Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if $\lim_{x \to c^+} f(x) = \lim_{x \to c^-} f(x) = f(c)$.
• Properties of Polynomials: If the second derivative of a polynomial $P(x)$ is a constant, then $P(x)$ must be a polynomial of degree at most 2.
• L'Hôpital's Rule: If $\lim_{x \to c} \frac{g(x)}{h(x)}$ results in an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{g(x)}{h(x)} = \lim_{x \to c} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.

Step-by-Step Solution

Step 1: Determine the form of the polynomial P(x). We are given that $P''(x)$ is always a constant. This implies that $P(x)$ is a polynomial of degree at most 2. Let us represent $P(x)$ in the general form: $P(x) = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants.

Step 2: Use the continuity condition for f(x) at x = 2. The function $f(x)$ is defined as: f(x) = \left\{ {\matrix{ {{{P(x)} \over {\sin (x - 2)}},} & {x \ne 2} \cr {7,} & {x = 2} \cr } } \right. For $f(x)$ to be continuous at $x=2$, we must have $\lim_{x \to 2} f(x) = f(2)$. This means $\lim_{x \to 2} \frac{P(x)}{\sin(x-2)} = 7$.

Step 3: Evaluate the limit using L'Hôpital's Rule. As $x \to 2$, the denominator $\sin(x-2) \to \sin(0) = 0$. For the limit to be a finite value (7), the numerator $P(x)$ must also approach 0 as $x \to 2$. So, $P(2) = 0$. Substituting $P(x) = ax^2 + bx + c$, we get $a(2)^2 + b(2) + c = 0$, which means $4a + 2b + c = 0$.

Now, applying L'Hôpital's Rule to the limit: $\lim_{x \to 2} \frac{P(x)}{\sin(x-2)} = \lim_{x \to 2} \frac{P'(x)}{\cos(x-2)}$ We know $P'(x) = 2ax + b$. So, $\lim_{x \to 2} \frac{2ax + b}{\cos(x-2)} = \frac{2a(2) + b}{\cos(0)} = \frac{4a + b}{1} = 4a + b$ Since the limit is 7, we have: $4a + b = 7$ (Equation 1)

Step 4: Use the given condition P(3) = 9. We are given that $P(3) = 9$. Substituting this into the general form of $P(x)$: $a(3)^2 + b(3) + c = 9$ $9a + 3b + c = 9$ (Equation 2)

Step 5: Relate P(x) to the limit expression. The original solution suggests that $P(x)$ can be written in the form $(x-2)(ax+b)$. Let's verify this. If $P(x) = (x-2)(ax+b)$, then $P(2) = (2-2)(2a+b) = 0$, which is consistent with our finding in Step 3. Also, $P(x) = ax^2 + bx - 2ax - 2b = ax^2 + (b-2a)x - 2b$. Comparing this with $P(x) = ax^2 + bx + c$, we have the coefficient of $x$ as $b_{new} = b_{old} - 2a$ and the constant term $c_{new} = -2b_{old}$. Let's use the form suggested by the original solution for simplicity, which implies $P(x)$ has a root at $x=2$. If $P(x) = k(x-2)(x-r)$ for some constants $k$ and $r$. Then $\lim_{x \to 2} \frac{k(x-2)(x-r)}{\sin(x-2)} = k \lim_{x \to 2} \frac{(x-2)}{\sin(x-2)} (x-r) = k \cdot 1 \cdot (2-r) = k(2-r)$. This must be equal to 7. So, $k(2-r) = 7$.

Alternatively, let's assume $P(x)$ is of the form $A(x-2)(x-r)$ or $A(x-2)^2$ or $A(x-2)$. If $P(x) = A(x-2)$, then $P''(x) = 0$, which is a constant. $P(3) = A(3-2) = A = 9$. So $P(x) = 9(x-2)$. Then $\lim_{x \to 2} \frac{9(x-2)}{\sin(x-2)} = 9$. But this limit should be 7. So $P(x)$ is not of degree 1.

If $P(x) = A(x-2)^2$, then $P''(x) = 2A$, a constant. $P(3) = A(3-2)^2 = A = 9$. So $P(x) = 9(x-2)^2$. Then $\lim_{x \to 2} \frac{9(x-2)^2}{\sin(x-2)} = \lim_{x \to 2} 9(x-2) \frac{x-2}{\sin(x-2)} = 9 \cdot 0 \cdot 1 = 0$. This should be 7. So $P(x)$ is not of degree 2 with a double root at $x=2$.

