
\[ f'(x) = \frac{d}{dx}(x^3 + 3x^2 + 4x + b \sin x + c \cos x) \]
Using the rules of differentiation, we find:
\[ f'(x) = 3x^2 + 6x + 4 + b \cos x - c \sin x \]

2. **Analyze the Quadratic Part:**
The quadratic part of the derivative is \( 3x^2 + 6x + 4 \). We will find its minimum value to ensure that the entire derivative \( f'(x) \) is always positive.

3. **Find the Vertex of the Quadratic:**
The vertex of a quadratic \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \):
\[ x = -\frac{6}{2 \cdot 3} = -1 \]

4. **Calculate the Minimum Value of the Quadratic:**
Substitute \( x = -1 \) into the quadratic:
\[ 3(-1)^2 + 6(-1) + 4 = 3 - 6 + 4 = 1 \]
Thus, the minimum value of \( 3x^2 + 6x + 4 \) is \( 1 \).

5. **Evaluate the Trigonometric Part:**
We need to ensure that the entire derivative \( f'(x) \) is always positive:
\[ f'(x) = 1 + b \cos x - c \sin x \]
To maintain the condition \( f'(x) \geq 0 \), we need:
\[ 1 + b \cos x - c \sin x \geq 0 \]

6. **Find the Minimum of the Trigonometric Expression:**
The expression \( b \cos x - c \sin x \) can be rewritten in the form \( R \cos(x + \phi) \), where:
\[ R = \sqrt{b^2 + c^2} \]
The minimum value of \( b \cos x - c \sin x \) is \( -R \). Therefore:
\[ 1 - R \geq 0 \implies R \leq 1 \]
This leads to:
\[ \sqrt{b^2 + c^2} \leq 1 \]

7. **Square Both Sides:**
Squaring both sides gives:
\[ b^2 + c^2 \leq 1 \]

### Conclusion:
The value of \( b^2 + c^2 \) must be less than or equal to \( 1 \) for the function \( f(x) \) to be a one-one function.