IIT-JEE 2012 · Paper 2 · Mathematics Q55

← IIT-JEE 2012 · Paper 2

MathematicsMultiple correct+4Question 55 of 60

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

a_{n-1}-b_{n-1}=0

a_n-b_n=1

a_n-b_{n+1}=1

a_{n-1}-b_n=-1

Answer and worked solution

Official answer: B, D

At $x=2n$ . From the left (on $(2n-1,2n)$ ): $b_n+\cos2n\pi=b_n+1$ . At the point: $a_n+\sin2n\pi=a_n$ . Continuity gives $a_n-b_n=1$ .

At $x=2n+1$ . At the point: $a_n+\sin(2n+1)\pi=a_n$ . From the right (on $(2n+1,2n+2)$ , which belongs to index $n+1$ ): $b_{n+1}+\cos(2n+1)\pi=b_{n+1}-1$ . So $a_n-b_{n+1}=-1$ , i.e. $a_{n-1}-b_n=-1$ .

(A) $a_{n-1}-b_{n-1}=1$ , not 0. False.
(B) True.
(C) It is −1, not 1. False.
(D) True.

(B), (D)

Topics:Continuity

← Previous question Next question →

IIT-JEE 2012 · Paper 2, Mathematics question 55