The Rolle-s theorem is applicable in the interval -1- x- 1 for the function

The correct option is C f( x ) = x ^ 2 Let nbsp; f( x ) = x ^ 2 , ∴ f( 1 ) =f( 1 ) =1 Also, f( x ) is continuous and differentiable in t ...

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Standard XII

Mathematics

Theorems for Differentiability

The Rolle's t...

Question

The Rolle's theorem is applicable in the interval $- 1 \leq x \leq 1$ for the function

f (x) = x

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f (x) = x^{2}

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f (x) = 2 x^{3} + 3

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f (x) = | x |

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Solution

The correct option is C $f (x) = x^{2}$
Let $f (x) = x^{2}$ ,
$∴ f (1) = f (- 1) = 1$
Also, $f (x)$ is continuous and differentiable in the given interval.

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Similar questions

f (x) = | x |

in the interval [-1, 1] Is Rolle's Theorem applicable?

Q. Rolle's theorem is not applicable for the function

f (x) = | x |

in the interval

[- 1, 1]

because:

Examine if Rolle's theorem is applicable to any of the following functions. Can you say something about the converse of Rolle's theorem from these example ?

(i) $f (x) = [x] f o r x ϵ [5, 9]$

(i) $f (x) = [x] f o r x ϵ [- 2, 2]$

(iii) $f (x) = x^{2} - 1 f o r x ϵ [5, 9]$ .

Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s Theorem from these examples?

(i)

(ii)

(iii)

f (x) = x^{2}

2 \leq x \leq 3

Is Rolle's theorem applicable?

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The Rolle's theorem is applicable in the interval −1≤x≤1 for the function

The correct option is C f(x)=x2Let f(x)=x2,∴f(1)=f(−1)=1Also, f(x) is continuous and differentiable in the given interval.

The Rolle's theorem is applicable in the interval $- 1 \leq x \leq 1$ for the function

The correct option is C $f (x) = x^{2}$
Let $f (x) = x^{2}$ ,
$∴ f (1) = f (- 1) = 1$
Also, $f (x)$ is continuous and differentiable in the given interval.