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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 |
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Answer» 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)` `=>g(-y)+g(y) = g(y)->(1)` If we replace, `y` with `-y`, then, `=>g(-y)+g(y) = g(-y)->(2)` From (1) and (2), we can see that `g(y) and g(-y)` are equal. It means `g(x)` is an even function. So, graph of `g` will be symmetrical about `y`-axis. `:.` Option `(b)` and `(d)` are the correct options. |
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