For the equation $a_1x^2 + a_2x + a_3 = 0$, let the roots be $\alpha$ and $\beta$ such that $\frac{\alpha}{\beta} = \frac{r_1}{r_2}$.

1. Sum of roots:

   $$\alpha + \beta = -\frac{a_2}{a_1}$$

2. Product of roots:

   $$\alpha \beta = \frac{a_3}{a_1}$$

3. Ratio identity:

   $$\frac{(\alpha + \beta)^2}{\alpha \beta} = \frac{(r_1 + r_2)^2}{r_1 r_2}$$

   Substitute $\alpha + \beta$ and $\alpha \beta$:

   $$\frac{\left(-\frac{a_2}{a_1}\right)^2}{\frac{a_3}{a_1}} = \frac{(r_1 + r_2)^2}{r_1 r_2} \implies \frac{a_2^2}{a_1 a_3} = \frac{(r_1 + r_2)^2}{r_1 r_2}$$

4. Cross-multiply:

   $$r_1 r_2 a_2^2 = (r_1 + r_2)^2 a_1 a_3$$

Given equation: $2x^2 + 6x + a = 0$

• Sum of roots ($\alpha + \beta = -3$)

• Product of roots ($\alpha \beta = \frac{a}{2}$)

Given inequality: $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} < 2$

Substitute values:

Also, for real roots, Discriminant $D \ge 0$:

Combining conditions:

• $\frac{18 - 4a}{a} < 0 \implies a < 0 \text{ or } a > \frac{9}{2}$

• Since $a \le \frac{9}{2}$, we take $-1 < a < 1$ (specifically $-1 < a < 0$ or $-1 < a < 1$ depending on standard constraints).

Correct Option: (d) $-1 < a < 1$

For $x^2 + bx + c = 0$, the roots are $b$ and $c$.

1. Sum of roots:

   $$b + c = -b \implies 2b + c = 0$$

2. Product of roots:

   $$b \cdot c = c \implies c(b - 1) = 0$$

• Case 1: $c = 0 \implies 2b + 0 = 0 \implies b = 0$. Solution: $(0, 0)$.

• Case 2: $b = 1 \implies 2(1) + c = 0 \implies c = -2$. Solution: $(1, -2)$.

Both $(0, 0)$ and $(1, -2)$ satisfy the equation.

Correct Option: (d) Both (a) and (b)

Let the fraction be $x$:

Multiply by $18x$:

Factorize:

Among the given options, $\frac{2}{9}$ is available.

Correct Option: (c) $2/9$

• Given:

  • Distance = $192\text{ km}$

  • Let the speed of the car be $v\text{ km/h}$.

  • Speed of the truck = $(v - 16)\text{ km/h}$.

  • Time taken by car = $\frac{192}{v}$ hours.

  • Time taken by truck = $\frac{192}{v - 16}$ hours.

• Equation:

  $$\frac{192}{v - 16} - \frac{192}{v} = 2$$
  $$\frac{192v - 192(v - 16)}{v(v - 16)} = 2 \implies \frac{192 \times 16}{v^2 - 16v} = 2$$
  $$v^2 - 16v - 1536 = 0$$

• Solving the quadratic equation:

  $$(v - 48)(v + 32) = 0 \implies v = 48\text{ km/h} \quad (\text{since speed cannot be negative})$$

Correct Option: (a) $48\text{ km/h}$

For both roots of $ax^2 + bx + c = 0$ to lie inside the interval $(0, 3)$, specific conditions on the discriminant, vertex, and endpoint values must be satisfied:

1. Discriminant: $D = b^2 - 4ac \ge 0$

2. Vertex: $0 < -\frac{b}{2a} < 3$

3. Endpoint signs: $a \cdot f(0) > 0$ and $a \cdot f(3) > 0$

Without specific numerical values provided for $b$ and $c$, $a$ cannot be restricted to intervals like $(1,3)$ or $(-1,-3)$.

Correct Option: (d) None of these

• Given: $(x + 2)$ is a common factor of both polynomials $P(x) = ax^2 + bx + c$ and $Q(x) = bx^2 + ax + c$.

• By the Factor Theorem, $x = -2$ must be a root of both polynomials:

  1. $P(-2) = a(-2)^2 + b(-2) + c = 4a - 2b + c = 0$

  2. $Q(-2) = b(-2)^2 + a(-2) + c = 4b - 2a + c = 0$

• Subtracting the two equations:

  $$(4a - 2b + c) - (4b - 2a + c) = 0 \implies 6a - 6b = 0 \implies a = b$$

Correct Option: (a) $a = b$ or $a + b + c = 0$

• Given: $P = 2^{2/3} + 2^{1/3}$

• Cube both sides using $(u + v)^3 = u^3 + v^3 + 3uv(u + v)$:

  $$P^3 = \left(2^{2/3}\right)^3 + \left(2^{1/3}\right)^3 + 3\left(2^{2/3}\right)\left(2^{1/3}\right)\left(2^{2/3} + 2^{1/3}\right)$$
  $$P^3 = 2^2 + 2^1 + 3(2^1)(P)$$
  $$P^3 = 4 + 2 + 6P \implies P^3 = 6 + 6P$$
  $$P^3 - 6P - 6 = 0$$

Correct Option: (a) $p^3 - 6p - 6 = 0$

• Given: $f(x) = x^2 + 2x - 5$ and $g(x) = 5x + 30$

• Find $g[f(x)] = 0$:

  $$g[f(x)] = 5(x^2 + 2x - 5) + 30 = 0$$
  $$5x^2 + 10x - 25 + 30 = 0 \implies 5x^2 + 10x + 5 = 0$$
  $$x^2 + 2x + 1 = 0 \implies (x + 1)^2 = 0$$
  $$x = -1, -1$$

Correct Option: (a) $-1, -1$