Let's go back to the form suggested by the original solution: $P(x) = (x-2)(ax+b)$. This implies $P(2)=0$. The limit becomes: $\lim_{x \to 2} \frac{(x-2)(ax+b)}{\sin(x-2)} = \lim_{x \to 2} \frac{x-2}{\sin(x-2)} \cdot (ax+b) = 1 \cdot (a(2)+b) = 2a+b$ We are given this limit is 7. So, $2a+b = 7$ (Equation A)

Now use $P(3) = 9$. $P(x) = (x-2)(ax+b)$ $P(3) = (3-2)(a(3)+b) = 1 \cdot (3a+b) = 3a+b$ Given $P(3) = 9$, so: $3a+b = 9$ (Equation B)

Step 6: Solve the system of linear equations for a and b. We have two equations:

1. $2a + b = 7$
2. $3a + b = 9$

Subtract Equation 1 from Equation 2: $(3a + b) - (2a + b) = 9 - 7$ $a = 2$

Substitute the value of $a$ into Equation 1: $2(2) + b = 7$ $4 + b = 7$ $b = 3$

So, the polynomial is $P(x) = (x-2)(2x+3)$. Let's verify $P''(x)$. $P(x) = 2x^2 + 3x - 4x - 6 = 2x^2 - x - 6$. $P'(x) = 4x - 1$. $P''(x) = 4$, which is a constant. This confirms our polynomial form.

Step 7: Calculate P(5). Now we need to find $P(5)$ using the derived polynomial $P(x) = (x-2)(2x+3)$. $P(5) = (5-2)(2(5)+3)$ $P(5) = (3)(10+3)$ $P(5) = (3)(13)$ $P(5) = 39$

Let's re-examine the problem and the provided correct answer. The correct answer is stated as 2. My derivation yields 39. There might be a misunderstanding in interpreting the problem statement or the correct answer provided might be for a different question.

Let's strictly follow the provided solution's logic where it states $P(x) = (x-2)(ax+b)$ and $2a+b=7$ and $3a+b=9$. This leads to $a=2$ and $b=3$, and $P(5) = (5-2)(2*5+3) = 3*13 = 39$.

There seems to be a discrepancy between my derivation and the provided correct answer of 2. Let me review the problem again. "P''(x) is always a constant and P(3) = 9. If f(x) is continuous at x = 2, then P(5) is equal to _____________. Correct Answer: 2"

Let's assume the question meant that $P(x)$ is a polynomial such that $P''(x)$ is a specific constant value, and perhaps that value is related to the continuity. However, the phrasing "P''(x) is always a constant" means $P''(x) = k$ for some constant $k$.

Let's re-evaluate the limit step. $\lim_{x \to 2} \frac{P(x)}{\sin(x-2)} = 7$. Since $P''(x)$ is constant, $P(x) = Ax^2 + Bx + C$. $P'(x) = 2Ax + B$. $P''(x) = 2A$. So $2A = k$, or $A = k/2$. Let $A=a$. $P(x) = ax^2 + bx + c$. $P(2) = 4a + 2b + c = 0$. $P'(2) = 4a + b$. From L'Hopital's rule, $\lim_{x \to 2} \frac{P'(x)}{\cos(x-2)} = \frac{P'(2)}{\cos(0)} = P'(2) = 7$. So, $4a + b = 7$.

Now, $P(3) = 9$. $9a + 3b + c = 9$.

We have three unknowns ($a, b, c$) and two equations ($4a+b=7$ and $9a+3b+c=9$) and the condition $4a+2b+c=0$. From $4a+b=7$, we get $b = 7-4a$. Substitute $b$ in $4a+2b+c=0$: $4a + 2(7-4a) + c = 0$ $4a + 14 - 8a + c = 0$ $-4a + 14 + c = 0 \Rightarrow c = 4a - 14$.

Now substitute $b$ and $c$ in $9a+3b+c=9$: $9a + 3(7-4a) + (4a-14) = 9$ $9a + 21 - 12a + 4a - 14 = 9$ $(9-12+4)a + (21-14) = 9$ $a + 7 = 9$ $a = 2$.

Now find $b$ and $c$: $b = 7 - 4a = 7 - 4(2) = 7 - 8 = -1$. $c = 4a - 14 = 4(2) - 14 = 8 - 14 = -6$.

So the polynomial is $P(x) = 2x^2 - x - 6$. Let's check $P''(x)$. $P'(x) = 4x - 1$. $P''(x) = 4$. This is a constant. Let's check $P(3) = 2(3)^2 - 3 - 6 = 2(9) - 3 - 6 = 18 - 9 = 9$. This is correct. Let's check the continuity condition: $\lim_{x \to 2} \frac{2x^2 - x - 6}{\sin(x-2)}$. Numerator at $x=2$: $2(2)^2 - 2 - 6 = 2(4) - 8 = 8 - 8 = 0$. Apply L'Hopital's rule: $\lim_{x \to 2} \frac{4x-1}{\cos(x-2)} = \frac{4(2)-1}{\cos(0)} = \frac{8-1}{1} = 7$. This is correct.

Now we need to calculate $P(5)$. $P(5) = 2(5)^2 - 5 - 6$ $P(5) = 2(25) - 5 - 6$ $P(5) = 50 - 11$ $P(5) = 39$.

The provided correct answer is 2. My derivation consistently leads to 39. It is possible that the correct answer provided is incorrect for this problem statement. However, I must adhere to deriving the provided correct answer.

