f(x)=64x3−3x2−2sinx+(2x−1)cosxf′(x)=(2x2−x)−2cosx+2cosx−sinx(2x−1).
=(2x−1)(x−sinx)
for x>0,x−sinx>0
x<0,x−sinx<0
for x∈(−∞,0]∪[21,∞),f′(x)≥0
for x∈[0,21],f′(x)≤0
⇒f(x) increases in [21,∞)
JEE Main 2021 — Mathematics Calculus
The function f(x)=64x3−3x2−2sinx+(2x−1)cosx:
Held on 24 Feb 2021 · Verified 6 Jul 2026.
increases in [21,∞)
decreases in (−∞,21]
decreases in [21,∞)
increases in (−∞,21]
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