`" If " g(x) = |{:(a^(-x),,e^(x log _(e)a),,x^(2)),(a^(-3x),,e^(3x log_(e)a),,x^(4)),(a^(-5x),,e^(5x log _(e)a),,1):}|` then

Options

A. graphs of g(x) is symmetrical about the origin
B. gpaphs of g(x) is symmertical about the y-axis
C. `(d^(4) g(x))/(dx^(4)) |_(x=0) =0`
D. `f(x) =g(x) xx log .((a-x)/(a+x))` is an odd function

Answer

A, C

Explanation

`g(x) = |{:(a^(-x) ,,e^(log_(e)a^(x)),,x^(2)),(a^(-3x),,e^(log_(e)a^(3x)),,x^(4)),(a^(-5x),,e^(log_(e)a^(5x)),,1):}|=|{:(a^(-x),,a^(x),,x^(2)),(a^(-3x),,a^(3x),,x^(4)),(a^(-5x),,a^(5x),,1):}| (e^(loga^(x)=a^(x)))`
`rArr g(-x) = |{:(a^(x),,a^(-x),,x^(2)),(a^(3x),,a^(-3x),,x^(4)),(a^(5x),,a^(-5x),,1):}|=-|{:(a^(-x),,a^(x),,x^(2)),(a^(-3x),,a^(3x),,x^(4)),(a^(-5x),,a^(5x),,1):}|`
[interchanging 1st the 2nd columns]
`=-g(x)`
`rArr g(x) +g(-x) =0`
`rArr g(x) ` is an odd function
Hence the graph is symmetrical about origin .Also `g_(4) (x)` is an odd function [where `g_(4)`(x) is fourth derivative of g(x) ].Hence
`g_(4)(x)=-g_(4) (-x)`
`rArr g_(4) (0)= -g_(4) (0)`
`rArr g_(4) (0)=0`