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: $$\frac{1}{\sqrt{3}}.$$