f(x)=sinx+3x−π2(x2+x)x∈[0,2π]f′(x)=cosx+3−π2(2x+1)>0f(x)↑f′(x)=−sinx+0−2π(2)=−sinx−π4<0f′(x)↓0<x<2π ⇒−π2(+10<+12x<+1π) +3−π2>π−2(+32x+1)>−π2(+3π+1)+ve3−π2>3−π2(2x+1)>3−π2(+veπ+1)
For the function f(x)=sinx+3x−π2(x2+x), where x∈[0,2π], consider the following two statements : (I) f is increasing in (0,2π). (II) f′ is decreasing in (0,2π).
Between the above two statements,
Held on 5 Apr 2024 · Verified 6 Jul 2026.
only (II) is true.
only (I) is true.
neither (I) nor (II) is true.
both (I) and (II) are true
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