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Question

The equation 2tan2x5secx=1 holds true for exactly eleven distinct values of x[0,nπ2], nN. The greatest value of n is 


A
21
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B
22
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C
24
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D
23
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Solution

The correct option is D 23
2sec2x25secx=1
2sec2x5secx3=0
secx=3
[secx1 and secx1] 
cosx=13
It has two solutions in [0,2π]
So, Eleven solutions in [0,21π2]

Now, we have to find the greatest interval for which the equation has 11 solutions
Eleven solutions in [0,23π2]

Hence the maximum value of n=23

Mathematics

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