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Question

Number of real roots of the equation esinxesinx4=0

A
None of these
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B
2
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C
1
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D
0
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Solution

The correct option is D 0
Let esinx=t
t1t4=0t=2±5esinx=2+5,25
But esinx cannot be negative
esinx=2+5
esinx>e
sinx>1, which is not possible.
So, the given equation has no real solution.

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