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(D) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\De

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$\frac{5}{2}$

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(D) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Comparing the given equation $2x^2 + 3xy - 2y^2 - 5x + 2fy - 3 = 0$ with the general form:$a = 2, h = \frac{3}{2}, b = -2, g = -\frac{5}{2}, f = f, c = -3$.Substituting these values into the condition $\Delta = 0$:$(2)(-2)(-3) + 2(f)(-\frac{5}{2})(\frac{3}{2}) - 2(f)^2 - (-2)(-\frac{5}{2})^2 - (-3)(\frac{3}{2})^2 = 0$$12 - \frac{15f}{2} - 2f^2 + 2(\frac{25}{4}) + 3(\frac{9}{4}) = 0$$12 - 7.5f - 2f^2 + 12.5 + 6.75 = 0$$-2f^2 - 7.5f + 31.25 = 0$Multiplying by $-2$ to simplify: $4f^2 + 15f - 62.5 = 0$,or $8f^2 + 30f - 125 = 0$.Using the quadratic formula $f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$f = \frac{-30 \pm \sqrt{900 - 4(8)(-125)}}{16} = \frac{-30 \pm \sqrt{900 + 4000}}{16} = \frac{-30 \pm \sqrt{4900}}{16} = \frac{-30 \pm 70}{16}$.Two possible values for $f$ are $f_1 = \frac{40}{16} = \frac{5}{2}$ and $f_2 = \frac{-100}{16} = -\frac{25}{4}$.Checking the options,$\frac{5}{2}$ matches option $D$.

If $2x^2 + 3xy - 2y^2 - 5x + 2fy - 3 = 0$ represents a pair of straight lines,then one of the possible values of $f$ is:

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