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Question

# Let f:[0,π2]→[0,1] be a differentiable function such that f(0)=0,f(π2)=1, then

A
f(α)=1(f(α))2for atleast one α(0,π2)
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B
f(α)=2πfor atleast one α(0,π2)
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C
f(α)f(α)=1πfor atleast one α(0,π2)
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D
f(α)=8απ2for atleast one α(0,π2)
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Solution

## The correct options are A f′(α)=√1−(f(α))2for atleast one α∈(0,π2) B f′(α)=2πfor atleast one α∈(0,π2) C f(α)f′(α)=1πfor atleast one α∈(0,π2) D f′(α)=8απ2for atleast one α∈(0,π2)Let : [0,π2]→[0,1] be a..... (A) Consider g(x)=sin−1f(x)−x Since g(0)=0,g(π2)=0 ∴ There is at least one value of α∈(0,π2) such that g′(α)=f′(α)√1−(f(α))2−1=0 i.e.f′(α)=√1−(f(α))2 for atleast one value of but may not be for all α∈(0,π2) ∴ True (B) Consider g(x)=f(x)−2xπ Since g(0)=0,g(π2)=0 ∴ there is at least one value of α∈(0,π2) such that g′(α)=f′(α)−2π=0 i.e.f′(α)=2π for atleast one value of but may not be for all α∈(0,π2) ∴ True (C) Consider g(x)=(f(x))2−2xπ Since g(0)=0,g(π2)=0 ∴ there is at least one value of α∈(0,π2) g′(α)=2f(α)f′(α)−2π=0∴f(α)f′(α)=1π ∴ True (D) Consider g(x)=f(x)−4x2π2 Since g(0)=0,g(π2)=0 ∴ there is at least one value of α∈(0,π2) such that g′(α)=f′(α)−8απ2=0∴f′(α)=8απ2 ∴ True

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