Let $h (x) = x^{4} - 8 x^{3} + 18 x^{2} - 8 x + 1$ on the closed interval $[1, 3]$ .

hard 4 marks Class 12 · Mathematics · Continuity and Differentiability

Answer

c = 2

The correct answer is $c = 2$ — 10. Hence the only $c$ in $(1, 3)$ with $h^{'} (c) = 0$ is $c = 2$ .

Step-by-step solution

1. Continuity and differentiability: $h$ is a polynomial, hence continuous and differentiable everywhere, including on $[1, 3]$ .
2. End–point values: $h (1) = 1 - 8 + 18 - 8 + 1 = 4$ and $h (3) = 81 - 216 + 162 - 24 + 1 = 4$ ; so $h (1) = h (3)$ .
3. Therefore the hypotheses of Rolle’s theorem are satisfied.
4. Compute derivative: $h^{'} (x) = 4 x^{3} - 24 x^{2} + 36 x - 8 = 4 (x^{3} - 6 x^{2} + 9 x - 2)$ . 5. Solve $h^{'} (x) = 0$ ⇔ $x^{3} - 6 x^{2} + 9 x - 2 = 0$ .
6. Check rational roots: try $x = 2 \to 8 - 24 + 18 - 2 = 0$ ; thus $x - 2$ is a factor.
7. Perform synthetic division: coefficients 1, -6, 9, -2 with root 2 gives quotient $x^{2} - 4 x + 1$ . 8. Solve $x^{2} - 4 x + 1 = 0 \to x = 2 p m s q r t 3$ which are $a pp r o x 3.732$ and $0.268$ . 9. Only root inside $(1, 3)$ is $x = 2$ .
10. Hence the only $c$ in $(1, 3)$ with $h^{'} (c) = 0$ is $c = 2$ .

Common mistake: Missing the root

x = 2

or including extraneous roots outside interval

Concept tested: Rolle’s theorem

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked