Q. If `x` is real, then `(x^2+2x+c)/(x^2+4x+3c)` can take all real values if a) `0<c<2` b) `-1<c<1` c) `-1<c<2` d) None of these

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Find all values of c such that `(x^2+2x+c)/(x^2+4x+3c)` can take all real values:

A. Assuming that the question means can accept all real values of the input x:

(a) Suppose c=0. Then the expression becomes `(x^2+2x)/(x^2+4x)` ; it is clear that x cannot be zero so answers b and c are incorrect.

(b) Suppose c=1. Then the expression becomes `(x^2+2x+1)/(x^2+4x+3)` or

`((x+1)^2)/((x+1)(x+3))=(x+1)/(x+3)` and we have x cannot be -1 or -3 so a is incorrect.

Therefore the answer is (d) none of these.

(B) If the question meant that the expression could take on all output values, the answer is also (d).

If c=0 the function cannot equal 1; if c=1 then the function cannot be 1.

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