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Given that f(x) is continuously differentiable on a ≤ x ≤ b where a < b , f a < 0 and f b > 0, which of the following are always true ?
(i) f(x) is bounded on a ≤ x ≤ b .
(ii) The equation f(x) = 0 has at least one solution in a < x < b .
(iii) The maximum and minimum values of f(x) on a ≤ x ≤ b occur at points where f ′ c = 0
(iv) There is at least one point c with a < c < b where f ′ c > 0.
(v) There is at least one point d with a < d < b where f ′ c < 0.
  • a)
    Only (ii) and (iv) are true
  • b)
    All but (iii) are true
  • c)
    All but (v) are true
  • d)
    Only (i), (ii) and (iv) are true
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Given that f(x) is continuously differentiable on a ≤ x ≤ b wher...
(i) This statement is true, every continuous function is bounded on a closed interval.
(ii) True again, by Intermediate Value Theorem.
(iii) Not ture, because maximum and/or minimum value could also occur at a or b, without the derivatives being O.
(iv) True By the Mean Value Theorem, there exists a point betweem a and b, where the derivative is exactly   a clearly positive value.
(v) Not always true, for example, the function might be strictly increasing guaranteeing the derivative to be always positive.
Thus, the true statements are (i), (ii) and (iv).
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Most Upvoted Answer
Given that f(x) is continuously differentiable on a ≤ x ≤ b wher...
The mean value theorem for integrals states that if f(x) is continuous on the interval [a,b], then there exists a number c in the interval (a,b) such that:

∫a^b f(x) dx = f(c) (b-a)

Using this theorem, we can prove that if f(x) is continuously differentiable on the interval [a,b], then there exists a number c in the interval (a,b) such that:

∫a^b f(x) dx = (b-a) f((a+b)/2)

Proof:

Since f(x) is continuously differentiable on the interval [a,b], it is also continuous on the interval [a,b]. Therefore, we can apply the mean value theorem for integrals to obtain:

∫a^b f(x) dx = f(c) (b-a)

for some c in the interval (a,b).

Now, let's define a new function g(x) as:

g(x) = f(x) - f((a+b)/2)

Since f(x) is continuously differentiable, g(x) is also continuously differentiable. Moreover, we have:

g(a) = f(a) - f((a+b)/2)

g(b) = f(b) - f((a+b)/2)

g((a+b)/2) = 0

Using these values, we can apply the mean value theorem for derivatives to g(x) on the interval [a,b]. This gives:

g'(d) = (g(b) - g(a))/(b-a)

for some d in the interval (a,b).

Substituting the expressions for g(a), g(b), and g'(d), we obtain:

f'(d) - f'((a+b)/2) = (f(b) - f(a))/(b-a)

Multiplying both sides by (b-a), we get:

∫a^b f'(x) dx - (b-a) f'((a+b)/2) = f(b) - f(a)

Finally, adding f((a+b)/2) to both sides, we arrive at:

∫a^b f(x) dx = (b-a) f((a+b)/2)

as desired. This completes the proof.
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Given that f(x) is continuously differentiable on a ≤ x ≤ b where a < b , f a < 0 and f b > 0, which of the following are always true ?(i) f(x) is bounded on a ≤ x ≤ b .(ii) The equation f(x) = 0 has at least one solution in a < x < b .(iii) The maximum and minimum values of f(x) on a ≤ x ≤ b occur at points where f ′ c = 0(iv) There is at least one point c with a < c < b where f ′ c > 0.(v) There is at least one point d with a < d < b where f ′ c < 0.a)Only (ii) and (iv) are trueb)All but (iii) are truec)All but (v) are trued)Only (i), (ii) and (iv) are trueCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Given that f(x) is continuously differentiable on a ≤ x ≤ b where a < b , f a < 0 and f b > 0, which of the following are always true ?(i) f(x) is bounded on a ≤ x ≤ b .(ii) The equation f(x) = 0 has at least one solution in a < x < b .(iii) The maximum and minimum values of f(x) on a ≤ x ≤ b occur at points where f ′ c = 0(iv) There is at least one point c with a < c < b where f ′ c > 0.(v) There is at least one point d with a < d < b where f ′ c < 0.a)Only (ii) and (iv) are trueb)All but (iii) are truec)All but (v) are trued)Only (i), (ii) and (iv) are trueCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given that f(x) is continuously differentiable on a ≤ x ≤ b where a < b , f a < 0 and f b > 0, which of the following are always true ?(i) f(x) is bounded on a ≤ x ≤ b .(ii) The equation f(x) = 0 has at least one solution in a < x < b .(iii) The maximum and minimum values of f(x) on a ≤ x ≤ b occur at points where f ′ c = 0(iv) There is at least one point c with a < c < b where f ′ c > 0.(v) There is at least one point d with a < d < b where f ′ c < 0.a)Only (ii) and (iv) are trueb)All but (iii) are truec)All but (v) are trued)Only (i), (ii) and (iv) are trueCorrect answer is option 'D'. Can you explain this answer?.
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