For every integer $n$, let $a_{n}$ and $b_{n}$ be real numbers. Let function $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x)=\left\{\begin{array}{ll}a_{n}+\sin \pi x, & \text { for } x \in[2 n, 2 n+1] \\ b_{n}+\cos \pi x, & \text { for } x \in(2 n-1,2 n)\end{array}\right.$ for all integers $n$. If $f$ is continuous, then which of the following hold(s) for all $n$ ?

(B) $a_{n}-b_{n}=1$; (D) $a_{n-1}-b_{n}=-1$

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Continuity and Differentiability

For every integer

n

, let

a_{n}

and

b_{n}

be real numbers. Let function

f : R \to R

be given by

f (x) = {a_{n} + sin π x, b_{n} + cos π x, for x \in [2 n, 2 n + 1] for x \in (2 n - 1, 2 n)

for all integers

n

.
If

f

is continuous, then which of the following hold(s) for all

n

Answer: (B), (D)

For continuity at

x = 2 n

:

Left limit:

lim_{x \to 2 n^{-}} (b_{n} + cos π x) = b_{n} + 1

Right limit:

lim_{x \to 2 n^{+}} (a_{n} + sin π x) = a_{n}

For continuity:

b_{n} + 1 = a_{n}

, so

a_{n} - b_{n} = 1

. ✓ (Option B)

For continuity at

x = 2 n + 1

:

Left limit:

lim_{x \to (2 n + 1)^{-}} (a_{n} + sin π x) = a_{n}

Right limit:

lim_{x \to (2 n + 1)^{+}} (b_{n + 1} + cos π x) = b_{n + 1} - 1

For continuity:

a_{n} = b_{n + 1} - 1

, so

a_{n} - b_{n + 1} = - 1

.

Replacing

n

n - 1

a_{n - 1} - b_{n} = - 1

. ✓ (Option D)

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · One or more correct options