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If f(x)=cosx, 0≤x≤(π/2), the value of 'c' for which f(x) satisfies Mean Value theorem is
- a)π/6
- b)π/4
- c)sin⁻1(2/π)
- d)cos⁻1(2/π)
Correct answer is option 'C'. Can you explain this answer?
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If f(x)=cosx, 0≤x≤(π/2), the value of c for which f(x) satisf...
Mean Value Theorem
The Mean Value Theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in the open interval (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
Given Function
In this case, f(x) = cosx on the interval [0, π/2].
Applying Mean Value Theorem
Now, we can apply the Mean Value Theorem to find the value of c:
f'(x) = -sinx
f'(π/2) = -sin(π/2) = -1
f(0) = cos(0) = 1
Therefore, by the Mean Value Theorem:
-1 = (1 - f(0))/(π/2 - 0)
-1 = (1 - 1)/(π/2)
-1 = 0
Solving for c, we get:
c = sin^(-1)(2/π)
Therefore, the correct answer is option C, sin^(-1)(2/π).
The Mean Value Theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in the open interval (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
Given Function
In this case, f(x) = cosx on the interval [0, π/2].
Applying Mean Value Theorem
Now, we can apply the Mean Value Theorem to find the value of c:
f'(x) = -sinx
f'(π/2) = -sin(π/2) = -1
f(0) = cos(0) = 1
Therefore, by the Mean Value Theorem:
-1 = (1 - f(0))/(π/2 - 0)
-1 = (1 - 1)/(π/2)
-1 = 0
Solving for c, we get:
c = sin^(-1)(2/π)
Therefore, the correct answer is option C, sin^(-1)(2/π).
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If f(x)=cosx, 0≤x≤(π/2), the value of c for which f(x) satisfies Mean Value theorem isa)π/6b)π/4c)sin1(2/π)d)cos1(2/π)Correct answer is option 'C'. Can you explain this answer?
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