(1)

$f\left(x\right)=3\sqrt{x-2}+\sqrt{4-x}$

$f\text{'}\left(x\right)=\frac{3}{2\sqrt{x-2}}-\frac{1}{2\sqrt{4-x}}$

$\therefore$     $f\text{'}\left(x\right)=0\Rightarrow 9(4-x)=(x-2)\Rightarrow x=\frac{19}{5}\in [2,4]$

Now, $f\left(2\right)=\sqrt{2}, f\left(\frac{19}{5}\right)=2\sqrt{5}, f\left(4\right)=3\sqrt{2}$

$\therefore$    $\alpha =\sqrt{2}, \beta =2\sqrt{5}\Rightarrow {\alpha }^2+2{\beta }^2=2+40=42$

(3)

(4)

$\text{We have, }f\left(x\right)=2x^3-9a{x}^2+12a^{2}x+1$

$\Rightarrow f\text{'}\left(x\right)=6x^2-18ax+12a^2=6(x^2-3ax+2a^2)$

$\Rightarrow 6(x-a)(x-2a)=0$

      $f\text{'}\left(x\right)=0\Rightarrow x=a,2a$

$\text{Now, }f\text{'}\text{'}\left(x\right)=12x-18a$

             $f\text{'}\text{'}\left(a\right)=12a-18a=-6a<0 \left(\text{maximum}\right)$

             $f\text{'}\text{'}\left(2a\right)=24a-18a=6a>0 \left(\text{minimum}\right)$

$\therefore f\left(x\right)\text{ has a local maximum at }x=a\text{ and minimum at }x=2a.$

$\therefore \alpha =a\text{ and }{\alpha }^2=2a$

$\Rightarrow a^2=2a\Rightarrow a=0,2 \Rightarrow a=2=\alpha (\because a>0)$

$\Rightarrow {\alpha }^2=4$

$\text{∴ Equation having roots }\alpha \text{ and }{\alpha }^2\text{ is }$

       $x^2-(\alpha +{\alpha }^2)x+\alpha ·{\alpha }^2=0$

$\Rightarrow x^2-(2+4)x+2\times 4=0\Rightarrow x^2-6x+8=0$

(4)

Equation of line $AB:\frac{x}{a}+\frac{y}{b}=1$

Since, it passes through (3, 5)

$\therefore$    $\frac{3}{a}+\frac{5}{b}=1$

$\Rightarrow 3b+5a=ab \Rightarrow 5a-ab=-3b \Rightarrow a(5-b)=-3b$

$\Rightarrow a=\frac{3b}{b-5}$

Area of triangle $=$$|\frac{1}{2}\times a\times b|$$=\frac{1}{2}\times \frac{3b}{b-5}\times b$

$\Rightarrow f\left(b\right)=\frac{3b^2}{2b-10}$

$\Rightarrow f\text{'}\left(b\right)=0 \Rightarrow (2b-10)6b-2\left(3b^2\right)=0$

$\Rightarrow 12b^2-60b-6b^2=0 \Rightarrow 6b^2-60b=0$

$\Rightarrow b^2-10b=0 \Rightarrow b(b-10)=0$

          $b=0 \text{or} b=10$

So for minimum area, $b=10$

$\text{Hence, minimum area}=\frac{3b^2}{2b-10}=\frac{3\times {\left(10\right)}^2}{2\times 10-10}$

                                                                        $=\frac{300}{10}=30 \text{sq. units}$

(1)

Let ABC be the required triangle whose area is A.

$\therefore$    $A=\frac{1}{2}$$\left|0\right(y-y)+x(y-0)-x(0-y\left)\right|$

$=\frac{1}{2}$$|xy+xy|$$=\frac{1}{2}\left(2xy\right)=xy$

$=x(-2x^2+54)=-2(x^3-27x)$

$\Rightarrow A=-2(x^3-27x)\Rightarrow \frac{dA}{dx}=-6x^2+54$

For critical point, $\frac{dA}{dx}=0$

$\Rightarrow 3x^2-27=0\Rightarrow x^2=9\Rightarrow x=\pm 3$

Now, $\frac{d^{2}A}{d{x}^2}=-12x,$$\frac{d^{2}A}{d{x}^2}{|}_{x=3}$$<0$

$\therefore$   Maximum area, $A=-2(27-(27\times 3\left)\right)$

                                         $=-2\times (-54)=108 \text{sq. units.}$

(4)

Given, $f\left(x\right)={(x+3)}^2{(x-2)}^3$                               ...(i)

Differentiate (i), w.r.t. $x,$ we get

$f\text{'}\left(x\right)=2(x+3){(x-2)}^3+3{(x+3)}^2{(x-2)}^2$

           $=(x+3){(x-2)}^2\left[2\right(x-2)+(x+3)\times 3]$

            $=(x+3){(x-2)}^2(5x+5)$

For maxima / minima, $f\text{'}\left(x\right)=0$

$\Rightarrow (x+3){(x-2)}^2(5x+5)=0\Rightarrow x=-3,-1,2$

Now find the value of (i) at $x=-4,-3,-1,2,4$

        $f(-4)={(-4+3)}^2{(-4-2)}^3=1\times (-216)=-216$

        $f(-3)=0$

        $f(-1)={(-1+3)}^2{(-1-2)}^3=4\times (-27)=-108$

        $f\left(2\right)=0$

         $f\left(4\right)={(4+3)}^2{(4-2)}^3=49\times 8=392$

$\therefore$    Maximum value, $M=392$

          Minimum value, $m=-216$

Hence, $M-m=392+216=608$