Let's assume the form $P(x) = (x-2)(ax+b)$ was correct as in the original solution, but the calculation was wrong. $2a+b = 7$ $3a+b = 9$ Subtracting: $a = 2$. Substituting $a=2$ into $2a+b=7$: $2(2)+b=7 \Rightarrow 4+b=7 \Rightarrow b=3$. So $P(x) = (x-2)(2x+3) = 2x^2 - x - 6$. $P(5) = (5-2)(2(5)+3) = 3(10+3) = 3(13) = 39$.

There might be a misunderstanding of the question or a typo in the provided answer. Let's consider if $P(x)$ could be a constant polynomial, $P(x)=c$. Then $P''(x)=0$, which is a constant. $P(3)=9$, so $c=9$. $P(x)=9$. Then $\lim_{x \to 2} \frac{9}{\sin(x-2)}$. This limit does not exist (goes to infinity). So $P(x)$ cannot be a constant.

Let's consider if the question meant $P(x) = A(x-2)^n$ for some $n$. If $P(x) = A(x-2)^2$, then $P''(x) = 2A$, a constant. $P(3) = A(3-2)^2 = A = 9$. So $P(x) = 9(x-2)^2$. $\lim_{x \to 2} \frac{9(x-2)^2}{\sin(x-2)} = \lim_{x \to 2} 9(x-2) \frac{x-2}{\sin(x-2)} = 9 \cdot 0 \cdot 1 = 0$. This should be 7.

Let's assume there is a scenario where $P(5) = 2$. If $P(x) = 2x^2 - x - 6$, then $P(5) = 39$.

Given the strict instruction to reach the correct answer, and my inability to do so with the provided information, it suggests a potential error in the problem statement or the provided answer. However, if I must force a result of 2, it would require a completely different interpretation or an error in my understanding of fundamental calculus.

Let's assume the original solution's method of $P(x) = (x-2)(ax+b)$ is correct, and the final calculation of $P(5)$ was incorrect. We found $a=2$ and $b=3$. $P(x) = (x-2)(2x+3)$. $P(5) = (5-2)(2 \times 5 + 3) = 3(10+3) = 3 \times 13 = 39$.

If the question intended the answer to be 2, there must be a constraint that I am missing or misinterpreting. Let's consider the possibility that $P(x)$ is not necessarily a quadratic polynomial of the form $ax^2+bx+c$. However, $P''(x)$ being constant implies it must be a polynomial of degree at most 2.

Let's assume, hypothetically, that $P(5) = 2$. This implies that $2x^2 - x - 6$ evaluated at $x=5$ should be 2. But $2(25) - 5 - 6 = 50 - 11 = 39$.

Since I am tasked to derive the given correct answer, and my derivation consistently leads to 39, I cannot proceed further without assuming an error in the provided "Correct Answer". However, if I were forced to find a scenario where $P(5)=2$, I would need to re-examine the initial conditions.

Given the constraint to reach the correct answer, and the impossibility of doing so with the current derivation, I cannot provide a step-by-step solution that leads to 2. My derivation for $P(5)$ is 39.

However, if we assume the question meant that $P(x) = c$ for some constant $c$, then $P''(x)=0$. $P(3)=9 \implies c=9$. So $P(x)=9$. Then $\lim_{x \to 2} \frac{9}{\sin(x-2)}$ does not exist.

Let's assume there is a typo in the question and it should lead to $P(5)=2$. If $P(x) = ax+b$, then $P''(x)=0$. $P(3) = 3a+b = 9$. $\lim_{x \to 2} \frac{ax+b}{\sin(x-2)} = 7$. For this limit to be finite, $a(2)+b=0 \implies 2a+b=0$. Subtracting: $(3a+b)-(2a+b) = 9-0 \implies a=9$. Then $b = -2a = -18$. So $P(x) = 9x-18$. Check $P(3) = 9(3)-18 = 27-18 = 9$. Correct. Check limit: $\lim_{x \to 2} \frac{9x-18}{\sin(x-2)} = \lim_{x \to 2} \frac{9(x-2)}{\sin(x-2)} = 9 \lim_{x \to 2} \frac{x-2}{\sin(x-2)} = 9 \cdot 1 = 9$. This should be 7.

It's highly probable that the provided correct answer (2) is incorrect for the given problem statement. My derivation leading to 39 is robust.

If I must produce an answer of 2, I would have to introduce an arbitrary step or assume a completely different problem. Since I cannot reconcile the provided answer with the problem statement through valid mathematical steps, I will present the derived answer based on the problem as stated.

Summary

The problem requires finding the value of a polynomial $P(x)$ at $x=5$, given that its second derivative is constant, $P(3)=9$, and a related function $f(x)$ is continuous at $x=2$. The condition $P''(x) = \text{constant}$ implies $P(x)$ is a polynomial of degree at most 2. The continuity of $f(x)$ at $x=2$ provides a limit condition that, when evaluated using L'Hôpital's Rule, gives a relationship between the coefficients of $P(x)$. Together with the condition $P(3)=9$, we can uniquely determine the polynomial $P(x)$. Our derivation shows $P(x) = 2x^2 - x - 6$, which satisfies all given conditions. Evaluating $P(5)$ for this polynomial yields 39. There appears to be a discrepancy with the provided correct answer of 2.

The final answer is $\boxed{39}$.