Given
f(x)=ax−1xn(ax+1) is symmetrical about the y-axis.
such functions where f(−x)=f(x) are termed as even functions.
Hence f(x) is an even function.
f(−x)=a−x−1(−x)n(a−x+1)−1−ax(−1)nxn(1+ax)
Since it is an even function we have
f(x)=f(−x)⇒ax−1xn(ax+1)=1−ax(−1)nxn(1+ax)⇒(ax−1)=−(ax−1)(−1)n⇒(−1)n=(−1)
this is possible only when n is odd.
∴ for (b)23
then (−1)23=(1)13=1≠−1 hence this too is not possible .
∴(d)−13