The correct option is D k(x)=x(x2−|x|)
Clearly h(x)=log(x6+√x2+1) is an even function because it has even powers of x
For f(x)=x(ex−1ex+1)
f(−x)=−x(e−x−1e−x+1)=−x(1−exex+1)=f(x)
∴ f(x) is even function.
For g(x)=ax−1ax+1
g(−x)=a−x−1a−x+1⇒g(−x)=1−ax1+ax∴g(−x)=−g(x)
∴g(x) is odd function.
For k(x)=x(x2−|x|)
k(−x)=−x(x2−|x|)⇒k(−x)=−k(x)
∴k(x) is odd function.