Let $p (x) = x^{3} - 3 x^{2} + 2 x$ be defined on $[0, 2]$ . (i) Verify that $p$ satisfies the hypotheses of Rolle’s theorem on $[0, 2]$ . (ii) Find all numbers $c$ in $(0, 2)$ such that $p^{'} (c) = 0$ .

hard 3 marks Class 12 · Mathematics · Continuity and Differentiability

Answer

c = 1 - \frac{3}{3}, c = 1 + \frac{3}{3}

The correct answer is $c = 1 - \frac{3}{3}, c = 1 + \frac{3}{3}$ — Both solutions lie in $(0, 2)$ , so the required numbers are $c = 1 - \frac{3}{3}$ and $c = 1 + \frac{3}{3}$ .

Step-by-step solution

Check continuity: $p (x)$ is a polynomial, hence continuous on $[0, 2]$ .
Check differentiability: polynomials are differentiable everywhere, so $p$ is differentiable on $(0, 2)$ .
Evaluate the endpoints: $p (0) = 0$ and $p (2) = 2^{3} - 3 \cdot 2^{2} + 2 \cdot 2 = 8 - 12 + 4 = 0$ . Since $p (0) = p (2)$ , the third hypothesis holds.
Thus all three hypotheses of Rolle’s theorem are satisfied.
Compute the derivative: $p^{'} (x) = 3 x^{2} - 6 x + 2$ .
Set the derivative to zero: $3 x^{2} - 6 x + 2 = 0$ .
Solve the quadratic: $x = \frac{6 \pm 36 - 24}{6} = \frac{6 \pm 12}{6} = \frac{6 \pm 2 3}{6} = 1 \pm \frac{3}{3}$ .
Both solutions lie in $(0, 2)$ , so the required numbers are $c = 1 - \frac{3}{3}$ and $c = 1 + \frac{3}{3}$ .

Common mistake: Forgetting to verify that

f (a) = f (b)

before applying the theorem.

Concept tested: Rolle’s theorem application

Published 9 September 2026 · How this solution was prepared and checked

← All Continuity and Differentiability questions

Let p(x)=x3−3x2+2x be defined on [0,2]. (i) Verify that p satisfies the hypotheses of Rolle’s theorem on [0,2]. (ii) Find all numbers c in (0,2) such that p′(c)=0.

Step-by-step solution

Let $p (x) = x^{3} - 3 x^{2} + 2 x$ be defined on $[0, 2]$ . (i) Verify that $p$ satisfies the hypotheses of Rolle’s theorem on $[0, 2]$ . (ii) Find all numbers $c$ in $(0, 2)$ such that $p^{'} (c) = 0$ .