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For x in R, the expression (x^2+2x+c)/(x...

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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`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`.
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