Question:

Step 1: Assign the given expression to a variable. Let the given expression be equal to $y$. This allows us to analyze the range of the expression by considering the possible values of $y$. $$y = \frac{2x^2 - 5x + 3}{4x - \lambda}$$ Step 2: Rearrange the equation into a quadratic form in terms of $x$. Multiply both sides by $(4x - \lambda)$ and rearrange the terms to form a quadratic equation in $x$. $$y(4x - \lambda) = 2x^2 - 5x + 3$$ $$4xy - \lambda y = 2x^2 - 5x + 3$$ $$2x^2 + (-5 - 4y)x + (3 + \lambda y) = 0$$ $$2x^2 - (5 + 4y)x + (3 + \lambda y) = 0$$ Step 3: State the condition for the expression to take all real values. For the expression to take all real values, for every real value of $y$, there must exist a real value of $x$ that satisfies the quadratic equation obtained in Step 2. This means the quadratic equation in $x$ must have real solutions for all $y \in \mathbb{R}$. Step 4: Apply the discriminant condition for real roots. For a quadratic equation $Ax^2 + Bx + C = 0$ to have real roots, its discriminant $\Delta = B^2 - 4AC$ must be greater than or equal to zero. Applying this to our quadratic in $x$: $$\Delta_x = (-(5 + 4y))^2 - 4(2)(3 + \lambda y) \ge 0$$ $$(5 + 4y)^2 - 8(3 + \lambda y) \ge 0$$ $$25 + 40y + 16y^2 - 24 - 8\lambda y \ge 0$$ $$16y^2 + (40 - 8\lambda)y + 1 \ge 0$$ This inequality must hold true for all real values of $y$. Step 5: Determine the condition for the quadratic in $y$ to be always non-negative. For a quadratic expression $ay^2 + by + c$ to be always non-negative (i.e., $ay^2 + by + c \ge 0$ for all $y \in \mathbb{R}$), two conditions must be met: 1. The leading coefficient $a$ must be positive ($a > 0$). In our case, $a = 16$, which is positive. 2. The discriminant $\Delta_y$ of the quadratic in $y$ must be less than or equal to zero ($\Delta_y \le 0$). Applying the second condition to $16y^2 + (40 - 8\lambda)y + 1 \ge 0$: $$\Delta_y = (40 - 8\lambda)^2 - 4(16)(1) \le 0$$ $$(40 - 8\lambda)^2 - 64 \le 0$$ Step 6: Solve the inequality for $\lambda$. Solve the inequality obtained in Step 5 for $\lambda$: $$(40 - 8\lambda)^2 \le 64$$ Take the square root of both sides: $$-8 \le 40 - 8\lambda \le 8$$ Subtract 40 from all parts: $$-8 - 40 \le -8\lambda \le 8 - 40$$ $$-48 \le -8\lambda \le -32$$ Divide all parts by -8 and reverse the inequality signs: $$\frac{-48}{-8} \ge \lambda \ge \frac{-32}{-8}$$ $$6 \ge \lambda \ge 4$$ $$4 \le \lambda \le 6$$ Thus, the range of $\lambda$ is $[4, 6]$. The final answer is $\boxed{[4, 6]}$.