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(B) Let $y = \frac{x+a}{2 x^2-3 x+1}$,where $y \in \mathbb{R}$.Since $y$ takes all real values,the equation $2 y x^2 - (3 y + 1) x + (y - a) = 0$ must

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$-1 < a < -\frac{1}{2}$

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(B) Let $y = \frac{x+a}{2 x^2-3 x+1}$,where $y \in \mathbb{R}$.Since $y$ takes all real values,the equation $2 y x^2 - (3 y + 1) x + (y - a) = 0$ must have real roots for $x$ for all $y \in \mathbb{R}$.For $y 
eq 0$,the discriminant $D \geq 0$:$D = (3 y + 1)^2 - 4(2 y)(y - a) \geq 0$$9 y^2 + 6 y + 1 - 8 y^2 + 8 a y \geq 0$$y^2 + (8 a + 6) y + 1 \geq 0$For this quadratic in $y$ to be non-negative for all $y \in \mathbb{R}$,the discriminant of this quadratic must be less than or equal to $0$ (since the coefficient of $y^2$ is $1 > 0$).$D_y = (8 a + 6)^2 - 4(1)(1) \leq 0$$(8 a + 6)^2 - 4 \leq 0$$(8 a + 6 - 2)(8 a + 6 + 2) \leq 0$$(8 a + 4)(8 a + 8) \leq 0$$4(2 a + 1) \cdot 8(a + 1) \leq 0$$32(2 a + 1)(a + 1) \leq 0$$(2 a + 1)(a + 1) \leq 0$Thus,$-1 \leq a \leq -\frac{1}{2}$.However,if $y=0$,then $x+a=0 \Rightarrow x=-a$. The denominator $2x^2-3x+1 = (2x-1)(x-1)$ must not be zero at $x=-a$. So $2(-a)-1 
eq 0$ and $-a-1 
eq 0$,which means $a 
eq -1/2$ and $a 
eq -1$. Therefore,$-1 < a < -\frac{1}{2}$.

For real values of $x$ and $a$,if the expression $\frac{x+a}{2 x^2-3 x+1}$ assumes all real values,then

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