Given `|px^(2) + qx + r| le |Px^(2) + Qx + R|AA x in R` and `d=q^(2) - 4pr gt 0` and `D =Q^(2) -4PR gt 0`
Which of the following must be ture ?

Options

A. (a) `|p| ge |P|`
B. (b) `|p| le |P|`
C. (c) `|p| = |P|`
D. (d) All of these

Explanation

To solve the problem, we need to analyze the given inequalities and the conditions provided.

### Step-by-Step Solution:

1. **Understanding the Given Inequality**:
We have the inequality:
\[ |px^2 + qx + r| \leq |Px^2 + Qx + R| \quad \forall x \in \mathbb{R} \]
This means that the quadratic polynomial \(px^2 + qx + r\) is always less than or equal to the polynomial \(Px^2 + Qx + R\) in absolute value for all real numbers \(x\).

2. **Discriminants**:
We are given that:
\[ d = q^2 - 4pr > 0 \quad \text{and} \quad D = Q^2 - 4PR > 0 \]
This indicates that both quadratics have real and distinct roots.

3. **Graphical Interpretation**:
The graphs of both quadratics will be parabolas. Since \(d > 0\) and \(D > 0\), both parabolas will intersect the x-axis at two distinct points.

4. **Roots of the Quadratics**:
Let the roots of \(px^2 + qx + r\) be \(\alpha\) and \(\beta\), and the roots of \(Px^2 + Qx + R\) also be \(\alpha\) and \(\beta\). Since the inequality holds for all \(x\), the two polynomials must be equal at their roots.

5. **Factoring the Quadratics**:
We can express the quadratics in factored form:
\[ px^2 + qx + r = p(x - \alpha)(x - \beta) \]
\[ Px^2 + Qx + R = P(x - \alpha)(x - \beta) \]

6. **Analyzing the Inequality**:
The given inequality can be simplified:
\[ |p(x - \alpha)(x - \beta)| \leq |P(x - \alpha)(x - \beta)| \]
Since \((x - \alpha)(x - \beta)\) is common in both sides, we can cancel it out, assuming it is not zero (which is valid except at the roots):
\[ |p| \leq |P| \]

7. **Conclusion**:
From the analysis, we conclude that:
\[ |p| \leq |P| \]
This means that the absolute value of \(p\) must be less than or equal to the absolute value of \(P\).

### Final Answer:
Thus, the statement that must be true is:
\[ |p| \leq |P| \]