Let `f(x)=x^3+kx^2+5x+4sin^2x` be an increasing function on `x in R.` Then domain of k is

Options

A. `(-sqrt2,sqrt2)`
B. `(-oo, sqrt3)`
C. `(sqrt3, oo)`
D. `(-sqrt3, sqrt3)`

Answer

To find the domain of \( k \) such that the function \( f(x) = x^3 + kx^2 + 5x + 4\sin^2 x \) is an increasing function for all \( x \in \mathbb{R} \), we will follow these steps:

### Step 1: Find the first derivative of the function
The first derivative \( f'(x) \) must be greater than or equal to zero for \( f(x) \) to be an increasing function.

\[ f'(x) = \frac{d}{dx}(x^3 + kx^2 + 5x + 4\sin^2 x) \]

Using the power rule and the chain rule, we differentiate:

\[ f'(x) = 3x^2 + 2kx + 5 + 4 \cdot 2\sin x \cos x \]

Recognizing that \( 4\sin x \cos x = 2\sin 2x \), we can rewrite the derivative as:

\[ f'(x) = 3x^2 + 2kx + 5 + 2\sin 2x \]

### Step 2: Set up the inequality
For \( f(x) \) to be increasing, we need:

\[ 3x^2 + 2kx + 5 + 2\sin 2x \geq 0 \]

### Step 3: Analyze the sine term
The term \( 2\sin 2x \) oscillates between -2 and 2. Therefore, we can analyze the worst-case scenario by considering the minimum value of \( 2\sin 2x \), which is -2.

Thus, we need to ensure:

\[ 3x^2 + 2kx + 5 - 2 \geq 0 \]

This simplifies to:

\[ 3x^2 + 2kx + 3 \geq 0 \]

### Step 4: Determine the conditions for the quadratic
For the quadratic \( 3x^2 + 2kx + 3 \) to be non-negative for all \( x \), its discriminant must be less than or equal to zero.

The discriminant \( D \) is given by:

\[ D = b^2 - 4ac = (2k)^2 - 4 \cdot 3 \cdot 3 \]

Calculating this gives:

\[ D = 4k^2 - 36 \]

Setting the discriminant less than or equal to zero:

\[ 4k^2 - 36 < 0 \]

### Step 5: Solve the inequality
Solving for \( k \):

\[ 4k^2 < 36 \]

Dividing both sides by 4:

\[ k^2 < 9 \]

Taking the square root of both sides gives:

\[ -\sqrt{9} < k < \sqrt{9} \]

Thus:

\[ -3 < k < 3 \]

### Conclusion
The domain of \( k \) such that \( f(x) \) is an increasing function on \( \mathbb{R} \) is:

\[ k \in (-3, 3) \]