Simple and short true-false tasks regarding Precalculus
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Here are few of the questions from the previous years' exams. I've chosen the ones I'm not sure about. It's a simple TRUE/FALSE task. Would anyone be able to verify my solution? Some of my answers are good, some are just random guess according to my intuition. I don't really need a detailed explanation... Thanks!
- Domain of $f'$ is contained within domain of $f$. - TRUE
- Boundary point of set A is also a cluster point of that set. - TRUE
- Every increasing sequence and bounded above is convergent. - TRUE
- Every increasing sequence and bounded below is convergent. - FALSE
- Every increasing sequence is always bounded below. - TRUE
- Every sequence is discontinuous function. - FALSE
- Every sequence is continuous function. - TRUE
- Every function integrable on $<a, b>$ is continuous on $<a, b>$. - FALSE
- Function $f(x) = ln{|x|}$ is discontinuous at $0$. - TRUE
- The continuity is necessary for differentiability. - TRUE
- Function $f(x) = frac{x}{|x|}$ is monotonic. - FALSE
real-analysis calculus algebra-precalculus derivatives education
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show 4 more comments
$begingroup$
Here are few of the questions from the previous years' exams. I've chosen the ones I'm not sure about. It's a simple TRUE/FALSE task. Would anyone be able to verify my solution? Some of my answers are good, some are just random guess according to my intuition. I don't really need a detailed explanation... Thanks!
- Domain of $f'$ is contained within domain of $f$. - TRUE
- Boundary point of set A is also a cluster point of that set. - TRUE
- Every increasing sequence and bounded above is convergent. - TRUE
- Every increasing sequence and bounded below is convergent. - FALSE
- Every increasing sequence is always bounded below. - TRUE
- Every sequence is discontinuous function. - FALSE
- Every sequence is continuous function. - TRUE
- Every function integrable on $<a, b>$ is continuous on $<a, b>$. - FALSE
- Function $f(x) = ln{|x|}$ is discontinuous at $0$. - TRUE
- The continuity is necessary for differentiability. - TRUE
- Function $f(x) = frac{x}{|x|}$ is monotonic. - FALSE
real-analysis calculus algebra-precalculus derivatives education
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1
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I would say you missed 2, 9, 11
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– MPW
Jan 29 at 19:15
1
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I do not know what is "cluster point" (language problem). For the rest: 1,3,4,5,6,7,8,10 are fine for me. 9 - for me is the function not defined at 0, so it cannot have a property there. 11 - f is monotonic!
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– user376343
Jan 29 at 19:20
1
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For 11: Constant functions are monotonic. Monotonicity is usually distinguished from strict monotonicity. For 9: it is neither continuous nor discontinuous at a point that is not in its domain; it is simply undefined there.
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– MPW
Jan 29 at 19:21
1
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Monotonic signifies "doesn´t change from increasing to decreasing and vice versa".
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– user376343
Jan 29 at 19:21
1
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@user376343 : Not quite, since constant functions can be regarded as increasing or decreasing anywhere you like (you would need to be more specific). A function $f:Xto Y$ is monotone if $f^{-1}(y)$ is connected (or empty) for each $yin Y$. In the case where $X$ and $Y$ are (subsets of) $mathbb R$, this means $f^{-1}(y)$ is a (possibly degenerate) interval for each $y$ in the image. We confirm this by noting that the image is ${-1,1}$ and the preimages are $f^{-1}(-1)=(-infty,0)$ and $f^{-1}(1)=(0,infty)$ which are both connected.
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– MPW
Jan 29 at 19:26
|
show 4 more comments
$begingroup$
Here are few of the questions from the previous years' exams. I've chosen the ones I'm not sure about. It's a simple TRUE/FALSE task. Would anyone be able to verify my solution? Some of my answers are good, some are just random guess according to my intuition. I don't really need a detailed explanation... Thanks!
