For every integer n, let a_n and b_n be real numbers. Let functio… - Vidyadip

MCQ

For every integer $n$, let $a_n$ and $b_n$ be real numbers. Let function $f: I R \rightarrow$ $IR$ 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 $n$ ?

$(A)$ $a_{n-1}-b_{n-1}=0$ $(B)$ $a_n-b_n=1$ $(C)$ $a_n-b_{n+1}=1$ $(D)$ $a_{n-1}-b_n=-1$

✓
$(B,D)$
B
$(B,C)$
C
$(A,D)$
D
$(C,D)$

✓

Answer

Correct option: A.

$(B,D)$

a
$\left.\begin{array}{c}f(2 n)=a_n \\ f\left(2 n^{+}\right)=a_n \\ f\left(2 n^{-}\right)=b_n+1\end{array}\right\} \quad \begin{array}{c}a_n=b_n+1 \\ a_n-b_n=1 \\ \text { So B is correct }\end{array}$

$\left.\begin{array}{c}f(2 n+1)=a_n \\ f\left((2 n+1)^{-}\right)=a_n \\ f\left((2 n+1)^{+}\right)=b_{n+1}-1\end{array}\right\} \quad \begin{array}{c}a_n=b_{n+1}-1 \\ a_n-b_{n+1}=-1 \\ a_{n-1}-b_n=-1\end{array}$

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