If D subseteq R and f: D rightarrow R defined | vedclass

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(C) Let $y = \frac{x^2+x+a}{x^2-x+a}$.Rearranging the terms, we get $y(x^2-x+a) = x^2+x+a$.$yx^2 - yx + ay = x^2 + x + a$.$(y-1)x^2 - (y+1)x + a(y-1)

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$(-\infty, 0)$

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(C) Let $y = \frac{x^2+x+a}{x^2-x+a}$.Rearranging the terms, we get $y(x^2-x+a) = x^2+x+a$.$yx^2 - yx + ay = x^2 + x + a$.$(y-1)x^2 - (y+1)x + a(y-1) = 0$.For $f$ to be a surjection, this quadratic equation in $x$ must have real roots for every $y$ in the range.Thus, the discriminant $D \geq 0$.$D = (-(y+1))^2 - 4(y-1)(a(y-1)) \geq 0$.$(y+1)^2 - 4a(y-1)^2 \geq 0$.$y^2 + 2y + 1 - 4a(y^2 - 2y + 1) \geq 0$.$(1-4a)y^2 + (2+8a)y + (1-4a) \geq 0$.For this to hold for all $y$, the coefficient of $y^2$ must be positive, i.e.,$1-4a > 0 \Rightarrow a Also, the discriminant of this quadratic in $y$ must be $\leq 0$.$(2+8a)^2 - 4(1-4a)^2 \leq 0$.$4(1+4a)^2 - 4(1-4a)^2 \leq 0$.$(1+4a-1+4a)(1+4a+1-4a) \leq 0$.$(8a)(2) \leq 0$ $\Rightarrow 16a \leq 0$ $\Rightarrow a \leq 0$.Combining $a < 1/4$ and $a \leq 0$, we get $a \in (-\infty, 0]$.

If $D \subseteq R$ and $f: D \rightarrow R$ defined by $f(x) = \frac{x^2+x+a}{x^2-x+a}$ is a surjection, then '$a$' lies in the interval

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