Question

If Rolle’s theorem is applicable for the function \( f(x) \) defined by \( f(x) = x^3 + Px - 12 \) on \( [0,1] \), then the value of \( C \) of the Rolle's theorem is:

• A. \( \pm \frac{1}{\sqrt{3}} \)
• B. \( -\frac{1}{\sqrt{3}} \)
• C. \( \frac{1}{\sqrt{3}} \)
• D. 3

Hint:

To apply Rolle's theorem, ensure \( f(a) = f(b) \), differentiate \( f(x) \), and solve \( f'(c) = 0 \) in the interval \( (a, b) \).

Answer 1

The Correct Option is C

Step 1: Checking conditions for Rolle’s theorem
Rolle’s theorem states that if a function \( f(x) \) satisfies the following conditions on \( [a, b] \): 1. \( f(x) \) is continuous on \( [a, b] \). 2. \( f(x) \) is differentiable on \( (a, b) \). 3. \( f(a) = f(b) \). Then, there exists some \( c \in (a, b) \) such that: \[ f'(c) = 0. \] Step 2: Checking \( f(0) = f(1) \)
Given \( f(x) = x^3 + P x - 12 \), we evaluate: \[ f(0) = (0)^3 + P(0) - 12 = -12. \] \[ f(1) = (1)^3 + P(1) - 12 = 1 + P - 12 = P - 11. \] For Rolle's theorem to hold: \[ f(0) = f(1) \Rightarrow -12 = P - 11. \] Solving for \( P \): \[ P = -1. \] Step 3: Finding \( c \) where \( f'(c) = 0 \)
Differentiating: \[ f'(x) = \frac{d}{dx} (x^3 - x - 12). \] \[ f'(x) = 3x^2 - 1. \] Setting \( f'(c) = 0 \): \[ 3c^2 - 1 = 0. \] \[ 3c^2 = 1. \] \[ c^2 = \frac{1}{3}. \] \[ c = \pm \frac{1}{\sqrt{3}}. \] Since \( c \) must be in \( (0,1) \), we take the positive root: \[ c = \frac{1}{\sqrt{3}}. \] Step 4: Conclusion
Thus, the required value of \( c \) is: \[ \frac{1}{\sqrt{3}}. \]

Answer 2

To apply Rolle's theorem to the function \( f(x) = x^3 + Px - 12 \) on the interval \([0,1]\), we need to verify three conditions:

1. \( f(x) \) is continuous on \([0,1]\).
2. \( f(x) \) is differentiable on \((0,1)\).
3. \( f(0) = f(1) \).

The polynomial \( f(x) \) is continuous and differentiable everywhere, thus conditions 1 and 2 are satisfied.

Checking condition 3:

\( f(0) = 0^3 + P \cdot 0 - 12 = -12 \)

\( f(1) = 1^3 + P \cdot 1 - 12 = 1 + P - 12 = P - 11 \)

For \( f(0) = f(1) \), \(-12 = P - 11 \) implies \( P = -1 \).

With \( P = -1 \), the function becomes:

\( f(x) = x^3 - x - 12 \).

The derivative is:

\( f'(x) = 3x^2 - 1 \).

Rolle's theorem guarantees an \( x = C \) in \((0,1)\) such that \( f'(C) = 0 \):

\( 3C^2 - 1 = 0 \)

\( 3C^2 = 1 \) \(\Rightarrow C^2 = \frac{1}{3} \)

\( C = \pm \frac{1}{\sqrt{3}} \).

Since \( C \) must be in \((0,1)\), we choose:

\( C = \frac{1}{\sqrt{3}} \).