Limit Continuity and Differentiability 4 Question 9
9. For every integer $n$, let $a _n$ and $b _n$ be real numbers. Let function $f: R \rightarrow R$ be given by
$f(x)=\begin{array}{ll}a _n+\sin \pi x, & \text { for } x \in[2 n, 2 n+1] \ b _n+\cos \pi x, & \text { for } x \in(2 n-1,2 n)\end{array}$
for all integers $n$.
If $f$ is continuous, then which of the following hold(s) for all $n$ ?
(2012)
(a) $a _{n-1}-b _{n-1}=0$
(b) $a _n-b _n=1$
(c) $a _n-b _{n+1}=1$
(d) $a _{n-1}-b _n=-1$
Fill in the Blank
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Answer:
Correct Answer: 9. $f(x)=\sqrt{4-x^{2}}$
Solution:
- We have, for $-1<x<1$
$\Rightarrow \quad 0 \leq x \sin \pi x \leq 1 / 2$
$\therefore \quad[x \sin \pi x]=0$
Also, $x \sin \pi x$ becomes negative and numerically less than 1 when $x$ is slightly greater than 1 and so by definition of $[x]$.
$$ f(x)=[x \sin \pi x]=-1 \text {, when } 1<x<1+h $$
Thus, $f(x)$ is constant and equal to 0 in the closed interval $[-1,1]$ and so $f(x)$ is continuous and differentiable in the open interval $(-1,1)$.
At $x=1, f(x)$ is discontinuous, since $\lim _{h \rightarrow 0}(1-h)=0$
and $\quad \lim _{h \rightarrow 0}(1+h)=-1$
$\therefore f(x)$ is not differentiable at $x=1$.
Hence, (a), (b) and (d) are correct answers.
