Let P(x) be a polynomial of degree 4 having a local maximum at x = 2 and $\underset{x\to 0}{\lim }(3-\frac{P\left(x\right)}{x})=27$ . If P(1) = –9 and  $P\text{'}\text{'}\left(x\right)$  has a local minimum at x = 2. If global maximum value of $y=P\text{'}\left(x\right)$  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)hasnocons\tan ttermletP\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx$$\Rightarrow 3-d=27\Rightarrow d=-24$

$P\left(x\right)=a{x}^4+b{x}^3+c{x}^2-24x;P\text{'}\left(2\right)=0,p\left(1\right)=-9,p\text{'}\text{'}\text{'}\left(2\right)=0$$\Rightarrow a+b+c=15\mathrm{........}\left(1\right)$

$P\text{'}\left(2\right)=0$

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

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

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

Solving (1), (2) and (3) 

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