∫1√sec2x+tan2xdx=log|sinx+√f(x)|+C, then which of the following is/are correct?
A
fmin=0
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B
fmin=1
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C
fmax=2
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D
f(0)=1
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Solution
The correct option is Df(0)=1 ∫cosx√1+sin2xdx⇒sinx=t⇒cosxdx=dt=∫dt√1+t2=log∣∣t+√1+t2∣∣+C=log∣∣sinx+√1+sin2x∣∣+C ⇒f(x)=1+sin2x
So, fmin=1,fmax=2 and f(0)=1