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Rolle's Theorem

In which of t...

Question

In which of the following functions is Rolle's theorem applicable

f (x) = {\begin{matrix} x, 0 \leq x < 1 0, x = 1 \end{matrix}

[0, 1]

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f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{sin x}{x}, - π \leq x < 0 0, x = 0 \end{matrix}

[- π, 0]

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f (x) = \frac{x^{2} - x - 6}{x - 1}

[- 2, 3]

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f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{x^{3} - 2 x^{2} - 5 x + 6}{x - 1}, i f x \neq 1 - 6, i f x = 1 \end{matrix}

[- 2, 3]

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Solution

The correct option is D $f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{x^{3} - 2 x^{2} - 5 x + 6}{x - 1}, i f x \neq 1 - 6, i f x = 1 \end{matrix}$ on $[- 2, 3]$
from lagranges theorem
only option D satisfies all condition.

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Similar questions

x = [- 2, 3]

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{x^{3} - 2 x^{2} - 5 x + 6}{x - 1}, & i f x \neq 1 - 6 & i f x = 1 \end{matrix}

x \in [- 2, 3]

f (x) = ⎧ ⎨ ⎩ \begin{matrix} - 6; x = 1 \frac{x^{3} - 2 x^{2} - 5 x + 6}{x - 1}; x \neq 1 \end{matrix}

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\frac{\sin x}{e^{x}}

{e^{1 - x}}^{2}

f (x) = \frac{x}{2} - \sin \frac{π x}{6} on [- 1, 0]

f (x) = \frac{6 x}{π} - 4 \sin^{2} x on [0, π / 6]

[0, \frac{π}{2}]

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{3} x}{3 {cos}^{2} x}, i f x < \frac{π}{2} a, i f x = \frac{π}{2} \frac{b (1 - sin x)}{{(π - 2 x)}^{2}}, i f x > \frac{π}{2} \end{matrix}

f (x)

x = \frac{π}{2}

a

b

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