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This question has multiple correct options

A .f′′(a)=0

B .f′(x)=0 for some x∈ (a, b)

C .f′′(x) $ \ne $0 for some x∈ (a, b)

D. f′′′(x)=0 for some x∈ (a, b)

Answer
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Hint: Proceed the solution of this question, by applying Rolle's theorem first on f(x) and obtain the results like the point where f’(x) will be zero, then use that obtained results to apply Rolle's theorem again on the derivative of f(x).

Complete step-by-step answer:

In the question it is given that, f is continuous and differentiable function on [a, b].

Also, f(a)= f(b) = 0

We know that According to Rolle's theorem if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some c with a ≤ c ≤ b

Therefore, By Rolle's theorem, there exists c∈ (a, b) such that f’(c)=0

Thus, there exists x∈ (a, b) such that f’(x)=0

Hence Option B is correct.

Let at x=c ∈ (a, b), f′(c)=0 ………. (1)

Now, f is continuous and differentiable on [a, b].

Also, f is twice differentiable on (a, b).

And we know that the necessary condition of the function f(x) to be twice differentiable is the first derivative of the function must be continuous and its 1st derivative will be differential in the open interval.

⇒ so f′ is continuous on [a, b].

∴f′ is differentiable on (a, b).

In question it is given f′(a)=0 and by expression (1) f′(c)=0

Now f′(a) = 0 = f′(c)

Here, we can also apply Rolle's theorem on f’(x) By Rolle's theorem, there exists k∈ (a, c) such that f"(k)=0

Hence we can say, f"(k)=0 and k is any point between x=an and x=c,

So to comment on option A,

Strictly we can’t say directly that x=k will be the same point as x=a, and f’’ will be zero at x=a. K is any general point. Therefore, f′′(a) $ \ne $0 Hence option A is wrong.

Thus, there exists x∈ (a, c) such that f′′(x)=0.

Hence option C is correct.

We can’t apply Rolle’s theorem on f’’(x) because there is no information about the 2nd point where f’’(x) will be equal to zero, so we can’t comment on option D.

So in This question option B and C both are correct.

Note-

When we came up of this types of question we should remember that the Rolle’s Theorem is a special case of Lagrange’s mean value theorem which states that-

If a function f is defined in the closed interval [a, b] in such a way that if it satisfies the conditions (1), (2), (3)

(1) The function f is continuous on the closed interval [a, b]

(2) The function f is differentiable on the open interval (a, b)

(3) Now if f (a) = f (b)

Then there exists at least one value of x, let us assume that value to be c, which lies between an and b that is (a < c < b) in such a way that f ‘(c) = 0.

Complete step-by-step answer:

In the question it is given that, f is continuous and differentiable function on [a, b].

Also, f(a)= f(b) = 0

We know that According to Rolle's theorem if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some c with a ≤ c ≤ b

Therefore, By Rolle's theorem, there exists c∈ (a, b) such that f’(c)=0

Thus, there exists x∈ (a, b) such that f’(x)=0

Hence Option B is correct.

Let at x=c ∈ (a, b), f′(c)=0 ………. (1)

Now, f is continuous and differentiable on [a, b].

Also, f is twice differentiable on (a, b).

And we know that the necessary condition of the function f(x) to be twice differentiable is the first derivative of the function must be continuous and its 1st derivative will be differential in the open interval.

⇒ so f′ is continuous on [a, b].

∴f′ is differentiable on (a, b).

In question it is given f′(a)=0 and by expression (1) f′(c)=0

Now f′(a) = 0 = f′(c)

Here, we can also apply Rolle's theorem on f’(x) By Rolle's theorem, there exists k∈ (a, c) such that f"(k)=0

Hence we can say, f"(k)=0 and k is any point between x=an and x=c,

So to comment on option A,

Strictly we can’t say directly that x=k will be the same point as x=a, and f’’ will be zero at x=a. K is any general point. Therefore, f′′(a) $ \ne $0 Hence option A is wrong.

Thus, there exists x∈ (a, c) such that f′′(x)=0.

Hence option C is correct.

We can’t apply Rolle’s theorem on f’’(x) because there is no information about the 2nd point where f’’(x) will be equal to zero, so we can’t comment on option D.

So in This question option B and C both are correct.

Note-

When we came up of this types of question we should remember that the Rolle’s Theorem is a special case of Lagrange’s mean value theorem which states that-

If a function f is defined in the closed interval [a, b] in such a way that if it satisfies the conditions (1), (2), (3)

(1) The function f is continuous on the closed interval [a, b]

(2) The function f is differentiable on the open interval (a, b)

(3) Now if f (a) = f (b)

Then there exists at least one value of x, let us assume that value to be c, which lies between an and b that is (a < c < b) in such a way that f ‘(c) = 0.

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