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Let f (x) be differentiable in (0, 4) and f (2) = f (3) and S = {c : 2 < c < 3, f’ (c) = 0} then
- a)S has exactly one point
- b)S = { }
- c)S has atleast one point
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Let f (x) be differentiable in (0, 4) and f (2) = f (3) and S = {c : 2...
Conditions of Rolle’s Theorem are satisfied by f(x) in [2,3].Hence there exist atleast one real c in (2, 3) s.t. f ‘(c) = 0 . Therefore , the set S contains atleast one element
Most Upvoted Answer
Let f (x) be differentiable in (0, 4) and f (2) = f (3) and S = {c : 2...
Explanation:
To prove that the set S has at least one point, we need to show that there exists a value of c in the interval (2, 3) such that f(c) = 0.
Given:
- f(x) is differentiable in the interval (0, 4)
- f(2) = f(3)
Mean Value Theorem:
The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the open interval (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
Proof:
Since f(x) is differentiable in the interval (0, 4), it is also continuous on the closed interval [2, 3]. Therefore, we can apply the Mean Value Theorem to the interval [2, 3].
Let a = 2 and b = 3. Since f(x) is continuous on [2, 3] and differentiable on (2, 3), there exists a value c in (2, 3) such that:
f'(c) = (f(3) - f(2))/(3 - 2)
Since f(2) = f(3), we have:
f'(c) = 0/1
f'(c) = 0
This means that the derivative of f(x) at the point c is equal to zero.
Conclusion:
Since the derivative of f(x) at the point c is zero, and f(x) is differentiable in the interval (0, 4), we can conclude that there exists at least one value of c in the interval (2, 3) such that f(c) = 0. Therefore, the set S has at least one point.
Hence, the correct answer is option 'C': S has at least one point.
To prove that the set S has at least one point, we need to show that there exists a value of c in the interval (2, 3) such that f(c) = 0.
Given:
- f(x) is differentiable in the interval (0, 4)
- f(2) = f(3)
Mean Value Theorem:
The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the open interval (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
Proof:
Since f(x) is differentiable in the interval (0, 4), it is also continuous on the closed interval [2, 3]. Therefore, we can apply the Mean Value Theorem to the interval [2, 3].
Let a = 2 and b = 3. Since f(x) is continuous on [2, 3] and differentiable on (2, 3), there exists a value c in (2, 3) such that:
f'(c) = (f(3) - f(2))/(3 - 2)
Since f(2) = f(3), we have:
f'(c) = 0/1
f'(c) = 0
This means that the derivative of f(x) at the point c is equal to zero.
Conclusion:
Since the derivative of f(x) at the point c is zero, and f(x) is differentiable in the interval (0, 4), we can conclude that there exists at least one value of c in the interval (2, 3) such that f(c) = 0. Therefore, the set S has at least one point.
Hence, the correct answer is option 'C': S has at least one point.
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