JEE Main 2021 (Online) 25th July Evening ShiftQuestion 1
Consider the function $f(x)=\begin{cases}\frac{P(x)}{\sin(x-2)}, & x\neq2\\ 7, & x=2\end{cases}$ 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 $\_\_\_\_\_\_$ .
Limits & Continuity
Answer$39$
Solution
Concept. The denominator $\sin(x-2)\to0$, so a finite limit needs $P(2)=0$; then the limit equals $P'(2)$. Formulas used. (1) $\lim_{t\to0}\frac{\sin t}{t}=1$(2) $P''$ constant $\Rightarrow$ $P$ has degree at most $2$Step 1 - Need $P(2)=0$. If $P(2)\ne0$, the fraction blows up at $2$. So $P(2)=0$ and $P(x)=(x-2)Q(x)$. Step 2 - Find the limit (1). $$\frac{P(x)}{\sin(x-2)}=\frac{x-2}{\sin(x-2)}\cdot Q(x)\to Q(2)$$ Continuity: $Q(2)=7$. Step 3 - Form of $P$ (2). $Q$ is linear: $Q(x)=7+c(x-2)$, so $$P(x)=7(x-2)+c(x-2)^2$$ Step 4 - Use $P(3)=9$. $7+c=9\Rightarrow c=2$. Step 5 - Compute. $P(5)=7\cdot3+2\cdot9=21+18=39$. Answer: 39