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Let g(x+y),g(x).g(y) and g(x-y) are in A...

Let g(x+y),g(x).g(y) and g(x-y) are in AP. For all x,y and `g(0)!=0` then a) g(2)=0 (b) the graph of g is symmetry about y-axis (c) g ia an odd unction (d) g(0)=-1

Text Solution

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As `g(x+y),g(x).g(y), g(x-y)` are in `AP`.
`:. g(x+y)+g(x-y) = 2g(x)*g(y)`
When `y = 0`,
`=>g(x)+g(x) = 2g(x)*g(0)`
`=>g(0) = 1`
It means, option `(d)` is correct.
When, `x = 0`,
`=>g(-y)+g(y) = 2g(0)*g(y)`
...
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