Standard Limits for Continuity

To find the value of the given expression, we systematically apply the conditions of continuity at $$x = 1$$ and differentiability at $$x = 2$$.

Step 1: Analyze continuity at x = 1

\For $$f(x)$$ to be continuous at $$x = 1$$, the left-hand limit (LHL), right-hand limit (RHL), and functional value must be equal:
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$

Left-Hand Limit (LHL):

As $$x \to 1^-$$, we use the first branch. Notice that $$x^2 - 3x + 2 = (x-1)(x-2)$$. For $$x < 1$$, both factors are negative, making the product positive, so the absolute value drops directly:

$$\lim_{x \to 1^-} \left( \alpha + \beta (x^2 - 3x + 2) \right) = \alpha + \beta(1 - 3 + 2) = \alpha$$

Right-Hand Limit (RHL) and Functional Value f(1):

As $$x \to 1^+$$, we use the second branch:

$$\lim_{x \to 1^+} \frac{\sin(\pi x)}{x - 1}$$

Let $$x = 1 + h$$, where $$h \to 0^+$$. Substituting this variable gives:

$$\lim_{h \to 0^+} \frac{\sin(\pi(1 + h))}{h} = \lim_{h \to 0^+} \frac{\sin(\pi + \pi h)}{h} = \lim_{h \to 0^+} \frac{-\sin(\pi h)}{h}$$

Multiply the denominator by $$\pi$$ to evaluate the standard limit format:

$$\lim_{h \to 0^+} -\pi \left( \frac{\sin(\pi h)}{\pi h} \right) = -\pi(1) = -\pi$$

Equating the limits yields our first key parameter constraint value:

$$\alpha = -\pi$$

Step 2: Analyze continuity at x = 2

Differentiability at a point requires the function to be continuous at that point first. Therefore, $$f(x)$$ must be continuous at $$x = 2$$:

$$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$$

Left-Hand Limit (LHL):

Using the second branch:

$$\lim_{x \to 2^-} \frac{\sin(\pi x)}{x - 1} = \frac{\sin(2\pi)}{2 - 1} = \frac{0}{1} = 0$$

Right-Hand Limit (RHL) and Functional Value f(2):

Using the third branch:

$$\lim_{x \to 2^+} \left( \gamma x^2 + \delta x + 1 \right) = \gamma(2)^2 + \delta(2) + 1 = 4\gamma + 2\delta + 1$$

Equating these two parameters yields our second system relationship:

$$4\gamma + 2\delta + 1 = 0 \implies 4\gamma + 2\delta = -1$$

Step 3: Solve for the parameters and target expression

From Step 2, we have the grouped variable value:

$$4\gamma + 2\delta = -1$$

From Step 1, we have: $$\alpha = -\pi$$

We evaluate the value of the target expression:

$$\text{Value} = 2\alpha + 4\gamma + 2\delta$$

$$\text{Value} = 2(-\pi) + (-1) = -2\pi - 1$$

Since

$$f(x)$$ is right continuous at $$x=0,$$

$$a=\lim_{x\to0^+}\frac{\ln(1+x)-\sin x+\frac{x^2}{2}}{x^3}$$

Using expansions,

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4)$$ and $$\sin x=x-\frac{x^3}{6}+O(x^5)$$

Therefore,

$$\ln(1+x)-\sin x+\frac{x^2}{2}$$

$$=\left(x-\frac{x^2}{2}+\frac{x^3}{3}\right)-\left(x-\frac{x^3}{6}\right)+\frac{x^2}{2}+O(x^4)$$

$$=\frac{x^3}{3}+\frac{x^3}{6}+O(x^4)$$

$$=\frac{x^3}{2}+O(x^4)$$

Hence,

$$a=\lim_{x\to0^+}\frac{\frac{x^3}{2}+O(x^4)}{x^3}$$

$$=\frac12$$

Now evaluate the left-hand limit.

For $$-1<x<0,$$

$$\lfloor x\rfloor=-1$$

Therefore,

$$\lim_{x\to0^-}\frac{1-\cos(x\lfloor x\rfloor)}{x^2}$$

$$=\lim_{x\to0^-}\frac{1-\cos(-x)}{x^2}$$

Since cosine is even,

$$=\lim_{x\to0^-}\frac{1-\cos x}{x^2}$$

Using

$$\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12,$$

we get

$$\text{LHL}=\frac12$$

Thus,

$$\text{LHL}=\text{RHL}=f(0)=\frac12$$

Hence,

$$f$$ is continuous at $$x=0.$$