The least integral value of 'k' If `(k-1) x ^(2) -(k+1) x+ (k+1)` is positive for all real value of x is:

Options

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

Answer

To find the least integral value of \( k \) such that the expression \( (k-1)x^2 - (k+1)x + (k+1) \) is positive for all real values of \( x \), we will analyze the quadratic expression.

### Step 1: Identify the quadratic expression
The given expression can be written as:
\[ p(x) = (k-1)x^2 - (k+1)x + (k+1) \]
Here, \( a = k-1 \), \( b = -(k+1) \), and \( c = k+1 \).

### Step 2: Conditions for positivity
For the quadratic \( p(x) \) to be positive for all real \( x \):
1. The coefficient \( a \) must be greater than 0:
\[ k - 1 > 0 \implies k > 1 \]
2. The discriminant \( D \) must be less than 0:
\[ D = b^2 - 4ac < 0 \]

### Step 3: Calculate the discriminant
Substituting \( a \), \( b \), and \( c \) into the discriminant:
\[ D = (-(k+1))^2 - 4(k-1)(k+1) \]
Calculating \( D \):
\[ D = (k+1)^2 - 4(k^2 - 1) \]
Expanding both terms:
\[ D = k^2 + 2k + 1 - 4k^2 + 4 \]
Combining like terms:
\[ D = -3k^2 + 2k + 5 \]

### Step 4: Set the discriminant less than 0
To ensure \( D < 0 \):
\[ -3k^2 + 2k + 5 < 0 \]
Multiplying through by -1 (and reversing the inequality):
\[ 3k^2 - 2k - 5 > 0 \]

### Step 5: Factor the quadratic
To find the roots of \( 3k^2 - 2k - 5 = 0 \), we can use the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{2 \pm \sqrt{4 + 60}}{6} = \frac{2 \pm \sqrt{64}}{6} = \frac{2 \pm 8}{6} \]
Thus, the roots are:
\[ k = \frac{10}{6} = \frac{5}{3} \quad \text{and} \quad k = \frac{-6}{6} = -1 \]

### Step 6: Determine intervals
The quadratic \( 3k^2 - 2k - 5 \) opens upwards (since the coefficient of \( k^2 \) is positive). The roots divide the number line into intervals:
1. \( k < -1 \)
2. \( -1 < k < \frac{5}{3} \)
3. \( k > \frac{5}{3} \)

Since we need \( 3k^2 - 2k - 5 > 0 \), we look at the intervals:
- The expression is positive in the intervals \( k < -1 \) and \( k > \frac{5}{3} \).

### Step 7: Combine conditions
From our earlier condition, we also have \( k > 1 \). Therefore, we need:
\[ k > \frac{5}{3} \]

### Step 8: Find the least integral value
The least integral value of \( k \) that satisfies \( k > \frac{5}{3} \) is:
\[ k = 2 \]

### Final Answer
Thus, the least integral value of \( k \) such that the expression is positive for all real \( x \) is:
\[ \boxed{2} \]