2621) f(x) = - f(-x) hence odd
2) f(x)=f(-x)
or [x2a]sinx + cosx = -[x2a]sinx + cosx
or [x2a]=0 , a>400
Q1:f(x)=(-1)[x] is even or odd?[.] is GINT.
Q2: If f:[-20,20]→R given by f(x)=[x2a]sin x+cos x is an even function.then find a.[.] is GINT.
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4 Answers
262
1708\hspace{-16}$(1)\;\;\; Here are two cases for $\mathbf{x}$\\\\ Either $\mathbf{x\in\mathbb{Z}}$ $O\mathbb{R}$ $\mathbf{x\notin\mathbb{Z}}$\\\\ $\bullet$ If $\mathbf{x\in\mathbb{Z}}$, Then $\mathbf{\left[-x\right]=-\left[x\right]}$\\\\ So $\mathbf{f(-x)=(-1)^{\left[-x\right]}=(-1)^{-\left[x\right]}=\frac{(-1)^\left[x\right]}{(-1)^{2\left[x\right]}}}$\\\\ So $\mathbf{f(-x)=(-1)^{\left[x\right]}=f(x)}$ (Even Function)\\\\ bcz $\mathbf{(-1)^{2\left[x\right]}=1}$\\\\ Similarly \\\\ $\bullet$ If $\mathbf{x\notin\mathbb{Z}}$, Then $\mathbf{\left[-x\right]=-\left[x\right]-1}$\\\\ So $\mathbf{f(-x)=(-1)^{\left[-x\right]}=(-1)^{-\left[x\right]-1}=-\frac{(-1)^\left[x\right]}{(-1)^{2\left[x\right]}}}$\\\\ So $\mathbf{f(-x)=-(-1)^{\left[x\right]}=-f(x)}$ (Odd Function)\\\\ bcz $\mathbf{(-1)^{2\left[x\right]}=1}$\\\\ So $\mathbf{f(x)=(-1)^{\left[x\right]}}$ is Even as well as Odd Function
