Question

# If cos2(x+y)−cos2(x−y)−4sinxsiny=limx→0+sgn(sgn(sgn x)), where sgn is signum function, then which of the following statements is/are correct?   x=nπ and y=mπ, where n,m are both even integersx=nπ and y=mπ, where n is even and m is an odd integer.x=nπ and y=mπ, where n,m are both odd integersx=nπ and y=mπ, where n is odd and m is an even integer.

Solution

## The correct options are A x=nπ and y=mπ, where n,m are both even integers C x=nπ and y=mπ, where n,m are both odd integerscos2(x+y)−cos2(x−y)−4sinxsiny=limx→0+sgn(sgn(sgn x)) ⇒cos2(x+y)−cos2(x−y)2(cos(x+y)−cos(x−y))=1 ⇒cos(x+y)+cos(x−y)2=1 ⇒cosxcosy=1 Now we know that, −1≤cosx≤1, and −1≤cosy≤1 So, cosx=1 and cosy=1 or cosx=−1 and cosy=−1 ⇒x=2nπ and y=2mπ (n,m∈Z) or x=(2n+1)π and y=(2m+1)π (n,m∈Z) x=nπ and y=mπ where n,m are both even integers or both odd integers.

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