Let `f(x)=x^(3)+ax^(2)+bx+5 sin^(2)x` be an increasing function on the set R. Then,

Options

A. `a^(2) -3b+15gt0`
B. `a^(2)-3b+5lt0`
C. `a^(2)-3b+15lt0`
D. `a^(2)-3b+5gt0`

Answer

To determine the conditions under which the function \( f(x) = x^3 + ax^2 + bx + 5\sin^2 x \) is an increasing function on the set of real numbers \( \mathbb{R} \), we will follow these steps:

### Step 1: Differentiate the function
To find whether the function is increasing, we first need to compute its derivative:

\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(ax^2) + \frac{d}{dx}(bx) + \frac{d}{dx}(5\sin^2 x) \]

Calculating each term, we have:

\[ f'(x) = 3x^2 + 2ax + b + 5 \cdot 2\sin x \cos x = 3x^2 + 2ax + b + 5\sin(2x) \]

### Step 2: Set the condition for increasing function
For \( f(x) \) to be an increasing function on \( \mathbb{R} \), the derivative \( f'(x) \) must be greater than or equal to zero for all \( x \):

\[ f'(x) \geq 0 \implies 3x^2 + 2ax + b + 5\sin(2x) \geq 0 \]

### Step 3: Analyze the sine term
The term \( 5\sin(2x) \) oscillates between -5 and 5. Therefore, the minimum value of \( f'(x) \) occurs when \( \sin(2x) = -1 \):

\[ f'(x) \geq 3x^2 + 2ax + b - 5 \]

### Step 4: Ensure the quadratic is non-negative
For \( f'(x) \) to be non-negative for all \( x \), the quadratic \( 3x^2 + 2ax + (b - 5) \) must have no real roots. This occurs when the discriminant is less than or equal to zero:

\[ D = (2a)^2 - 4 \cdot 3 \cdot (b - 5) \leq 0 \]

### Step 5: Simplify the discriminant condition
Calculating the discriminant:

\[ D = 4a^2 - 12(b - 5) \leq 0 \]
\[ 4a^2 - 12b + 60 \leq 0 \]

Dividing the entire inequality by 4 gives:

\[ a^2 - 3b + 15 \leq 0 \]

### Conclusion
Thus, for the function \( f(x) \) to be increasing on \( \mathbb{R} \), the condition that must be satisfied is:

\[ a^2 - 3b + 15 < 0 \]