Given: f(29)=92 and f(x) is continuous.
x→0limf(x21−cos3x)=f(x→0limx21−cos3x)
=f(x→0limx22sin2(23x)) ∵1−cos2θ=2sin2θ
=f(x→0lim94(23x)22sin2(23x))
=f(29x→0lim(23x)2sin2(23x))
=f(29×12) ∵x→0limxsinx=1
=f(29)
=92
JEE Main 2014 — Mathematics Calculus
If f(x) is continuous and f(29)=92, then x→0limf(x21−cos3x) equals to
Held on 9 Apr 2014 · Verified 6 Jul 2026.
98
0
92
29
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