Question:
Consider the function \(f(x)=\begin{cases}\dfrac{P(x)}{\sin(x-2)}, & x\ne 2\\ 7, & x=2\end{cases}\) 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 _____________.
Consider the function \(f(x)=\begin{cases}\dfrac{P(x)}{\sin(x-2)}, & x\ne 2\\ 7, & x=2\end{cases}\) 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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\(P\) is quadratic; continuity forces \(P(2)=0\) and \(P'(2)=7\).
Updated On: Nov 4, 2026
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Correct Answer: 39
Solution and Explanation
\(P''\) constant \(\Rightarrow P\) is a polynomial of degree at most 2.
Continuity at 2: \(\lim_{x\to2}\frac{P(x)}{\sin(x-2)}=7\). The limit is finite only if \(P(2)=0\); then by L'Hospital it equals \(P'(2)=7\).
So \(P(x)=7(x-2)+a(x-2)^2\). \(P(3)=7+a=9\Rightarrow a=2\).
\(P(5)=7(3)+2(9)=21+18=39\).
Answer: 39
Continuity at 2: \(\lim_{x\to2}\frac{P(x)}{\sin(x-2)}=7\). The limit is finite only if \(P(2)=0\); then by L'Hospital it equals \(P'(2)=7\).
So \(P(x)=7(x-2)+a(x-2)^2\). \(P(3)=7+a=9\Rightarrow a=2\).
\(P(5)=7(3)+2(9)=21+18=39\).
Answer: 39
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