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(C) Since $P''(x)$ is a constant,$P(x)$ must be a polynomial of degree $2$. Let $P(x) = ax^2 + bx + c$.Given that $f(x)$ is continuous at $x = 2$,we h

Vedclass · Accepted Answer

$39$

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(C) Since $P''(x)$ is a constant,$P(x)$ must be a polynomial of degree $2$. Let $P(x) = ax^2 + bx + c$.Given that $f(x)$ is continuous at $x = 2$,we have $\lim_{x 	o 2} f(x) = f(2) = 7$.This implies $\lim_{x 	o 2} \frac{P(x)}{\sin(x-2)} = 7$.For the limit to exist and be finite,$P(2)$ must be $0$ because $\sin(x-2) 	o 0$ as $x 	o 2$. Thus,$(x-2)$ is a factor of $P(x)$.Let $P(x) = (x-2)(ax + k)$.Then $\lim_{x 	o 2} \frac{(x-2)(ax+k)}{\sin(x-2)} = \lim_{x 	o 2} \frac{x-2}{\sin(x-2)} \cdot (ax+k) = 1 \cdot (2a+k) = 7$.So,$2a + k = 7$.We are given $P(3) = 9$. Substituting $x=3$ into $P(x) = (x-2)(ax+k)$ gives $(3-2)(3a+k) = 9$,so $3a + k = 9$.Subtracting the two equations: $(3a+k) - (2a+k) = 9 - 7$,which gives $a = 2$.Substituting $a=2$ into $2a+k=7$ gives $4+k=7$,so $k=3$.Thus,$P(x) = (x-2)(2x+3)$.Finally,$P(5) = (5-2)(2(5)+3) = 3(13) = 39$.

Consider the function $f(x) = \begin{cases} \frac{P(x)}{\sin(x-2)}, & x \neq 2 \\ 7, & x = 2 \end{cases}$ where $P(x)$ is a polynomial such that $P''(x)$ is always a constant and $P(3) = 9$. If $f(x)$ is continuous at $x = 2$,then $P(5)$ is equal to:

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