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If `f:R -> R, g:R -> R` defined as `f(x) = sin x and g(x) = x^2`, then find the value of `(gof)(x) and (fog)(x) `and also prove that `gof != fog`. |
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Answer» (fog) (x) = f[(g(x)] = `f(x^(2)) = sin x^(2)`Ans. and (gof)(x)= g[f(x)] =g(sinx) = `(sinx)^(2)= sin^(2) x` Ans. Therefore, fog `ne` gof . Hence Proved. |
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