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For `x in R`, the expression `(x^2+2x+c)/(x^2+4x+3c)` can take all real values if `c in` |
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Answer» `f(x) = (x^2+2x+c)/(x^2+4x+3c)` When `c = 0`, `f(x) = (x^2+2x)/(x^2+4x) = (x(x+2))/(x(x+4))` But, it is given that `x in R`, so `x !=0`. When `c = 1`, `f(x) = (x^2+2x+1)/(x^2+4x+3) = ((x+1)^2)/((x+3)(x+1))` As, `x in R`, so `x!=-1 and x! = -3`. So, `c` can take any values between `0` and `1`. `:. c in (0,1)`. |
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