Let $P\left(x\right)$   be a polynomial of degree 4 having a local maximum at $x=2 \text{and }\underset{x\to 0}{lim}\left(3-\frac{P\left(x\right)}{x}\right)=27\text{. If }P\left(1\right)=-9$   and  $P^{\prime \prime }\left(\mathrm{x}\right)$  has a local minimum at $x=2$. If global maximum value of  $\mathrm{y}={\mathrm{P}}^{\mathrm{\prime }}\left(\mathrm{x}\right)$ on the set    $A=\{x:x^2+12\le 7x\}$ is 4M, then the value of M is

$\underset{x\to 0}{lim}(3-\frac{P\left(x\right)}{x})=27\Rightarrow P\left(x\right)$  has no constant term let $P\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx$

$\Rightarrow 3-d=27\Rightarrow d=-24$

$\mathrm{P}\left(\mathrm{x}\right)={\mathrm{ax}}^4+b{\mathrm{x}}^3+{\mathrm{cx}}^2-24\mathrm{x}; \mathrm{P}\left(2\right)=0,\mathrm{p}\left(1\right)=-9,{\mathrm{p}}^{\text{'}\text{'}\text{'}}\left(2\right)=0$

$P\left(x\right)=4a{x}^3+30x^2+2cx-24; p^{\prime \prime }\left(x\right)=12a{x}^2+6bx+2c$

$p^{\prime \prime \mathrm{\prime }}\left(x\right)=24ax+6b; a+b+c-24=-9$

$\Rightarrow a+b+c=15$  .........(1)

$p^{\mathrm{\prime }}\left(2\right)=0$

$\Rightarrow 4a\left(8\right)+3b\left(4\right)+2c\left(2\right)-24=0$

$\Rightarrow 8a+3b+c=6$ .......(2)

${\mathrm{p}}^{\prime \prime \mathrm{\prime }}\left(2\right)=0$

$\Rightarrow 24\mathrm{a}\left(2\right)+6\left(\mathrm{b}\right)=0\Rightarrow 8\mathrm{a}+\mathrm{b}=0$ .........(3)

Solving (1), (2) and (3)

$\mathrm{a}=1, \mathrm{b}=-8, \mathrm{c}=22$

$\Rightarrow P\left(x\right)=x^4-8x^3+22x^2-24x;$

$P^{\mathrm{\prime }}\left(x\right)=4x^3-24x^2+44x-24=4\left[x^3-6x^2+11x-6\right]$

$P^{\mathrm{\prime }}\left(x\right)=4\left[(x-1)(x-2)(x-3\right]$

$P^{\prime \prime }\left(x\right)=4\left[3x^2-12x+11]>0\mathrm{\forall }x\in [3,4]=4\left[\right(3\left)\right(2\left)\right(1)\right]=24=4\mathrm{M}\Rightarrow \mathrm{M}=6$