Rational Solver

+1 vote

Best answer

The set in Option 1 is countable. It follows from the fact that the logarithm is monotonically increasing and one-to-one function. Thus the preimage of a countable set ($\mathbb Q$) can at most countable.

Certainly, **option 2** is correct. Note that $\cos^2x+\sin^2 x=1$ is true for all $x\in\mathbb R$. Thus the set is uncountable, as $\mathbb R$ is uncountable.

The set in option 3 is countable as an image of a countable set under a function that can at most countable.

The set in option 4 is also countable it is a trickier one to answer. First note that $\cos :[k\pi,(k+1)\pi]\to [-1,1]$ is a bijection for each $k\in\mathbb Z$. Thus we can have at most countable number of $x\in [k\pi,(k+1)\pi]$ that can map to a countable set $\mathbb Q\cap [-1,1]$. Thus the following set is countable $$S_k=\left\{x\in [k\pi ,(k+1)\pi ]\mid \cos (x)=\frac pq \quad \text{ for some } p,q\in\mathbb N \right\}$$

It follows that the given set can be written as a countable union of countable set $\bigcup_{k\in\mathbb Z}S_k$.

- All categories
- JAM 10
- WBJEEE 7
- Linear Algebra 8
- Calculus 3
- Discrete Mathematics 10
- Differential Equation 10
- Functional Analysis 1
- Abstract Algebra 3
- Topology 1
- Complex Analysis 5
- Probability 5
- Real Analysis 10
- Anonymous 0
- Chemistry 0
- Physics 0
- Spam 0

76 questions

33 answers

2 comments

1,801 users