Application of DerivativeshardFree

Rolle's Theorem for a Piecewise Rational Function on [-2,3] | JEE

JEE Maths question with a full step-by-step solution.

Question
Consider the function, for x∈[−2,3]x \in [-2, 3],
f(x)={x3−2x2−5x+6x−1if x≠1−6if x=1f(x) = \begin{cases} \dfrac{x^{3}-2x^{2}-5x+6}{x-1} & \text{if } x \ne 1 \\[2mm] -6 & \text{if } x = 1 \end{cases}
 then
Aff is discontinuous at x=1x = 1, so Rolle's theorem is not applicable in [−2,3][-2,3]
Bf(−2)≠f(3)f(-2) \ne f(3), so Rolle's theorem is not applicable in [−2,3][-2,3]
Cff is not derivable in (−2,3)(-2,3), so Rolle's theorem is not applicable
DRolle's theorem is applicable as ff satisfies all the conditions, and cc of Rolle's theorem is 1/21/2correct
Solution
Step 1: Test whether x−1x-1 divides the numerator, by substituting x=1x = 1:
1−2−5+6=0.1 - 2 - 5 + 6 = 0 .
 So (x−1)(x-1) is a factor and the fraction simplifies. Step 2: Carry out the division.
x3−2x2−5x+6=(x−1)(x2−x−6).x^{3}-2x^{2}-5x+6 = (x-1)\left(x^{2}-x-6\right).
 (Check by expanding: x3−x2−6x−x2+x+6=x3−2x2−5x+6x^{3}-x^{2}-6x-x^{2}+x+6 = x^{3}-2x^{2}-5x+6 .) Step 3: Simplify ff for x≠1x \ne 1.
f(x)=(x−1)(x2−x−6)x−1=x2−x−6.f(x) = \frac{(x-1)\left(x^{2}-x-6\right)}{x-1} = x^{2}-x-6 .
 Step 4: Check the value at x=1x = 1 against that formula.
1−1−6=−6=f(1).1 - 1 - 6 = -6 = f(1) .
 So in fact
f(x)=x2−x−6for every x∈[−2,3],f(x) = x^{2}-x-6 \quad \text{for every } x \in [-2,3] ,
 a polynomial - hence continuous on [−2,3][-2,3] and differentiable on (−2,3)(-2,3). That already disposes of (1) and (3). Step 5: Check the endpoint values, the third condition of Rolle's theorem.
f(−2)=4+2−6=0,f(3)=9−3−6=0,f(-2) = 4+2-6 = 0, \qquad f(3) = 9-3-6 = 0 ,
 so f(−2)=f(3)f(-2) = f(3), which disposes of (2). Step 6: All three conditions hold, so Rolle's theorem applies and guarantees some c∈(−2,3)c \in (-2,3) with f′(c)=0f'(c) = 0. Step 7: Find it.
f′(x)=2x−1=0  ⟹  c=12,f'(x) = 2x-1 = 0 \implies c = \frac12 ,
 and 12\dfrac12 does lie in (−2,3)(-2,3). Answer: (4).
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