- Domain of $f'$ is contained within domain of $f$. - TRUE
- Boundary point of set A is also a cluster point of that set. - TRUE
- Every increasing sequence and bounded above is convergent. - TRUE
- Every increasing sequence and bounded below is convergent. - FALSE
- Every increasing sequence is always bounded below. - TRUE
- Every sequence is discontinuous function. - FALSE
- Every sequence is continuous function. - TRUE
- Every function integrable on $<a, b>$ is continuous on $<a, b>$. - FALSE
- Function $f(x) = ln{|x|}$ is discontinuous at $0$. - TRUE
- The continuity is necessary for differentiability. - TRUE
- Function $f(x) = frac{x}{|x|}$ is monotonic. - FALSE
real-analysis calculus algebra-precalculus derivatives education
$endgroup$
Here are few of the questions from the previous years' exams. I've chosen the ones I'm not sure about. It's a simple TRUE/FALSE task. Would anyone be able to verify my solution? Some of my answers are good, some are just random guess according to my intuition. I don't really need a detailed explanation... Thanks!
- Domain of $f'$ is contained within domain of $f$. - TRUE
- Boundary point of set A is also a cluster point of that set. - TRUE
- Every increasing sequence and bounded above is convergent. - TRUE
- Every increasing sequence and bounded below is convergent. - FALSE
- Every increasing sequence is always bounded below. - TRUE
- Every sequence is discontinuous function. - FALSE
- Every sequence is continuous function. - TRUE
- Every function integrable on $<a, b>$ is continuous on $<a, b>$. - FALSE
- Function $f(x) = ln{|x|}$ is discontinuous at $0$. - TRUE
- The continuity is necessary for differentiability. - TRUE
- Function $f(x) = frac{x}{|x|}$ is monotonic. - FALSE
real-analysis calculus algebra-precalculus derivatives education
real-analysis calculus algebra-precalculus derivatives education
asked Jan 29 at 19:11
wenoweno
37011
37011
1
$begingroup$
I would say you missed 2, 9, 11
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– MPW
Jan 29 at 19:15
1
$begingroup$
I do not know what is "cluster point" (language problem). For the rest: 1,3,4,5,6,7,8,10 are fine for me. 9 - for me is the function not defined at 0, so it cannot have a property there. 11 - f is monotonic!
$endgroup$
– user376343
Jan 29 at 19:20
1
$begingroup$
For 11: Constant functions are monotonic. Monotonicity is usually distinguished from strict monotonicity. For 9: it is neither continuous nor discontinuous at a point that is not in its domain; it is simply undefined there.
$endgroup$
– MPW
Jan 29 at 19:21
1
$begingroup$
Monotonic signifies "doesn´t change from increasing to decreasing and vice versa".
$endgroup$
– user376343
Jan 29 at 19:21
1
$begingroup$
@user376343 : Not quite, since constant functions can be regarded as increasing or decreasing anywhere you like (you would need to be more specific). A function $f:Xto Y$ is monotone if $f^{-1}(y)$ is connected (or empty) for each $yin Y$. In the case where $X$ and $Y$ are (subsets of) $mathbb R$, this means $f^{-1}(y)$ is a (possibly degenerate) interval for each $y$ in the image. We confirm this by noting that the image is ${-1,1}$ and the preimages are $f^{-1}(-1)=(-infty,0)$ and $f^{-1}(1)=(0,infty)$ which are both connected.
$endgroup$
– MPW
Jan 29 at 19:26
|
show 4 more comments
1
$begingroup$
I would say you missed 2, 9, 11
$endgroup$
– MPW
Jan 29 at 19:15
1
$begingroup$
I do not know what is "cluster point" (language problem). For the rest: 1,3,4,5,6,7,8,10 are fine for me. 9 - for me is the function not defined at 0, so it cannot have a property there. 11 - f is monotonic!
