A) \[f\]is even and non-periodic
B) \[f\]is odd and periodic
C) \[f\]is odd and non-periodic
D) \[f\]is even and periodic
Correct Answer: B
Solution :
Given \[f\left( \lambda +x \right)=f\left( \lambda -x \right)\] | ?(1) |
\[f\left( 2\lambda +x \right)=-f\left( 2\lambda -x \right)..(2)\] | |
For \[\lambda >0\] | |
Replacing \[x\,\operatorname{by}\,\lambda -x\]in (1), we get \[f\left( 2\lambda -x \right)=f\left( x \right)..(3)\] | |
\[\therefore \,\]from (2) and (3), \[f\left( x \right)=-f\left( 2\lambda +x \right)\] |
\[\Rightarrow f\left( x \right)=-\left[ -f\left( 2\lambda +2\lambda \right) \right]\]\[\Rightarrow f\left( x \right)=f\left( x+4\lambda \right)..(4)\] |
\[\Rightarrow f\left( x \right)\] is periodic with period \[4\lambda \] |
Further from (3), replacing \[x\]by \[-x\], we get \[f\left( 2\lambda +x \right)=f\left( -x \right)..(5)\] |
From (2),(3), and (5), we have \[f\left( -x \right)=f\left( 2\lambda +x \right)=-f\left( 2\lambda -x \right)=-f\left( x \right)\]\[i.e\,f\left( -x \right)=-f\left( x \right)\] |
\[\Rightarrow f\left( x \right)\]is odd function |
Thus, f is odd and periodic function. |
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