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Limits, Continuity & Differentiability — JEE Main Previous Year Questions · PaperYodha
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← JEE Main topics
Limits, Continuity & Differentiability
· JEE Main previous year questions
208 questions asked across 19 exam years, newest first
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208
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16
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5
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3
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1
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2014
1 question
Q19
2014 · 6 Apr 2014 · Shift 1
lim
x
→
0
x
2
s
i
n
(
π
c
o
s
2
x
)
is equal to :
A.
−
π
B.
π
C.
2
π
D.
1
Show answer
Answer: B
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