The limit $\mathrm{p}={\displaystyle \underset{\mathrm{x}\to \pi }{lim}\left(\frac{{\mathrm{x}}^2+\alpha \mathrm{x}+2{\pi }^2}{\mathrm{x}-\pi +2\sin \mathrm{x}}\right)}$ has a finite value for a real α. The value of α and the corresponding limit p are

The correct option is: α = −3π, and p = π

Let us analyze the limit:

$$p=\underset{x\to \pi }{lim}\left(\frac{x^2+\alpha x+2{\pi }^2}{x-\pi +2\sin x}\right)$$

First, we examine the behavior of the denominator as $x\to \pi$:
$$\underset{x\to \pi }{lim}(x-\pi +2\sin x)=\pi -\pi +2\sin (\pi )=0$$

Since the limit $p$ is given to be finite, the numerator must also approach $0$ as $x\to \pi$. If the numerator approached a non-zero constant, the limit would be undefined (tending to infinity).

Therefore, we set the limit of the numerator to $0$:
$$\underset{x\to \pi }{lim}(x^2+\alpha x+2{\pi }^2)=0$$

Substituting $x=\pi$ into the numerator expression:
$${\pi }^2+\alpha \pi +2{\pi }^2=0$$

Simplifying the terms:
$$3{\pi }^2+\alpha \pi =0$$
$$\alpha \pi =-3{\pi }^2$$
$$\alpha =-3\pi$$

Now, we substitute $\alpha =-3\pi$ back into the original limit expression to evaluate $p$:
$$p=\underset{x\to \pi }{lim}\left(\frac{x^2-3\pi x+2{\pi }^2}{x-\pi +2\sin x}\right)$$

This limit represents an indeterminate form of type $\frac{0}{0}$. Therefore, we can apply L'Hôpital's Rule, which states that we can differentiate the numerator and denominator with respect to $x$ to find the limit:

Differentiating the numerator:
$$\frac{d}{dx}(x^2-3\pi x+2{\pi }^2)=2x-3\pi$$

Differentiating the denominator:
$$\frac{d}{dx}(x-\pi +2\sin x)=1+2\cos x$$

Now, apply L'Hôpital's Rule:
$$p=\underset{x\to \pi }{lim}\frac{2x-3\pi }{1+2\cos x}$$

Evaluating the limit by substituting $x=\pi$:

Since $\cos (\pi )=-1$:

$$p=\frac{-\pi }{1-2}$$
$$p=\frac{-\pi }{-1}=\pi$$

Thus, we have successfully determined that:
α = −3π, and p = π