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Question

# If 2tan2x−5secx is equal to 1 for exactly 7 distinct values of x∈[0, nπ2],n∈N, then the greatest value of n is

A
4
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B
10
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C
13
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D
15
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Solution

## The correct option is A 42tan2x−5secx=12(sec2x−1)−5secx=12sec2x−5secx−3=0(2secx+1)(secx−3)=0 secx=−1/2 or 3, but ∣secx∣≥1hence secx=3 is only solution or, cosx=1/3 therefore general solution is given asx=2nπ±cos−1(1/3), suppose cos−1(1/3)=αor, x=.....,α,2π−α,2π+α,4π−α,4π+α,6π−α,6π+α so there are two value of nϵN for which there is 7 distinct roots in given intervalthese are, n=13 and 15. (This can be easily seen by drawing cosx=1/3 graph)Therefore, the value of n is 15

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