JEE Advanced 2012 Paper 2Question 1
For every integer $n$, let $a_n$ and $b_n$ be real numbers. Let function $f: \mathbb{R} \to \mathbb{R}$ be given by $$f(x) = \begin{cases} a_n + \sin\pi x, & \text{for } x \in [2n, 2n+1] \\ b_n + \cos\pi x, & \text{for } x \in (2n-1, 2n) \end{cases}$$ for all integers $n$. If $f$ is continuous, then which of the following hold(s) for all $n$?
Limits & Continuity
Solution
Concept. Check the joins at $x=2n$ (left piece is the cosine one) and at $x=2n+1$ (right piece is the next cosine one). Formulas used. (1) $\sin k\pi=0$, $\cos2n\pi=1$, $\cos(2n+1)\pi=-1$(2) continuity: one-sided limit $=$ valueStep 1 - At $x=2n$. Left: $(2n-1,2n)$ piece $\to b_n+\cos2n\pi=b_n+1$. Value: $a_n+\sin2n\pi=a_n$. So $a_n-b_n=1$. (B) true; then (A) $a_{n-1}-b_{n-1}=1\ne0$, false. Step 2 - At $x=2n+1$. Value: $a_n+\sin(2n+1)\pi=a_n$. Right: $(2n+1,2n+2)$ is the piece with index $n+1$: $\to b_{n+1}+\cos(2n+1)\pi=b_{n+1}-1$. So $a_n-b_{n+1}=-1$. (C) false. Step 3 - Shift $n\to n-1$. $a_{n-1}-b_n=-1$. (D) true. Answer: B,D