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Question

Question: The period of \(\lambda\), \(- \lambda\) is....

The period of λ\lambda, −λ- \lambda is.

A

sin⁡λ+cos⁡pλ=1\sin\lambda + \cos p\lambda = 1

B

−sin⁡λ+cos⁡pλ=1- \sin\lambda + \cos p\lambda = 1

C

sin⁡λ=0\sin\lambda = 0

D

None of these

Answer

sin⁡λ=0\sin\lambda = 0

Explanation

Solution

f(x)=sin⁡(πxn−1)+cos⁡(πxn)f ( x ) = \sin \left( \frac { \pi x } { n - 1 } \right) + \cos \left( \frac { \pi x } { n } \right)

Period of sin⁡(πxn−1)=2π(πn−1)=2(n−1)\sin \left( \frac { \pi x } { n - 1 } \right) = \frac { 2 \pi } { \left( \frac { \pi } { n - 1 } \right) } = 2 ( n - 1 )

and period of cos⁡(πxn)=2π(πn)=2n\cos \left( \frac { \pi x } { n } \right) = \frac { 2 \pi } { \left( \frac { \pi } { n } \right) } = 2 n

Hence period of f(x)f ( x ) is L.C.M. of 2n2 n and 2(n−1)2 ( n - 1 )

⇒2n(n−1)\Rightarrow 2 n ( n - 1 ).