The set of values of ‘a’ for which f(x) = ax2 + 2x(1 – a) – 4 is negative for exactly three integral values of x, is

To determine the set of values of $a$ for which the quadratic function $f(x) = ax^2 + 2x(1 - a) - 4$ is negative for exactly three integral values of $x$, we need to analyze the behavior of the quadratic function. Here are the steps to solve this problem:

1. Identify the quadratic function: $f(x) = ax^2 + 2x(1 - a) - 4$

2. Determine the general form of a quadratic function: A quadratic function $f(x) = ax^2 + bx + c$ is a parabola. The sign of $a$ determines whether the parabola opens upwards ($a > 0$) or downwards ($a < 0$).

3. Find the roots of the quadratic function: To find the roots, we set $f(x) = 0$: $ax^2 + 2x(1 - a) - 4 = 0$ This is a standard quadratic equation of the form $ax^2 + bx + c = 0$, where $a = a$, $b = 2(1 - a)$, and $c = -4$.

4. Calculate the discriminant: The discriminant $\Delta$ of a quadratic equation $ax^2 + bx + c = 0$ is given by: $\Delta = b^2 - 4ac$ Substituting $b = 2(1 - a)$ and $c = -4$: $\Delta = [2(1 - a)]^2 - 4a(-4)$ Simplify the expression: $\Delta = 4(1 - a)^2 + 16a$ $\Delta = 4(1 - 2a + a^2) + 16a$ $\Delta = 4 + 8a + 4a^2$ $\Delta = 4a^2 + 8a + 4$ $\Delta = 4(a^2 + 2a + 1)$ $\Delta = 4(a + 1)^2$

5. Analyze the discriminant: Since $\Delta = 4(a + 1)^2$, the discriminant is always non-negative. This means the quadratic equation always has real roots.

6. Determine the vertex of the parabola: The vertex of the parabola $f(x) = ax^2 + bx + c$ is given by: $x = -\frac{b}{2a}$ Substituting $b = 2(1 - a)$: $x = -\frac{2(1 - a)}{2a} = -\frac{1 - a}{a} = \frac{a - 1}{a}$

7. Determine the conditions for $f(x)$ to be negative for exactly three integral values of $x$: For $f(x)$ to be negative for exactly three integral values of $x$, the parabola must intersect the x-axis at two points, and the vertex must lie between two of these integral values. This implies that the parabola must open upwards ($a > 0$) and the roots must be such that there are exactly three integer values between them.

8. Find the range of $a$: Since the parabola opens upwards ($a > 0$), the roots must be such that there are exactly three integer values between them. This means the distance between the roots must be slightly more than 3 but less than 4.

   Let the roots be $x_1$ and $x_2$. The distance between the roots is given by: $|x_2 - x_1| = \frac{\sqrt{\Delta}}{a}$ Substituting $\Delta = 4(a + 1)^2$: $|x_2 - x_1| = \frac{2|a + 1|}{a}$ For the distance to be slightly more than 3 but less than 4: $3 < \frac{2|a + 1|}{a} < 4$ Solving these inequalities: $3a < 2|a + 1| < 4a$

   For $a > 0$: $3a < 2(a + 1) < 4a$ $3a < 2a + 2 < 4a$ $a < 2 \quad \text{and} \quad 2 < 2a$ $a < 2 \quad \text{and} \quad 1 < a$

   Combining these inequalities: $1 < a < 2$

Therefore, the set of values of $a$ for which $f(x) = ax^2 + 2x(1 - a) - 4$ is negative for exactly three integral values of $x$ is: $\boxed{(1, 2)}$