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. |
You need to login to perform this action.
You will be redirected in
3 sec
