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if tan^(-1)(2x)+tan^(-1)(3x)=npi+(pi)/(4...

if `tan^(-1)(2x)+tan^(-1)(3x)=npi+(pi)/(4),nepsilonI` then number of order pair (s) `(n,x)` is (are)`

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The correct Answer is:
2
`(2x+3x)/(1-6x^(2))=1,6x^(2)+5x-1=0x=-1,(1)/(6)`
`x=-1,-3(pi)/(4)=npi+(pi)/(4)`
`impliesn=-1,x=(1)/(6),n=0`
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