Limit Continuity And Differentiability Ques 38
- Match the conditions/expressions in Column I with statement in Column II
| Column I | Column II | |||
|---|---|---|---|---|
| A. | $\sin (\pi[x])$ | p. | differentiable everywhere | |
| B. | $\sin \{\pi(x-[x])\}$ | q. | no where differentiable | |
| r. | not differentiable at 1 and -1 |
(1992, 2M)
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Answer:
Correct Answer: 38.$(A) \rightarrow \mathrm{p}$ ; $ (B) $ $\rightarrow \mathrm{r}$
Solution: We know, $[x] \in I, \forall x \in R$.
Therefore, $\sin (\pi[x])=0, \forall x \in R$.
By theory, we know that $\sin (\pi[x])$ is differentiable everywhere, therefore (A) $\leftrightarrow$ (p).
Again, $f(x)=\sin \{\pi(x-[x])\}$
Now, $ x-[x]=\{x\}$
then $\pi(x-[x])=\pi\{x\}$
which is not differentiable at $x \in I$.
Therefore, $(B) \leftrightarrow(r)$ is the answer.
BETA
