Discriminant

Given:

$$x^{2}+2ax+\left(3a+10\right)=0$$

Now $$\alpha$$ and $$\beta$$ be the roots given that $$\alpha < 1 < \beta$$.

Thus roots are distinct and real.

$$\Rightarrow$$ $$D > 0$$

$$\Rightarrow$$ $$4a^2 - 4(3a + 10) > 0$$

$$\Rightarrow$$ $$a^2 - 3a - 10 > 0$$

$$\Rightarrow$$ $$(a - 5)(a + 2) > 0$$

$$\Rightarrow$$ $$a < - 2$$ OR $$ a > 5$$   ----- (i)

The graph of the equation will be an upward parabola as the coefficient of $$x^2$$ is $$1$$, which is greater than $$0$$

Thus, we if we put $$x = 1$$ in the given quadratic equation, the value should be negative.

$$\Rightarrow$$ $$1 + 2a + (3a + 10) < 0$$

$$\Rightarrow$$ $$ 5a < -11$$

$$\Rightarrow$$ $$a < -\dfrac{11}{5}$$ ----(ii)

Thus by taking the intersection of (i) and (ii), we get the final interval for $$a$$ as,

$$a \in \left(-\infty, -\dfrac{11}{5}\right)$$

Hence, option A is the correct choice.

The equation given to us is $$x\lvert x+4 \rvert + 3\lvert x+2 \rvert + 10 = 0$$

Critical points for this are $$ x = -4, -2$$

Case 1: $$ x \ge -2$$

Then, $$ |x+4| = x+4,\quad |x+2| = x+2 $$

The equation becomes, 

$$ x(x+4) + 3(x+2) + 10 = 0$$
or, $$ x^2 + 4x + 3x + 6 + 10 = 0$$
or, $$  x^2 + 7x + 16 = 0 $$

For this equation, the discriminant is 49 - 64 = -15 < 0. So, this has no real solution.

Case 2: $$ -4 \le x < -2$$

Then, $$ |x+4| = x+4,\quad |x+2| = -(x+2) $$

The equation becomes,

$$ x(x+4) + 3(-(x+2)) + 10 = 0$$
or, $$ x^2 + 4x - 3x - 6 + 10 = 0 $$
or, $$ x^2 + x + 4 = 0 $$

For this equation too, the discriminant is  1 - 16 = -15 < 0. So this too has no real solution.

Case 3: $$ x < -4$$

Then, $$ |x+4| = -(x+4),\quad |x+2| = -(x+2) $$

The equation becomes, 

$$x(-(x+4)) + 3(-(x+2)) + 10 = 0$$
or, $$ x^2 - 4x - 3x - 6 + 10 = 0 $$
or, $$ x^2 - 7x + 4 = 0 $$

or, $$ x = \dfrac{-7 \pm \sqrt{49 + 16}}{2} = \dfrac{-7 \pm \sqrt{65}}{2} $$

Now we check if the solution values lie in the domain x < -4:

• $$ \frac{-7 + \sqrt{65}}{2} \approx 0.53$$ , which is not a valid solution
• $$ \frac{-7 - \sqrt{65}}{2} \approx -7.53$$ , which is a valid solution. 

Hence, the number of distinct real solutions is 1. 

Given:

$$x^{4} - ax^{2} + 9 = 0$$

Let $$t = x^2$$ $$\Rightarrow$$ $$t^2 = x^4$$

$$\Rightarrow$$ $$t^2 - at + 9 = 0$$

Now all roots are real and distinct,

$$\therefore$$ $$ D > 0$$

$$\Rightarrow$$ $$a^2 - 36 > 0$$

$$\Rightarrow$$ $$ a > 6$$ OR $$ a < -6$$

We want smallest positive integral value of $$a$$

$$\therefore$$ $$a > 6$$

Smallest positive integral value of $$a$$ is $$7$$

Thus, option B is the correct choice.