if allvalues of x which satisfies the inequality `log _((1//3))(x ^(2) +2px+p^(2) +1) ge 0` also satisfy the inequality `kx ^(2)+kx- k ^(2) le 0` for all real values of k, then all possible values of p lies in the interval :

Options

A. `[-1,1]`
B. `[0,1]`
C. `[0,2]`
D. `[-2,0]`

Answer

To solve the problem step by step, we need to analyze the two inequalities given in the question.

### Step 1: Analyze the first inequality
The first inequality is given as:
\[ \log_{(1/3)}(x^2 + 2px + p^2 + 1) \geq 0 \]
This implies that:
\[ x^2 + 2px + p^2 + 1 \leq 1 \]
because the logarithm is non-negative when its argument is less than or equal to 1 for a base less than 1.

### Step 2: Simplify the first inequality
Rearranging the inequality:
\[ x^2 + 2px + p^2 + 1 - 1 \leq 0 \]
This simplifies to:
\[ x^2 + 2px + p^2 \leq 0 \]
This is a quadratic inequality in \(x\).

### Step 3: Determine the conditions for the quadratic inequality
The quadratic \(x^2 + 2px + p^2\) can be factored as:
\[ (x + p)^2 \leq 0 \]
The expression \((x + p)^2\) is a perfect square and is non-negative. It equals zero only when:
\[ x + p = 0 \implies x = -p \]
Thus, the only solution to the inequality \(x^2 + 2px + p^2 \leq 0\) is:
\[ x = -p \]

### Step 4: Analyze the second inequality
The second inequality is:
\[ kx^2 + kx - k^2 \leq 0 \]
Factoring this gives:
\[ k(x^2 + x - k) \leq 0 \]
This is a quadratic in \(x\) which will have real roots if the discriminant is non-negative.

### Step 5: Find the discriminant of the second inequality
The discriminant \(D\) of the quadratic \(x^2 + x - k\) is given by:
\[ D = 1^2 - 4(1)(-k) = 1 + 4k \]
For the quadratic to have real roots, we need:
\[ 1 + 4k \geq 0 \implies k \geq -\frac{1}{4} \]

### Step 6: Substitute \(x = -p\) into the second inequality
Substituting \(x = -p\) into the second inequality:
\[ k((-p)^2 + (-p) - k) \leq 0 \]
This simplifies to:
\[ k(p^2 - p - k) \leq 0 \]

### Step 7: Analyze the quadratic in \(k\)
The quadratic \(p^2 - p - k\) must be non-positive for all \(k\). This means the quadratic must open upwards (coefficient of \(k\) is positive) and have its vertex above or on the x-axis.

### Step 8: Find the conditions for \(p\)
The vertex of the quadratic \(p^2 - p - k\) occurs at:
\[ k = \frac{p^2 - p}{2} \]
For the quadratic to be non-positive for all \(k\), the discriminant must be non-positive:
\[ (-1)^2 - 4(1)(p^2) \leq 0 \]
This gives:
\[ 1 - 4p^2 \leq 0 \implies 4p^2 \geq 1 \implies p^2 \geq \frac{1}{4} \implies |p| \geq \frac{1}{2} \]

### Step 9: Determine the interval for \(p\)
Thus, the possible values of \(p\) lie in the intervals:
\[ p \leq -\frac{1}{2} \quad \text{or} \quad p \geq \frac{1}{2} \]

### Conclusion
The final answer is that all possible values of \(p\) lie in the intervals:
\[ (-\infty, -\frac{1}{2}] \cup [\frac{1}{2}, \infty) \]