A real valued differentiable function defined on [1,∞) where f(1)=1. If f′(x)=1x2+f2(x) then the maximum value of [f(x)] is (where [.] is greatest integer function)
A
2
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B
1
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C
3
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D
0
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Solution
The correct option is B1 Since, f′(x)>0 ∴f(x) is increasing function. ∵f(1)=1,∴f(x)>1 ⇒1x2+f2(x)<1x2+1 f(t)−f(1)=t∫11x2+f2(x)dx ⇒f(t)−f(1)<∞∫11x2+1dx ⇒f(t)−f(1)<tan−1x∣∣∞1 ⇒f(t)−π4<1 ⇒f(x)−π4<1⇒f(x)<1+π4 Hence [f(x)]=1