Sum of roots

We have $$(x-1)^2- 5|x-1| +6=0$$

Expanding and simplifying, we get,

$$x^2+1-2x - 5|x-1| + 6 = 0$$

$$\Rightarrow x^2 - 2x - 5|x-1|+7 = 0$$

Case A: $$x\geq 1$$

The equation simplifies to

$$x^2-2x-5x+5+7=0$$

$$\Rightarrow x^2 - 7x + 12 = 0$$

The sum of roots of the quadratic equation above is $$\dfrac{-b}{a} = \dfrac{-(-7)}{1} = 7$$

Case B: $$x<1$$

The equation simplifies to

$$x^2-2x+5x-5+7 = 0$$

$$\Rightarrow x^2+3x+2=0$$

The sum of roots of the quadratic equation above is $$\dfrac{-b}{a} = \dfrac{-3}{1} = -3$$

Thus, the sum of all roots of the given equation is $$7+(-3) = 4$$. 

Option A is the correct answer.

We have the equation: $$12x^{2}-20x+3\lambda=0, \lambda\in Z$$

Sum of the roots: $$\alpha+\beta\ =-\dfrac{b}{a}=-\ \dfrac{\left(-20\right)}{12}=\dfrac{5}{3}$$

Product of the roots: $$\alpha\ \beta\ =\dfrac{c}{a}=\dfrac{3λ}{12}=\dfrac{λ}{4}$$

Now, we know that: $$\left(\beta\ -\alpha\ \right)^2=\left(\beta\ +\alpha\ \right)^2-4\alpha\ \beta\ $$

$$\left(\beta\ -\alpha\ \right)^2=\left(\dfrac{5}{3}\right)^2-4\left(\dfrac{λ}{4}\right)$$

$$\left(\beta\ -\alpha\ \right)^2=\dfrac{25}{9}-λ$$

$$\left(\beta\ -\alpha\ \right)^2=\dfrac{25-9λ}{9}$$

Taking the square root of both sides:

$$\left|\beta\ -\alpha\right|=\dfrac{\sqrt{\left(25-9λ\right)}}{3}$$

Now, we are given that: 

$$\dfrac{1}{2}\le\ \left|\beta\ -\alpha\right|\le\ \dfrac{3}{2}$$

Substituting:

$$\dfrac{1}{2}\le\ \dfrac{\sqrt{\left(25-9λ\right)}}{3}\le\dfrac{3}{2}$$

Multiplying by 3:

$$\dfrac{3}{2}\le\ \sqrt{\left(25-9λ\right)}\le\dfrac{9}{2}$$

Or, $$\dfrac{9}{4}\le\ 25-9λ\le\dfrac{81}{4}$$

Subtracting 25 from all three:

$$-\dfrac{91}{4}\le\ -9λ\le-\dfrac{19}{4}$$

Dividing all three by -9:

$$\dfrac{91}{36}\ge\ \ λ\ge\dfrac{19}{36}$$

Or, we can say, $$0.527 \leq \lambda \leq 2.527$$

The only integers that fall within the range [0.527, 2.527] are: 1 and 2. 

Hence, the sum of all possible values of $$\lambda$$ is 3.

Let $$y=\ln\ x$$

Since $$x>0$$, this substitution is valid.

$$\therefore$$ $$x=e^y$$

and $$x^8=\left(e^y\right)^8=e^{8y}$$

Substitute into the equation 

$$e^{5\left(\ln\ x\right)^2-3}=e^{8y}$$

Since $$\left(\ln\ x\right)=y$$

$$\therefore$$ $$e^{5\left(y\right)^2-3}=e^{8y}$$

Equate the exponents as the bases are same (e)

$$5y^2-3=8y$$

$$\therefore$$ $$5y^2-8y-3=0$$

Sum of the quadratic equation =  $$y_1+y_2$$ which is equal to $$-\dfrac{b}{a}$$

$$\therefore$$ $$y_1+y_2=8/5$$------(i)

To find: $$x_1\times\ x_2$$

We know  $$x=e^y$$ 

$$\therefore$$ $$x_1\times\ x_2=e^{y_1+y_2}$$

Substitute the value of  $$y_1+y_2$$ from---(i)

$$\therefore$$  $$x_1\times\ x_2=e^\dfrac{8}{5}$$.

Hence, option A is the correct answer