If x is real,the function f(x) = frac{(x - a) | vedclass

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(D) Let $y = \frac{(x - a)(x - b)}{(x - c)}$.Then $y(x - c) = x^2 - (a + b)x + ab$,which rearranges to $x^2 - (a + b + y)x + (ab + cy) = 0$.For $x$ to

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$a < c < b$

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(D) Let $y = \frac{(x - a)(x - b)}{(x - c)}$.Then $y(x - c) = x^2 - (a + b)x + ab$,which rearranges to $x^2 - (a + b + y)x + (ab + cy) = 0$.For $x$ to be real,the discriminant $D$ must be $\ge 0$:$D = (a + b + y)^2 - 4(ab + cy) \ge 0$.Expanding this,we get $y^2 + 2y(a + b - 2c) + (a - b)^2 \ge 0$.For the function to assume all real values $y$,this quadratic in $y$ must be non-negative for all $y$. However,since the coefficient of $y^2$ is positive,this quadratic will take all real values if its discriminant $D_y $D_y = [2(a + b - 2c)]^2 - 4(a - b)^2 $4(a + b - 2c)^2 - 4(a - b)^2 $(a + b - 2c)^2 - (a - b)^2 Using $A^2 - B^2 = (A-B)(A+B)$:$(a + b - 2c - a + b)(a + b - 2c + a - b) $(2b - 2c)(2a - 2c) $4(b - c)(a - c) $(c - b)(c - a) This inequality holds if $c$ lies between $a$ and $b$,i.e.,$a < c < b$ or $b < c < a$.

If $x$ is real,the function $f(x) = \frac{(x - a)(x - b)}{(x - c)}$ will assume all real values,provided

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