$endgroup$
– user376343
Jan 29 at 19:20
1
$begingroup$
For 11: Constant functions are monotonic. Monotonicity is usually distinguished from strict monotonicity. For 9: it is neither continuous nor discontinuous at a point that is not in its domain; it is simply undefined there.
$endgroup$
– MPW
Jan 29 at 19:21
1
$begingroup$
Monotonic signifies "doesn´t change from increasing to decreasing and vice versa".
$endgroup$
– user376343
Jan 29 at 19:21
1
$begingroup$
@user376343 : Not quite, since constant functions can be regarded as increasing or decreasing anywhere you like (you would need to be more specific). A function $f:Xto Y$ is monotone if $f^{-1}(y)$ is connected (or empty) for each $yin Y$. In the case where $X$ and $Y$ are (subsets of) $mathbb R$, this means $f^{-1}(y)$ is a (possibly degenerate) interval for each $y$ in the image. We confirm this by noting that the image is ${-1,1}$ and the preimages are $f^{-1}(-1)=(-infty,0)$ and $f^{-1}(1)=(0,infty)$ which are both connected.
$endgroup$
– MPW
Jan 29 at 19:26
1
1
$begingroup$
I would say you missed 2, 9, 11
$endgroup$
– MPW
Jan 29 at 19:15
$begingroup$
I would say you missed 2, 9, 11
$endgroup$
– MPW
Jan 29 at 19:15
1
1
$begingroup$
I do not know what is "cluster point" (language problem). For the rest: 1,3,4,5,6,7,8,10 are fine for me. 9 - for me is the function not defined at 0, so it cannot have a property there. 11 - f is monotonic!
$endgroup$
– user376343
Jan 29 at 19:20
$begingroup$
I do not know what is "cluster point" (language problem). For the rest: 1,3,4,5,6,7,8,10 are fine for me. 9 - for me is the function not defined at 0, so it cannot have a property there. 11 - f is monotonic!
$endgroup$
– user376343
Jan 29 at 19:20
1
1
$begingroup$
For 11: Constant functions are monotonic. Monotonicity is usually distinguished from strict monotonicity. For 9: it is neither continuous nor discontinuous at a point that is not in its domain; it is simply undefined there.
$endgroup$
– MPW
Jan 29 at 19:21
$begingroup$
For 11: Constant functions are monotonic. Monotonicity is usually distinguished from strict monotonicity. For 9: it is neither continuous nor discontinuous at a point that is not in its domain; it is simply undefined there.
$endgroup$
– MPW
Jan 29 at 19:21
1
1
$begingroup$
Monotonic signifies "doesn´t change from increasing to decreasing and vice versa".
$endgroup$
– user376343
Jan 29 at 19:21
$begingroup$
Monotonic signifies "doesn´t change from increasing to decreasing and vice versa".
$endgroup$
– user376343
Jan 29 at 19:21
1
1
$begingroup$
@user376343 : Not quite, since constant functions can be regarded as increasing or decreasing anywhere you like (you would need to be more specific). A function $f:Xto Y$ is monotone if $f^{-1}(y)$ is connected (or empty) for each $yin Y$. In the case where $X$ and $Y$ are (subsets of) $mathbb R$, this means $f^{-1}(y)$ is a (possibly degenerate) interval for each $y$ in the image. We confirm this by noting that the image is ${-1,1}$ and the preimages are $f^{-1}(-1)=(-infty,0)$ and $f^{-1}(1)=(0,infty)$ which are both connected.
$endgroup$
– MPW
Jan 29 at 19:26
$begingroup$
@user376343 : Not quite, since constant functions can be regarded as increasing or decreasing anywhere you like (you would need to be more specific). A function $f:Xto Y$ is monotone if $f^{-1}(y)$ is connected (or empty) for each $yin Y$. In the case where $X$ and $Y$ are (subsets of) $mathbb R$, this means $f^{-1}(y)$ is a (possibly degenerate) interval for each $y$ in the image. We confirm this by noting that the image is ${-1,1}$ and the preimages are $f^{-1}(-1)=(-infty,0)$ and $f^{-1}(1)=(0,infty)$ which are both connected.
