If ` f(x)=x^(3)+x^(2)+kx+4` is always increasing then least positive integral value of k is

Options

A. 4
B. 3
C. 2
D. 1

Answer

To determine the least positive integral value of \( k \) such that the function \( f(x) = x^3 + x^2 + kx + 4 \) is always increasing, we will follow these steps:

### Step 1: Find the derivative of \( f(x) \)
The first step is to find the derivative of the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 + x^2 + kx + 4) \]
Calculating the derivative:
\[ f'(x) = 3x^2 + 2x + k \]

### Step 2: Set the condition for the function to be always increasing
For the function \( f(x) \) to be always increasing, its derivative \( f'(x) \) must be greater than or equal to zero for all \( x \):
\[ f'(x) \geq 0 \quad \text{for all } x \in \mathbb{R} \]

### Step 3: Analyze the quadratic equation
The expression \( f'(x) = 3x^2 + 2x + k \) is a quadratic equation in \( x \). A quadratic function is always positive if its discriminant is less than zero. The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]
For our derivative:
- \( a = 3 \)
- \( b = 2 \)
- \( c = k \)

Thus, the discriminant \( D \) is:
\[ D = 2^2 - 4 \cdot 3 \cdot k = 4 - 12k \]

### Step 4: Set the discriminant less than zero
To ensure that \( f'(x) \) is always greater than zero, we set the discriminant less than zero:
\[ 4 - 12k < 0 \]

### Step 5: Solve the inequality
Rearranging the inequality:
\[ 4 < 12k \]
Dividing both sides by 12:
\[ \frac{4}{12} < k \quad \Rightarrow \quad \frac{1}{3} < k \]

### Step 6: Determine the least positive integral value of \( k \)
The least positive integral value of \( k \) that satisfies \( k > \frac{1}{3} \) is:
\[ k = 1 \]

### Final Answer
Thus, the least positive integral value of \( k \) is:
\[ \boxed{1} \]