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Question

If f(x)=x13t1+t2dt, where x>1, then which of the following is/are correct


A

For 1<α<β,f(α)<f(β)

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B

For 1<α<β,f(α)>f(β)

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C

f(x)+π4<tan1x, x1

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D

f(x)+π4>tan1x, x>1

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Solution

The correct option is D

f(x)+π4>tan1x, x>1


Given: f(x)=x13t1+t2dt
f(x)=3x1+x2>0 x>1
f(x) is increasing.
Hence for 1<α<β,f(α)<f(β)
f(x)=3x1+x2>11+x2x>1
x1f(x)dx>x111+x2dx
f(x)>tan1xtan11 [f(1)=0]f(x)+π4>tan1x

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