$endgroup$
– MPW
Jan 29 at 19:26
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
- Right
- Wrong (What is the boundary of ${0}$ in $mathbb R$? What is its set of cluster points?)
- Right
- Right
- Right
- Right
- Right
- Right
- Wrong (it isn't defined there)
- Right
- Wrong (think about what this function is)
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1
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It's up to you.
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– José Carlos Santos
Jan 29 at 19:23
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
- Right
- Wrong (What is the boundary of ${0}$ in $mathbb R$? What is its set of cluster points?)
- Right
- Right
- Right
- Right
- Right
- Right
- Wrong (it isn't defined there)
- Right
- Wrong (think about what this function is)
$endgroup$
1
$begingroup$
It's up to you.
$endgroup$
– José Carlos Santos
Jan 29 at 19:23
add a comment |
$begingroup$
- Right
- Wrong (What is the boundary of ${0}$ in $mathbb R$? What is its set of cluster points?)
- Right
- Right
- Right
- Right
- Right
- Right
- Wrong (it isn't defined there)
- Right
- Wrong (think about what this function is)
$endgroup$
1
$begingroup$
It's up to you.
$endgroup$
– José Carlos Santos
Jan 29 at 19:23
add a comment |
$begingroup$
- Right
- Wrong (What is the boundary of ${0}$ in $mathbb R$? What is its set of cluster points?)
- Right
- Right
- Right
- Right
- Right
- Right
- Wrong (it isn't defined there)
- Right
- Wrong (think about what this function is)
$endgroup$
- Right
- Wrong (What is the boundary of ${0}$ in $mathbb R$? What is its set of cluster points?)
- Right
- Right
- Right
- Right
- Right
- Right
- Wrong (it isn't defined there)
- Right
- Wrong (think about what this function is)
answered Jan 29 at 19:18
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
1
$begingroup$
It's up to you.
$endgroup$
– José Carlos Santos
Jan 29 at 19:23
add a comment |
1
$begingroup$
It's up to you.
$endgroup$
– José Carlos Santos
Jan 29 at 19:23
1
1
$begingroup$
It's up to you.
$endgroup$
– José Carlos Santos
Jan 29 at 19:23
$begingroup$
It's up to you.
$endgroup$
– José Carlos Santos
Jan 29 at 19:23
add a comment |
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1
$begingroup$
I would say you missed 2, 9, 11
$endgroup$
– MPW
Jan 29 at 19:15
1
$begingroup$
I do not know what is "cluster point" (language problem). For the rest: 1,3,4,5,6,7,8,10 are fine for me. 9 - for me is the function not defined at 0, so it cannot have a property there. 11 - f is monotonic!
$endgroup$
– user376343
Jan 29 at 19:20
1
$begingroup$
For 11: Constant functions are monotonic. Monotonicity is usually distinguished from strict monotonicity. For 9: it is neither continuous nor discontinuous at a point that is not in its domain; it is simply undefined there.
$endgroup$
– MPW
Jan 29 at 19:21
1
$begingroup$
Monotonic signifies "doesn´t change from increasing to decreasing and vice versa".
$endgroup$
– user376343
Jan 29 at 19:21
1
$begingroup$
@user376343 : Not quite, since constant functions can be regarded as increasing or decreasing anywhere you like (you would need to be more specific). A function $f:Xto Y$ is monotone if $f^{-1}(y)$ is connected (or empty) for each $yin Y$. In the case where $X$ and $Y$ are (subsets of) $mathbb R$, this means $f^{-1}(y)$ is a (possibly degenerate) interval for each $y$ in the image. We confirm this by noting that the image is ${-1,1}$ and the preimages are $f^{-1}(-1)=(-infty,0)$ and $f^{-1}(1)=(0,infty)$ which are both connected.
$endgroup$
– MPW
Jan 29 at 19:26