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Question

If limx02asinxsin2xtan3x exists and is equal to 1, then the value of a is

A
2
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B
1
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C
0
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D
1
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Solution

The correct option is C 1
limx02asinxsin2xtan3x

=limx02a(xx33!+x55!)(2x(2x)33!+(2x)55!)(x+x33+215x5+)3

=limx0(2a2)x+(2a3!+233!)x3+(2a5!255!)x5+x3(1+x23+215x4+)3

Since, it is given that given limit is exist, then
2a2=0a=1

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