Question

If f(x) = \(x\)³ −4\(x\) \(+\)P and f(0) and f(1) are of opposite signs, Which one of the following is necessarily true?

• A. -1 < p < 2
• B. 0 < p < 3
• C. -2 < p < 1
• D. -3 < p < 0

Answer 1

The Correct Option is B

 To solve the given problem, we need to determine the range of the constant \( P \) for which the polynomial function \( f(x) = x^3 - 4x + P \) has \( f(0) \) and \( f(1) \) of opposite signs.

First, let's calculate the values of \( f(0) \) and \( f(1) \): 

• The value of the function at \( x = 0 \) is:

\( f(0) = 0^3 - 4 \cdot 0 + P = P \)

• The value of the function at \( x = 1 \) is:

\( f(1) = 1^3 - 4 \cdot 1 + P = 1 - 4 + P = P - 3 \)

We are given that \( f(0) \) and \( f(1) \) must be of opposite signs. This implies one of the following scenarios:

• \( f(0) > 0 \) and \( f(1) < 0 \)
• \( f(0) < 0 \) and \( f(1) > 0 \)

Now, let's analyze these scenarios:

1. If \( P > 0 \) (i.e., \( f(0) > 0 \)), then for \( f(1) < 0 \) we need \( P - 3 < 0 \) which gives \( P < 3 \). Hence, in this case, \( 0 < P < 3 \).
2. If \( P < 0 \) (i.e., \( f(0) < 0 \)), then for \( f(1) > 0 \) we need \( P - 3 > 0 \) which gives \( P > 3 \). But since \( P \) is less than 0, this scenario is not possible.

Thus, the only feasible range for \( P \) where \( f(0) \) and \( f(1) \) are of opposite signs is \( 0 < P < 3 \).

Therefore, the necessarily true statement in this context is:

Correct Answer: \( 0 < P < 3 \)