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B
f1(x)>f2(x)
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C
f1(|x|)<f2(|x|)
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D
none of these
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Solution
The correct option is Bf1(x)>f2(x) Let f(x)=f1(x)−f2(x) f(x)=2x−3sinx+xcosx ⇒f′(x)=2−2cosx−xsinx ⇒f′′(x)=sinx−xcosx=cosx(tanx−x) ⇒f′′(x)>0∀x∈(0,π2). Thus f′(x) is increasing in (0,π2);f′(0)=0 ⇒f′(x)>0∀x∈(0,π2). ⇒f(0)=0 ⇒f(x)>0∀x∈(0,π2). ⇒f1(x)>f2(x) ⇒f(x) is increasing in (0,π2)