Verify Rolle's theorem for f(x) = x² - 5x + 6 on [2,3]

Question

Verify Rolle’s theorem for $f(x) = x^2 - 5x + 6$ on the interval $[2, 3]$ .

(NCERT Class 12, Exercise 5.8)

Solution — Step by Step

Condition 1: $f$ is continuous on $[2, 3]$ . $f(x) = x^2 - 5x + 6$ is a polynomial — polynomials are continuous everywhere. ✓

Condition 2: $f$ is differentiable on $(2, 3)$ . Polynomials are differentiable everywhere. ✓

Condition 3: $f(2) = f(3)$ . $f(2) = 4 - 10 + 6 = 0$ $f(3) = 9 - 15 + 6 = 0$ $f(2) = f(3) = 0$ ✓

All three conditions are satisfied.

f'(x) = 2x - 5

Set $f'(c) = 0$ : $2c - 5 = 0 \Rightarrow c = \frac{5}{2} = 2.5$

$c = 5/2 = 2.5 \in (2, 3)$ ✓

\boxed{c = \frac{5}{2} \text{ satisfies Rolle's theorem on } [2, 3]}

Why This Works

Rolle’s theorem says: if a function starts and ends at the same height (on a closed interval), and is smooth in between, then there must be at least one point where the slope is zero — a horizontal tangent.

For $f(x) = x^2 - 5x + 6 = (x-2)(x-3)$ , the graph is an upward parabola with roots at $x = 2$ and $x = 3$ . Since it starts at 0 and ends at 0, it must dip down in between and come back up. The lowest point of this dip is the vertex at $x = 5/2$ , where the tangent is horizontal.

Alternative Method — Factor first

Since $f(x) = (x-2)(x-3)$ , we immediately see $f(2) = f(3) = 0$ . The vertex of the parabola $y = x^2 - 5x + 6$ is at $x = -(-5)/(2 \cdot 1) = 5/2$ . The vertex of a parabola always has zero slope, so $c = 5/2$ .

For CBSE boards, verification of Rolle’s theorem is a structured 4-mark question. Follow this exact template: (1) state continuity, (2) state differentiability, (3) verify $f(a) = f(b)$ , (4) find $c$ with $f'(c) = 0$ and check $c \in (a, b)$ . The marking scheme maps directly to these steps.

Common Mistake

Students sometimes forget to verify that $c$ lies in the open interval $(a, b)$ . Finding $f'(c) = 0$ is not enough — you must confirm $a < c < b$ . If $c$ falls outside the interval, Rolle’s theorem is not verified for that interval. In this problem $c = 2.5$ clearly lies between 2 and 3, but in harder problems, $f'(c) = 0$ might give multiple roots, and you need to pick the one inside $(a, b)$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Factor first

Common Mistake

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